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UNIT –I Analytical conditions of Equilibrium, Virtual

Work and Catenary

STRUCTURE

1.1 Introduction

1.2 Objectives

1.3 Analytical conditions of equilibrium of Coplanar forces

1.4 Virtual Work

1.5 Catenary

1.6 Unit Summary

1.7 Assignments

1.8 References

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1.1 Introduction:

This unit introduces the basics of Analytical conditions of equilibrium of

Coplanar forces, Virtual Work, Catenary. It also help us to understand the basic concepts

of above topics.

In this unit we shall study the basic ideas of Analytical conditions of equilibrium

of Coplanar forces, Virtual Work, Catenary. It is hopped the unit help students in

studying.

1.2 Objectives:

At the end of the unit the students would be able to understand the concept of:

Analytical conditions of equilibrium of Coplanar forces

Equivalent Force and Couple

Equilibrium of a Rigid Body acted on by three forces

Trigonometrical Theorems

General Condition of Equilibrium: Analytical Method

Virtual Work

Positions of Equilibrium

Method of Virtual Work

Principal of virtual work

Forces, which may be omitted

Tension and Thrust

Roberval‟s Balance

Catenary

Equation of Catenary

Relation between x, y and

Cartesian Equation of the catenary

Relation between x and s

Tension at a point

Geometrical Properties of Catenary

Approximations to the common catenary

Sag of a tightly stretched wire

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1.3 Analytical conditions of equilibrium of Coplanar forces

Theorem1:

A system of forces acting in one plane at different points of a rigid body can

always be reduced to a single force through any given point and a couple.

Proof:

To proving this theorem, let us consider the Y

forces F1, F2, F3, …., act at points (x1, y1), (x2, y2),

(x3, y3), …., the coordinates of the points being given Y1

with reference to any rectangular axes OX and OY F1

through the given point O.

Firstly consider the force F1 acting at A1(x1, y1). A1

Let it be resolved into two forces X1 and Y1 X1

parallel to the coordinate axes. At O introduce two

equal and opposite forces X1, one along OX and another Y2

along OX. This will have no effect on the body. F2

Now the forces X1 at A1 and X1 at O along OX form A2 X2

a couple of moment –X1 (the negative sign is prefixed |

since the tendency of the couple is to rotate the body | X

clockwise). A1M1 or –X1y1 and we are left with the force O M1

X1 along OX. Hence the force X1 at A1 is equivalent to

a force X1 at O along OX and a couple of moment –X1y1.

Similarly if we introduce at O equal force Y1 along OY and OY, it is easy to see

that the force Y1 at A1 is equivalent to a force Y1 at O along OY and a couple of moment

Y1x1.

It follows therefore that the force F1 at A1 is equivalent to forces X1 and Y1 at O

along the axes OX and OY respectively and a couple of moment Y1x1 – X1y1.

Proceeding in the same manner with the force F2 (whose components along the

coordinate axes are X2 and Y2, say) at (x2,y2), we will see that the force F2 at (x2,y2) is

equivalent to forces X2 and Y2 along OX and OY at O and a couple of moment Y2x2 –

X2y2.

Applying this process again and again we see that the given system of forces is

equivalent to forces

Rx = X1 + X2 + X3 + ……, i.e. X1 along OX,

Ry = Y1 + Y2 + Y3 + ……, i.e. Y1 along OY

and a couple of moment

G = (Y1x1 – X1y1) + (Y2x2 – X2y2) + …….

= (Y1x1 – X1y1) . . . (1)

The forces Rx and Ry can be compounded into a single force through O of

magnitude R given by

R2 = (Rx)

2 + (Ry)

2, . . . (2)

and acting at an angle to the axis of X, given by

= tan-1

(Ry/Rx) . . . (3)

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Hence the system of forces can be reduced to a single force R through O and

couple of moment G given by the equation (2) and (1) respectively. It is evident that G

depends upon the position of the origin O while R does not.

Theorem2:

A system of forces acting in one plane at different points of a rigid body can be

reduced to a single force, or a couple.

Proof:

We have just seen that a system of forces Y

F1, F2, F3, ….. can be reduced to a single force R

and a couple of moment G given by (2) and (1) in

above Th.1.

If R = 0, the forces reduce to a couple. But R

if R 0, we shall show that the force R and the

couple G can be reduced to a single force R acting

in the direction given by (3) of Th.1 but in a different O

line. To prove this replace the couple G by two equal X

and opposite forces of intensity R, one along OB in R R C

the direction opposite to R and the other along OC, B Owhere OO is perpendicular to OB, O lies to the right

or left of OB as required by the sign of G and OO.R = G, . . (1)

or, (Y1x1 – X1y1)

OO = ---------------------

{(Rx)2 + (Ry)

2}

The forces at O balance each other and we are left with the force R acting at O

along C.

Since tan = Ry/Rx,

we have

sin = Ry/R and cos = Rx/R. (2)

The equation of the line C is therefore

x cos{-(/2 - )} + y sin{-(/2 - )} = OO,

or x sin – y cos = OO,

or, by (1) and (2),

xRy – yRx = G. . . (3)

Equivalent Force and Couple:

If G = 0, the given forces reduce to the single force R. Hence in every case the

forces can be reduced either to a single force, or a couple.

If the coordinates of any point Q be given by (a, b) and we are required to reduce

the system of forces to a force through the point Q and a couple, the value of the

corresponding couple G may be obtained as follows:

In the notation of Th.1 the moment of F1 about (a, b) is evidently

Y1(x1 – a) – X1(y1 – b),

or (Y1x1 – X1y1) – Y1a + X1b.

The moment of F2 about (a, b) is

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(Y2x2 – X2y2) – Y2a + X2b,

and similarly for the other forces. Hence the total moment, G., of the forces about the

point is given by the equation

G = G – aY + bX1 = G – aRy + bRx.

If the point Q lies on resultant, then

G – aRy + bRx = 0,

Since the algebraic sum of the moments about any point on the line of action of the

resultant is zero. Change the current coordinate for a, b, we have equation (3) of Th.2 for

the line of action of the resultant.

Ex.1 If a system of forces in one plane reduces to a couple whose moment is G and when

each force is turned round its point of application through a right-angle it reduces to a

couple H; prove that when each force is turned through an angle ; the system is

equivalent to a couple whose moment is G cos + H sin

Sol.

Let the forces be given by Fr acting through the point (xr, yr) inclined at r to axis

of x (where r = 1, 2, 3, ….), then

Rx = Fr cosr = 0, . . (1)

Ry = Fr sinr = 0; . . . (2)

and G = (xrsinr – yrcosr) Fr . (3)

at the origin.

If each force be rotated through a right angle in the positive sense, say, and the

resultant is R, having components Rx and Ry, then

Rx = Fr cos(r + /2) = - Frsinr = 0, by (2)

Ry = Frsin(r + /2) = Frcosr = 0, by (1)

and the resultant moment of the forces in their positions about O is given by

H = {xrsin(r + /2) – yrcos(r + /2)}Fr

= (xrcosr + yr sinr)Fr . (4)

Now if the directions are changed by in the same sense and the resultant is R,

with components Rx and Ry, then

Rx = Frcos(r +)

= Fr(cosrcos – sin rsin)

= cosFrcosr – sinFrsinr = 0, . by (1) and (2)

and

Ry = Frsin(r +)

= Fr(sinrcos + cosrsin)

= cosFrsinr + sinFrcosr = 0, . by (1) and (2)

and the resultant moment of the forces in their final positions is given by

G = {xr sin(r +) – yrcos(r +)}Fr

= {cos(xrsinr – yrcosr) + sin(xrcosr + yrsinr)}Fr

= Gcos + Hsin, . . by (3) and (4)

Ex.2 The algebraic sums of the moments of a system of coplanar forces about points

whose coordinates are (1, 0), (0, 2) and (2, 3) referred to rectangular axes are G1, G2 and

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G3 respectively. Find the tangent of the angle, which the direction of the resultant force

makes with the axis of x.

Sol.

Let R represent the magnitude of the resultant force and let its equation be

y = mx + c,

where c is positive.

Then (– m – c)R

G1 = -----------------,

(1 + m2)

(2 – c)R

G2 = --------------,

(1 + m2)

(3 – 2m – c)R

and G3 = --------------------

(1 + m2)

These equations give

G1 m + c G2 c - 2

---- = -------- and ---- = ------------ ,

G2 c – 2 G3 2m + c - 3

i.e. G2m + c (G2 – G1) + 2G1 = 0

and 2G2m + c (G2 – G3) – 3G2 + 2G3 = 0

solving we get

- 2G1 (G2 – G3) – (3G2 – 2G3)(G2 – G1)

m = ----------------------------------------------------------

G2(G2 – G3) – 2G2(G2 – G1)

G1G2 – 3G22 + 2G2G3

= ------------------------------

2G1G2 – G22 – G2G3

G1 – 3G2 + 2G3

= ---------------------- ,

2G1 – G2 – G3

which gives the tangent of the angle which the direction of the single force makes with

the axis of x.

Ex.3 If six forces, of relative magnitudes 1, 2, 3, 4, 5 and 6 act along the sides of a

regular hexagon, taken in order, show that the single equivalent force is of relative

magnitude 6 and that it acts along a line parallel to the force 5 at a distance from the

centre of hexagon 3½ times the distance of the side from the centre.

Sol.

Let ABCDEF be a regular hexagon of side 2a. Suppose that forces of magnitudes

k, 2k, 3k, 4k, 5k and 6k act along the sides AB, BC, CD, DE, EF and FA respectively.

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Let O be the center of the hexagon. Choose through O a set of rectangular axes OMX and

OCY. Y

It is easy to see that the six forces inclined to C

OMX at the angles, 3/6, 5/6, 7/6, 9/6, 11/6 and 3k 2k

13/6. Hence the sum of their resolved parts along the D B

axes of x and y are

Rx = k[cos/2 + 2cos5/6 + 3cos7/6 + 4cos9/6 4k O M X

+ 5cos11/6 + 6cos13/6] k

= , E A

and Ry = k[sin/2 + 2sin5/6 + 3sin7/6 + 4sin9/6 5k 6k

+ 5sin11/6 + 6sin13/6] = -3k F

Therefore the magnitude of the resultant force

= {(k)2 + (- 3k)

2} = 6k,

and it makes an angle

tan– 1

(-3k/k) = tan– 1

(-1/3) = 11/6

with OMX. It is, therefore, parallel to the force 5k acting along EF.

Further since all the sides of the hexagon are at a distance OM from O, the

algebraic sum of the moments of the forces about O

= (k + 2k + 3k + 4k + 5k + 6k)OM = 21k.OM

Hence the distance of the resultant force from the center

= (21k.OM)/6k = 7/2.OM

Ex.4 A system of forces in one plane is equivalent to a couple of moment G. If the line of

action of each force is turned about its point of application in the same direction through

a right angle, prove that the new system is also equivalent to a couple. Also prove that if

the moment of this couple be H and if the lines of action of the original forces be each

turned through 2tan-1

[H/G], they would still be equivalent to a couple of moment G.

Sol.

Let P1, P2, …….. be the forces acting at points (x1, y1), (x2, y2),….. and making

angles a1, a2, ….. with OX, then

X = P1cosa1

Y = P1sina1

G = (P1sina1x1 – P1cosa1y1) = P1(x1sina1 – y1cosa1) …(1)

Since the system is equivalent to a couple so X = 0 i.e.

P1cosa1 = 0 ….(2)

Y = 0 i.e.

P1sina1 = 0 ….(3)

(i) When each force is turned about its point of application through a right angle in the

+ve direction,

X = P1cos(a1 + /2) = (-P1sina1) = 0 by (3)

Y = P1sin(a1 + /2) = P1cosa1 = 0 by (2)

The system again reduces to a couple. If H be its moment

H = [P1cosa1x1 – (- P1sina1)y1] = P1(x1cosa1 + y1sina1)

(ii) Let each original force be now turned through an angle b

X = P1cos(a1 + b) = P1(cosa1cosb – sina1sinb)

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= cosb P1cosa1 – sinb P1sina1 = cosb x 0 – sinb x 0 = 0

Y = P1sin(a1 + b) = P1(sina1cosb + cosa1sinb)

= cosb P1sina1 + sinb P1cosa1 = cosb x 0 + sinb x 0 = 0

The system again reduces to a couple. If H be the moment

H = [P1sin(a1 + b)x1 – P1cos(a1 + b)y1]

= P1[x1(sina1cosb + cosa1sinb) – y1(cosa1cosb – sina1sinb)]

= cosb P1(x1sina1 – y1cosa1) + sinb P1(x1cosa1 + y1sina1)

= cosb.G + sinb.H = G cosb + H sinb

Now H = G

if G cosb + H sinb = G

i.e. H sinb = G(1- cosb)

H 1 – cosb

or --- = ------------- = tanb/2

G sinb

i.e. b = 2tan-1

[H/G]

Ex.5 Forces equal to 3P, 7P and 5P act along the sides AB, BC, and CA of an equilateral

triangle ABC; find the magnitude, direction, and line of action of the resultant.

Sol.

Let the side of the triangle be a, and let the

resultant force meet the side BC in Q. Then the sum

of the moments of the forces about Q vanishes.

3P x (QC + a)sin600 = 5P x Qcsin60

0

QC = 3a/2 A

The sum of the components of the forces

perpendicular to BC 3P 5P

= 5Psin600 – 3Psin60

0 = P3 7P C

Also the sum of the components in the B Q

direction BC

= 7P – 5Pcos600 – 3Pcos60

0 = 3P

Hence the resultant is P12 inclined at an

angle tan-13/3, i.e. 30

0, to BC and passing through Q

where CQ = (3/2) BC.

Equilibrium of a Rigid Body acted on by three forces:

Theorem:

If three forces, acting in one plane upon a rigid body, keep it in equilibrium, they

must either meet in a point or be parallel.

Proof:

To prove this, suppose that the given forces P, Q and R are not all parallel. Then

at least two of them, say, P and Q must meet in a point O. R must balance the resultant of

P and Q as the three forces P, Q and R are in equilibrium. Since the resultant of P and Q

passes through O, R must also pass through O, i.e. the three forces are concurrent.

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But if two of them are parallel, their resultant is parallel to them, and therefore so

is the force, which is to balance them, i.e. the third force is also parallel to them.

Ex.1 A sphere of given weight rests on two smooth planes inclined to the horizon at

given angles; determine the pressures on the planes.

Sol.

Let the sphere ABC, of weight W, rest on the planes OA and OB, inclined to the

horizon at angles and respectively, A and B being the points of contact of the sphere

with the planes.

Since there are three forces acting on

the sphere, namely the reaction R and S at

A and B and the weight W, therefore, for C

equilibrium they must be in the same plane

and their lines of action must meet in a point.

The figure represents a vertical G

section of the sphere and the planes, containing S R

the lines of action of the reactions and the W

weight. The reactions R and S, being B

perpendicular to the planes, pass through the

centre G of the sphere through which the weight A

W acts vertically downwards.

Then for the equilibrium of the sphere, O

since the three forces meet at G, we have,

by Lami‟s theorem

R S W .

sin (S, W) sin (W, R) sin (R, S)

or R S W .

sin () sin () sin ()

giving W sin W sin

sin () sin ()

But the reactions of the planes are equal to the pressures on the planes. Therefore

the pressure on the plane OA whose inclination to the horizon is is

W sin

sin ()

and the pressure on the plane OB whose inclination is is

W sin

sin ()

Ex2. Three uniform rods AB, BC and CD, whose weights are proportional to their

lengths a, b and c are jointed at B and C and are in a horizontal position resting on two

pegs P and Q; find the actions at the joints B and C, and show that distance between the

pegs must be

a2 c

2

------- + ------- + b

2a + b 2c + b

= =

= =

R = and S =

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Sol. Let w be the weight of a rod per unit length,

and let R and S be the vertical actions on BC at A P B C Q D

the points B and C respectively.

For the equilibrium of BC,

R + S = bw R S

and symmetry demands that R must be equal to S,

so that

R = S = ½ bw B C

This gives that the action at each joint is bw

equal to half the weight of the middle rod. Now

consider the equilibrium of the rod AB. The

pressure of the peg P vertically upwards A B

= aw + R = aw + ½ bw P

Moments about B of the forces acting on AB give that aw

aw .a/2 = (aw + ½ bw).BP. Therefore

BP = a2/(2a + b)

Also the horizontal action at B is zero. Similarly we can show that

CQ = c2/(2c + b)

Hence a2 c

2

PQ = ------- + --------- + b

2a + b 2c + b

Trigonometrical Theorems:

Theorem: If a straight line CD drawn from the vertex C of a triangle ABC to the

opposite side AB divides it into two segments in the ratio of m: n, then

(m + n) cot = m cot – n cot and (m + n) cot = n cotA – m cotB,

where and are the angles which CD makes with CA and CB and the angle which

CD makes with the base AB.

Proof:

We have

m AD AD DC

---- = ------ = ----- -----

n DB DC DB

sin ACD sin DBC

= ----------- --------- (1)

sin DAC sin DCB

sin sin() cot + cot

= --------- ---------------- = ----------------

sin() sin cot – cot

Therefore

(m + n) cot = m cot – n cot

Again from (1) we have also

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m sin( A) sin B

---- = ------------- ------------

n sin A sin(+ B)

cot A – cot

= ----------------

cot B + cot

Hence (m + n) cot = n cot A – m cot B.

Ex.1 A heavy uniform rod, of length 2a, rests partly within and partly without a fixed

smooth hemispherical bowl, of radius r; the rim of the bowl is horizontal, and one point

of the rod is in contact with the rim; if be the inclination of the rod to the horizon, show

that 2r cos2 = a cos.

Sol.

Suppose the given figure represent

that vertical section of the hemisphere, which

passes through the rod. Let AB be the rod, G D

its centre of gravity, and C the point where the

rod meets the edges of the bowl. S B

The reaction at A is along the line to the O

centre, O, of the bowl; for AO is the only line R G C

through A which is perpendicular to the surface W

of the bowl at A. Also the reaction at C is

perpendicular to the rod; for this is the only A E

direction that is perpendicular to both the rod and the rim of the bowl.

These two reactions meet in a point D, which lies on the geometrical sphere of

which the bowl is a portion. Hence the vertical line through G, the middle point of the

rod, must pass through D.

Through A draw AE horizontal to meet DG in E and join OC.

Then OAC = OCA = CAE = .

So a cos = AE = AD cos2 = 2r cos2.

Also, by Lami‟s Theorem, if R and S be the reactions at A and C, we have

R S W .

sin sin ADG sin ADC

or R S W .

sin cos2 cos

Ex.2 A hemispherical bowl, of radius r, rests on a smooth table and partly inside it rests a

rod of length 2l and of weight equal to that of the bowl. Show that the position of

equilibrium is given by the equations

l sin() = r sin = – 2r cos(),

where is the inclination of the base of the hemisphere to the horizon, and 2 is the

angle subtended at the centre by the part of the rod within the bowl.

= =

= =

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Sol.

Let O be the centre of the base of the bowl and A and B the points of contact of

the rod AH with the bowl. The angle AOB is 2 and OFE is , E being the point of

contact of the bowl with the table. Let G be the middle point of the rod. The centre of

gravity, G, of the bowl lies at the mid-point of the central radius OJ.

The rod is in equilibrium under the action of three forces viz., its weight W acting

vertically through G, the reaction R at A

acting normally to the hemisphere, i.e., D

through O and the reaction S at B acting

at right-angles to AB. The lines of action

of these three forces must meet at a point, D,

say, and since ABD is a right-angle, D must

lie on the circle AEB and AD = 2r.

Also AG = l. O N

The system consisting of the bowl

and the rod is in equilibrium under the G S

action of the reaction at E and the weights W R

acting along GK and GL; since the J G B H

reaction passes through O, the weights must A

have equal and opposite moments about E so

that KE = EL; (K and L being the points where K E L F

the vertical lines through G and G meet the

table). Since further AO = OD, it is easy to see W W

that GK passes through A.

We shall now find in three ways the length AM, the perpendicular from A on the

vertical DG.

For this we require the angles ADB, DON and OAB, ON being the perpendicular

from O on the vertical DG.

Since AOB = 2, ADB = ,

DON = 1800 – (AOB + BON) = 180

0 – (2),

and OAB = OBA = 900 – .

Hence AM = AD cos DAM = 2r cos DON

= 2r cos(180 ). = – 2r cos()

Also AM = AG cos GAM = l cos(900 ) = l sin(),

and lastly AM = 2KE = 2GI, where GI is at right angles to OE,

= 2OG sin = r sin.

It follows therefore that

l sin() = r sin = – 2r cos(),

Ex.3 Equal weight P and P are attached to two strings ACP and BCP passing over a

smooth peg C. AB is a heavy beam, of weight W, whose centre of gravity is a feet from

A and b feet from B; show that AB is inclined to the horizon at an angle

a – b W

a + b 2P

tan sin– 1

tan– 1

M

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Sol.

Let AB be the rod and G its centre of C

gravity dividing AB in the ratio a : b. Let C

be the smooth peg. The tension in the strings

BC and AC are each P; since W balances the P

resultant of two equal forces, CA and CB are b B

equally inclined to the vertical. Suppose we a G

denote ACG or BCG by .

Resolving vertically, we have A

2P cos – W = 0, i.e. = cos – 1

(W/2P). W

Therefore

(a + b) cot BGC = a cot – b cot.

Also BGC = ½ where is the inclination of AB to the horizon.

It follows that

a – b a – b

a + b a + b

a – b

a + b

Hence

a – b W

a + b 2P

Ex.4 A square, of side 2a, is placed with its plane vertical between two smooth pegs,

which are in same horizontal line and at a distance c, show that it will be in equilibrium

when the inclination of one of its edges to the horizon is either

450

or ½ sin– 1

[(a2 – c

2)/c

2].

Sol.

Let OABC be the square, its side OA

being inclined at an angle to the horizontal

through O and let DE be the horizontal line

joining the pegs. The reactions at D and E are B

perpendicular respectively to OC and OA.

Since the only other force acting on the square

is weight, these three forces must meet in a C

point, say, O. The figure OEOD is obviously a G

rectangle. We can now solve the problem by O A

two methods. R

R

Method1: D E

Let the reactions at D and E are R and

R respectively and W is the weight of the O N

square. Resolving horizontally, we get

R cos = R sin. (1) W

Taking moments about G, we have

R (a – c cos) = R (a – c sin). (2)

tan = cot = tan(½ )

= tan [½ – cos– 1

(W/2P)]

tan sin– 1

tan– 1

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Eliminating R and R between (1) and (2), we get

cos (a – c cos) = sin (a – c sin).

or c (sin2 – cos

2) = a (sin – cos).

Hence, either

sin – cos = 0, where = 450,

or c (sin + cos) = a. (3)

Squaring either side of (3), we have

c2 (1 + sin2) = a

2.

Therefore

½ sin– 1

[(a2 – c

2)/c

2].

Method2:

Let ON be the perpendicular from O on the vertical through G. Then we can write

the value of ON in two ways.

ON = OG cos( ¼)

= a2 (cos¼ cos – sin¼ sin) = a (cos – sin).

Also ON = OO cos OON.

But OO = DE = c,

and OON = OOE + EON = = 2.

Hence ON = c cos2.

It follows therefore that

a (cos – sin) = c cos2,

or a (cos – sin) = c (cos2 – sin

2), etc.

Ex.5 A beam whose centre of gravity divides it into two portions, a and b, is placed

insides a smooth sphere; show that, if be its inclination in the horizon in the position of

equilibrium and 2a be the angle subtended by the beam at the centre of the sphere, then

b – a

b + a

Sol.

In this case both the reactions, R and

S, at the ends of the rod pass through the

centre, O, of the sphere. Hence the centre

of gravity, G, of the rod must be vertically

below O. Let OG meet the horizontal line O

through A in N. Draw OD perpendicular to AB. R S

Then AOD = BOD = , a b B

and DOG = 900 – DGO = DAN = .

G D

Therefore N W

(a + b) cot OGB = b cot OAB – acot OBA,

i.e. (a + b) tan = (b – a) tan.

Also by Lami‟s Theorem,

R S W .

sin BOG sin AOG sin AOB

tan = tan

= =

15

or R S W . ,

sin() sin() sin2

giving the reactions.

General Condition of Equilibrium: Analytical Method:

Theorem1:

The resultant R and the couple G must separately vanish.

Proof:

We know that the system of forces acting at different points may be reduced to a

force R through an arbitrary point O and a couple G. Since a finite force R cannot balance

a couple G, it is necessary for equilibrium that the resultant force R and the couple G

should separately vanish. The vanishing of R involves the condition that Rx = 0 and Ry =

0.

Note: Now we arrive at the following necessary and sufficient conditions of equilibrium.

A system of forces in a plane will be in equilibrium if the algebraic sums of their

resolved part in any two perpendicular directions vanish, and if the algebraic sum of their

moments about any point also vanishes.

Theorem 2:

A system of forces in a plane will be in equilibrium if the algebraic sum of the

moments of all the forces with respect to each of three non-collinear points is zero.

Proof:

To proving this theorem, consider the origin at one of the three points and let the

coordinates of the other points be (x1, y1) and (x2, y2). If we denotes the algebraic sum of

the moments of the forces about the above points by G, G‟, G‟‟ then by

G = 0,

G‟ = G – x1Ry + y1Rx = 0,

and G‟‟ = G – x2Ry + y2Rx = 0.

These reduce to

- x1Ry + y1Rx = 0

and - x2Ry + y2Rx = 0.

Since the three points are not collinear, y1/x1 y2/x2 and the above conditions

reduce to

Rx = 0, Ry = 0 and G = 0

a set of conditions involved in Theorem 1.

Theorem 3:

A system of forces in a plane will be in equilibrium if the algebraic sum of the

moments about each of any two different points is zero and the algebraic sum of the

resolved parts of the forces in any given direction not perpendicular to the line joining the

given points is zero.

= =

16

Proof:

Let the two different points be A and B. As before we take origin at A and x-axis

along the given direction. Let the coordinates of B be (x1, y1). Then the conditions are

G = 0,

G‟ = G – x1Ry + y1Rx = 0,

Rx = 0,

which lead to

Rx = 0, Ry = 0 and G = 0

provided that x1 is not zero, a set of conditions involved in Theorem 1.

Theorem4: Resultant forces and couple corresponding to any base point O of a system of

coplanar forces.

Proof:

Through O take any pair of rectangular axes Ox, and Oy. At P1, the point (x1, y1),

let there act a force whose components parallel to the axes are X1 and Y1.

Then X1 at P1 is equivalent to a parallel force X1 at O together with a couple

y1X1 . So Y1 at P1 is equivalent to a parallel force Y1 at O together with a couple x1Y1 .

Hence the force at P1 is equivalent to components X1, Y1 along Ox, Oy and a

couple x1Y1 – y1X1 .

So for the other forces at P2, P3, etc.

Hence the system of forces is equivalent to components X, Y along Ox, Oy, and a

couple G about O, such that y

X = X1 + X2 + X3 + … = X1, Y1 . P2

Y = Y1 + Y2 + Y3 + … = Y1, P1 .P3

and G = (x1Y1 – y1X1) + (x2Y2 – y2X2) + … x1 y1 X1

= (x1Y1 – y1X1). O x

X and Y compound into a single force R acting at O.

Theorem5: Equation to the resultant of a system of forces in one plane.

Proof:

We know that the system can be reduced to components X and Y along any two

rectangular axes Ox and Oy, and a couple G about O .

Let Q be any point (h, k), which

lies on the resultant of the given system. y

The moment of the system about it is

equal to the moment of the resultant about Y Q

it and is therefore zero. G

Now the moment of the system about Q

= G + X.NQ – Y.ON O x

= G – hY + kX, X N

so that G – hY + kX = 0.

Hence the locus of (h, k), i.e. the resultant, is the straight line

G – xY + yX = 0.

17

Ex.1 Two equal uniform rods, AB, AC, each of weight W, are freely joined at A and rest

with the extremities B and C on the inside of a smooth circular hoop, whose radius is

greater than the length of either rod, the whole being in a vertical plane, and the middle

points of the rods being joined by a height string; show that, if the string is stretched, its

tension is W(tan – 2 tan), where 2 is the angle between the rods, and the angle

either rod subtends at the centre.

Sol.

Let O be the centre of the circular hoop O

BFC, AB, AC the equal rods and DE the string

joining their middle points.

If R be the pressure of the hoop on either

rod, then resolving vertically for the system A

consisting of the two rods

W = R cos (1)

Taking moments about A for either rod, D E

suppose of length 2l, we have B C

R.2l sin() = W.l sin + T.l cos, F

i.e. 2R sin () – W sin

cos

2W sin () – W sincos

cos on using (1)

= W (tan2 tan)

Ex.2 Three uniform equal heavy cylinders, each of which touches the other two, are tied

together by a string passing round them and laid with their axes horizontally upon a

horizontal plane. The tension of the string being given, find the pressure between the

cylinders, the string is supported to lie in the vertical plane through the centres of gravity.

Sol.

Consider the section through the string

and A, B, C, the centres of gravity of the three

cylinders. Let W be the weight of each cylinder,

and T the tension of the string. The reaction

between the higher and each of the two lower T A T

cylinders will by symmetry be equal. Let us

denote it by R. Let S be the reaction between R R

the two lower cylinders and R the reaction of T W T

the plane XY upon each of the lower cylinders. R R R R

It is clear that ABC forms an equilateral triangle, B C

and the lines of the action of the tension will S S

also forms a similar triangle.

Resolving vertically the forces acting on

the upper cylinder alone, we have X T T Y

2T cos 300 + W = 2R cos 30

0, W W

giving R = T + W/3 (1)

Again resolving vertically the forces on either of the two lower cylinders, we get

T cos 300 + R = R cos 30

0 + W,

T =

=

18

hence R = ½3(T + W/3) + W – T.½ 3 = (3/2)W (2)

If we resolve horizontally for either of the two lower cylinders, we have

T cos 600 + T = S + R cos 60

0,

and therefore

S = (3/2)T – ½ (T + W/3) = (T – W/23) (3)

The equations (1) and (3) give the required reactions R and S. It should be noted

that the pressure on the plane by both the lower cylinders is, by (2), equal to 3W, i.e. to

the sum of the weights of the three cylinders.

Ex.3 An elliptic lamina is acted upon at the extremities of pairs of conjugate diameters by

forces in its own plane tending outwards and normal to its edge; show that there will be

equilibrium if the force at the end of each diameter is proportional to the conjugate

diameter.

Sol.

Suppose that the forces acting at the

ends P, P and D, D of the pair of conjugate B

diameters PP and DD be respectively X1,

X2, and Y1 and Y2. Since the normals at P

and D are parallel to the normals at P and D P

D respectively, the forces X1 and X2 at P

and P are acting in opposite senses along A S O S A

parallel lines and so the forces Y1 and Y2 at

D and D respectively. Obviously the lines

of action of X1 and Y1 are not parallel. P D

Moreover the resultant of X1 at P and X2 at P B

and that of Y1 at D and Y2 at D will not act

in the same straight line even though the

magnitudes of these resultants may be equal. Hence for equilibrium, we must have X1 =

X2 and Y1 = Y2, so that the forces at P and P form a couple, as also the forces at D and

D. Since the tendencies of rotation of these couples are opposite, for equilibrium we

must have the magnitudes of the couples to be equal.

Suppose that the angle POD is equal to . Since tangent at D is parallel to OP, the

normal at D is perpendicular to OP, hence the moment of the couple formed by the equal

and opposite forces Y1 at D and D is equal to Y1.2OD cos(), as the perpendicular

from D on OP is OD cos().

Similarly the moment of the couple formed by the equal and opposite forces X1 at

P and P is equal to X1.2OP cos.

Magnitudes of these moments will be equal if X1/OD = Y1/OP. Therefore

X1/DD = Y1/PP.

Hence there will be equilibrium if the force at the end of each diameter is

proportional to the conjugate diameter.

Ex.4 Three equal uniform rods, each of weight W, are smoothly jointed so as to form an

equilateral triangle. If the system be supported at the middle point of one of the rods,

19

show that the action at the lowest angle is 3W/6, and that at each of the others is

W(13/12).

Sol.

Let ABC be the triangle formed by the rods, and D the middle point of the side

AB at which the system is supported.

Let the action of the hinge at A on the rod AB consist of two components,

respectively equal to Y and X, acting in vertical and horizontal directions; hence the

action of the hinge on AC consists of components equal and opposite to these. Since the

whole system is symmetrical about the vertical line through D, the action at B will consist

of components, also equal to Y and X, as in the figure.

Let the action of the hinge C on CB consist of Y1 vertically upward, and X1

horizontally to the right, so that the action of the same hinge on CA consists of two

components opposite to these, as the figure.

S Y S Y

A D B X X

A D B

W X Y W Y X

Y1

W W W W

C X1 C X1

Y1

For AB, resolving vertically, we have

S = W + 2Y (1),

where S is the vertical reaction of the peg at D.

For CB, resolving horizontally and vertically, and taking moments about C, we

have

X + X1 = 0 (2),

W = Y + Y1 (3),

and W.a cos600 + X.2a sin60

0 + Y.2a cos60

0 (4).

For CA, by resolving vertically, we have

W = Y – Y1 (5).

Solving these equations, we have

X1 = – 3W/6, Y1 = 0, Y = W, X = 3W/6 and S = 3W.

Hence the action of the hinge at B consists of a force (X2 + Y

2) [i.e. W(13/12)],

acting at an angle tan– 1

(Y/X) [i.e. tan– 1

23], to the horizon; also the action of the hinge

at C consists of a horizontal force equal to 3W/6.

Ex.5A heavy uniform bar, of length 2a and weight w is movable in a vertical plane round

a smooth hinge fixed at one extremity; a heavy smooth sphere of weight W and radius r is

attached to the hinge, by a cord of length l; the two bodies rest in contact. If and be

the inclinations of the bar and the cord to the vertical, show that

sin() = r/(l + r) and W (l + r) sin = wa sin.

Find the pressure between the bar and the sphere.

20

Sol.

Let AB be the bar hinged at A, G its middle point, and D the point at which it is in

contact with the sphere. Let E be the point of the sphere to which the cord AE is attached.

Let C be the centre of the sphere. AB and AE make angle and respectively with the

vertical.

The forces acting on the sphere only are: N A L

(i)its weight W acting vertically downwards

through its centre C, T

(ii)reaction R of the bar on the sphere along DC,

(iii)tension T of the string along EA. R

Since forces (i) and (ii) meet at C. So for

equilibrium, the third force must also pass through C D

C i.e., AEC must be a straight line. R

Now from CAD,

sin.CAD = CD/AC = CD/(EA + CE)

Therefore sin () = r/(l + r). G

Now consider the equilibrium of the sphere

and the bar as a whole. In this case, the two equal

and opposite reactions of the sphere on the rod and W w B

of the rod on the sphere will not come in the picture,

and the forces will be

(i)weight W at C acting vertically downwards,

(ii)weight w at G acting vertically downwards,

(iii)tension T along EA,

(iv)reaction of the hinge at A on the bar.

To avoid the reaction of the hinge, we take moment of the forces about A.

Therefore W.AN = w.AL

where AN, AL are perpendiculars from A on vertical lines drawn through C and G.

W.AC sin = w.AG sin,

W.(l + r) sin = wa sin,

Finally to get R, the reaction of the sphere on the bar, we consider the force on the

bar only. They are

(i)weight w of the bar at G vertically downward,

(ii)reaction R of the sphere along CD,

(iii)reaction of the hinge at A.

Therefore taking moments about A, w.AL = R.AD

R = w.AL /AD

Now AL = AG sin = a sin.

AD = AC cos () = (l + r) cos ()

Therefore

wa sin

(l + r) cos ()

W (l + r) sin

(l + r) cos ()

W sin

cos ()

R =

=

=

E

21

1.4 Virtual Work

Positions of Equilibrium:

Now we come to a very powerful method

for attacking problems on equilibrium. Consider a Y

heavy particle on a smooth curve. If the axis of Y

is vertical it is obvious that the particle can rest n B

equilibrium at the point A, B or C. These are points A

of maxima or minima on the curve, i.e. , points for C

which dy/dx = 0. The physical interpretation of the

positions of equilibrium may be given as follows: O X

If a small displacement be given along the curve to the particle from a position of

equilibrium, the work done is zero, for, if W be the weight of the particle, the work done

is Wy, i.e., W(dy/dx)x (to a first approximation). Again to find the position of

equilibrium we have to find those positions on the curve for which Wy = 0, i.e., those

points for which dy/dx = 0.

It should be noted that y is equal to zero to a first approximation only. For, if y =

f(x) be the equation of the curve,

y + y = f(x + x) = f(x) + xf (x) + [(x)2/2!]f (x) + ….

As f (x) = 0 at A, B, or C,

y = [(x)2/2!]f (x) + ….

i.e., y = 0 if squares and higher powers of x are neglected. Hence y = 0 to a first

approximation only. Thus it is necessary that the displacement given be small.

Method of Virtual Work:

In any given problem, we imagine the body to be displaced a little and find out

the work done during the displacement. The condition of equilibrium is obtained by

equating to zero the sum of the work done. Since the body is not actually displaced, the

work done is called virtual work. The virtual work that is calculated is the amount of

work that would have been done if the displacements had actually been made.

We shall now formally enunciate the principle of virtual work and establish the

same. Since the proof is simpler for the case of a particle acted upon by a number of

forces, we consider this first.

Principal of virtual work:

If a system of forces acting on a body be in equilibrium and the body undergo a

slight displacement consistent with the geometrical conditions of the system, the

algebraic sum of the virtual works is zero; and conversely, if this algebraic sum be zero,

the forces are in equilibrium. In other words, if each force P have a virtual displacement

p in the direction of its line of action, then, to the first order of small quantities, (P.p)

= 0; also conversely, if (P.p) be zero, the forces are in equilibrium.

22

Theorem:

The necessary and sufficient condition that a rigid body acted upon by a number

of coplanar forces be in equilibrium is that the algebraic sum of the virtual works done by

the forces in any small displacement consistent with the geometrical conditions of the

system is zero.

Proof:

Let any number of coplanar forces F1, Y F2 R

F2, F3, … act at points A1, A2, A3, … of a F3 F1

rigid body. Through an arbitrary chosen point P Q

O, take two fixed rectangular axes OX and OY A2

in the plane of the forces. Let the coordinates A3 A1

of the points A1, A2, A3, … be (x1, y1), (x2, y2), O X

(x3, y3), .. and suppose that the component of A4 A5

the forces F1, F2, F3, … along the coordinate

axes are X1 and Y1, X2 and Y2, X3 and Y3, … F4 F5

respectively.

If the body undergoes a certain displacement parallel to a fixed plane, it may be

brought from its old position to its new position by

(i) a motion of rotation which has the same magnitude and sense for all the points of the

rigid body, and

(ii) a motion of translation which has also the same magnitude and direction for all its

points.

The latter displacement can, in general, be decomposed into a motion of

translation along each of the two perpendicular axes OX and OY and these component

displacements will also be shared by all the points of the body. Let us denote by and

respectively the displacements of rotation and translations along the axes of x and y

corresponding to a slight displacement of the body.

If we assume that the polar coordinates of A1 are (r1, 1), the Cartesian

coordinates of R, the new position of A1, can be written as

r1 cos(1 + ) + and r1 sin(1 + ) +

i.e. r1 cos1 - r1 sin1 + [ is small, so sin = , cos = 1]

and r1 sin1 + r1 cos1 +

neglecting squares and higher powers of .

The relative coordinates of R with respect to A1 are therefore

[(r1 cos1 - r1 sin1 + ) - r1 cos1]

and [(r1 sin1 + r1 cos1 + ) - r1 sin1],

i.e., - r1 sin1 and + r1 cos1,

i.e., - y1 and + x1.

Since the work done by a force is equal to the sum of the works done by its

components, the virtual work of the force F1 is

X1( - y1) + Y1( + x1),

i.e., X1 + Y1 + (x1Y1 - y1X1).

- y1 and + x1 being the displacements of the point A1 of the application of the

force F1 along the axes of x and y respectively.

Similarly the virtual work of the force F2 acting at the point (x2, y2) is

23

X2 + Y2 + (x2Y2 - y2X2)

and we get similar expression for the other forces.

The algebraic sum of the virtual works is, therefore,

X1 + Y1 + (x1Y1 - y1X1).

Since the body is in equilibrium under the action of the forces,

X1 = 0, Y1 = 0 and (x1Y1 - y1X1) = 0.

This gives that

X1 + Y1 + (x1Y1 - y1X1) = 0.

Hence the condition is necessary; in other words, if a body in equilibrium under

the action of a number of coplanar forces undergo a small displacement, then the

algebraic sum of the virtual works done by different forces is zero.

Now we shall show that the condition is sufficient. It is given that the algebraic

sum of the virtual works of the various forces is zero for all displacements and has to

show that there is equilibrium. Since the equation

X1 + Y1 + (x1Y1 - y1X1) = 0. (1)

is true for all displacements, and are independent of each other. Suppose then that

and are the translations along OX and OY and rotation respectively for a new

displacement of the rigid body. We still have

X1 + Y1 + (x1Y1 - y1X1) = 0. (2)

Subtracting (2) from (1), we have

( - ) (x1Y1 - y1X1) = 0.

But since ( - ) is not equal to zero, therefore

(x1Y1 - y1X1) = 0.

Similarly by varying alone, we prove that Y1 = 0 and by making alone vary,

X1 = 0. But X1 = 0, Y1 = 0 and (x1Y1 - y1X1) = 0 are the analytical conditions for

equilibrium and hence the body is in equilibrium.

Forces, which may be omitted:

The equation (1) is known as the equation of virtual work. In a given problem of

equilibrium, we write down this equation corresponding to a suitable displacement. The

displacement should be such as to exclude the forces, which are not required and to

include those, which are required in the final result. There are, however, certain forces,

which may be omitted in forming the equation of virtual work. We shall now enumerate

them.

1.If the distance between two particles of a system is invariable, the work done by the

mutual action and reaction between the two particles is zero.

To show this, suppose that T is the

force acting between the particles A, B B

which suffer displacements to A, B

respectively such that AB = AB. If be A

the small angle between AB and AB and

M, N the projections of A, B in AB, then A M T T B N

the work done by the two forces

= T.AM – T.BN = T(AM + MB) – T(BN + MB)

24

= T(AB – MN)

= T.(AB – AB cos)

= T.AB(½2 – higher powers of )

= 0, to the first order of small quantities.

2.The reaction R of any smooth surface with which the body is in contact does no work.

For, if the surface is smooth, the reaction R on the point A of the body is normal

to the surface. If A moves to a neighbouring point A, then AA is at right angles to the

forces and accordingly the work done by R is zero.

In the case of a rough surface, however, the work done by the frictional force F,

viz., F.(-AA) must come into the equation, since it is not in general zero.

3.If the force between two bodies of a system is a mutual pressure at a point of contact,

then no work is done in any virtual displacement of the system by the action and reaction

if the same points of the bodies remain in contact during the displacement.

For clearly the work done by the action balances that done by the reaction.

4.When a body rolls without sliding on any fixed surface the work done in a small

displacement by the reaction of the surface on the rolling body is zero.

For the point of contact P of the body is momentarily at rest and so its

displacement is zero. Hence the normal reaction at P and the frictional force F there have

zero displacements.

5.If any body is constrained to run round a point or on an axis fixed in space, the virtual

work of the reaction at the point or on the axis is zero.

For the displacement of the point of application of the force is zero.

Tension and Thrust:

In many cases we are required to find the tension of a string or the thrust of a rod.

The geometrical conditions of the problem often make it impossible to give a virtual

displacement, which will be consistent with the constraints. Thus, if we have a

quadrilateral ABCD, consisting of heavy rods AB, BC, CD, DA suspended from the

point A and the points B and D connected by a rigid rod, it is impossible to give a virtual

displacement which will make thrust of the rod do work. To get over this problem, we

replace the system under consideration by another in which there are forces T and T

acting at B and D in the direction DB and BD respectively. If T in the second system is

equal to the thrust of the rod in the first system, the equation is evidently not affected. We

can now give a virtual displacement to the second system, which will involve T in the

equation of virtual work. T will thus be determined.

Similarly the equations of equilibrium of the surrounding bodies are not altered if

we replace the string joining A and B by a force T acting at A along AB and a force T

acting at B along BA.

We shall show that it is not necessary to consider the separate virtual

displacements of the ends of the string or the rod, but merely its increment in length.

25

Taking the case of a string joining A and B where AB = l, and considering a

displacement of A to A and B to B where AB = l + l, we have that the work done by

the two forces introduced as above

= T.AM – T.BN,

where M and N are the projections of A and B on AB, B

= T(AB –MN) = T(AB – AB cos), A

where is the angle between AB and AB,

= T(AB – AB)

to the first order of approximation. A M T T B N

Hence the work done is equal to – T l.

Similarly, the work done by the thrust in an extension of a rod from length l to l +

l is +T l, where T is the thrust in the rod.

Roberval’s Balance:

This balance, which is a common form of letter- weigher, consists of four rods

AB, BE, ED and DA freely jointed at the corners A, B, E and D so as to form a

parallelogram, whilst the middle points, C and F of AB and ED are attached to fixed

points C and F which are in vertical straight line. The rods AB and DE can freely turn

about C and F.

To the rods AD and BE are attached equal scale-pans. In one of these is placed

the substance W, which is to be weighted and in the other the counterbalancing weight P.

A . C . B

D . F

. E

We shall apply the Principal of Virtual work to prove that it is immaterial on what

part of the scale-pans the weights P and W are placed.

Since CBEF and CADF are parallelograms it follows that, whatever be the angle

through which the balance is turned, the rods BE and AD are always parallel to CF and

therefore are always vertical.

If the rod AB be turned through a small angle the point B rises as much as the

point A falls. The rod BE therefore rises as much as AD falls, and the right hand scale-

pan rises as much as the left-hand one falls. In such a displacement the virtual work of the

weights of the rod BE and its scale-pan is therefore equal and opposite to the virtual work

of the weights of AD and its scale-pan. These Virtual work therefore cancel one another

in the equation of virtual work.

Also if the displacement of the right-hand scale-pan be p upwards, that of the left-

hand one is p downwards. The equation of virtual work therefore gives

26

P.p + W(- p) = 0, i.e. P = W.

Hence, if the machine balance in any position whatever, the weights P and W are

equal and this condition is independent of the position of the weights in the scale-pans.

The weights therefore may have any position on the scale-pans. It follows that the scale-

pans need not have the same shape, nor be similarly attached to the machine, provided

only that their weights are the same.

Ex.1 Five weightless rods of equal length are jointed together so as to form a rhombus

ABCD with one diagonal BD. If a weight W be attached to C and the system be

suspended from A, show that there is a thrust in BD equal to W/3.

Sol.

Let the five rods AB, BC, CD, DA, A

and BD form the rhombus ABCD and the

diagonal BD. Since the system is suspended

from A and there is a weight W attached to

C, AC will be vertical and consequently BD

horizontal. B T T D

Suppose that the rod AB or AD makes

an angle with the horizontal through A. let us

give a small symmetrical displacement such C

that changes to + . Since A is fixed, we

measure the depth of C, the point of application W

of W, below A. Now AC is 2a sin where a is the

length of a rod and hence the work done by W during the small displacement is W (2a

sin), the point C moves down and hence the plus sign. The length BD is 2a cos and the

work done by the thrust T in the rod AD is T (2a cos).

By the principle of virtual work,

T (2a cos) + W (2a sin) = 0,

or 2a (W cos - T sin) = 0.

Since 0, it follows that

W cos - T sin = 0,

i.e. T = W cot.

In the equilibrium position, ABC is an equilateral triangle and is 600. Therefore

T = W/3.

Ex.2 A quadrilateral ABCD, formed of four uniform rods freely jointed to each other at

their ends, the rods AB, AD being equal and also the rods BC, CD, is freely suspended

from the joint A. A string joins A to C and is such that ABC is right angle. Apply the

principle of virtual work to show that the tension of the string is

(W+W )sin2 + W,

where W is the weight of an upper rod and W of a lower rod and 2 is equal to the angle

BAD.

27

Sol.

If the angle BCA be equal to , then

in the position of equilibrium is the A

complement of . Let AB = AD = 2a and

CB = CD = 2b.

Then if we assume from the very

beginning that = /2 - , we shall have

b = a tan and it would be impossible to make

an increment in without altering b or a. This B W W D

would introduce into the equation of virtual

work the unknown stress in the sides of the

quadrilateral with which we are not concerned.

We, therefore, use an independent symbol for

the angle ACB and imagine a displacement

which alters the angles of the figure and the W C W

length AC but not the lengths of the sides of

the quadrilateral.

Now AC = 2a cos + 2b cos, the depth of the centre of gravity of AB or AD is a

cos and the depth of the centre of gravity of BC or DC is 2a cos + b cos

The equation of virtual work is then

- T(2a cos + 2b cos + 2W(a cos) + 2W(2a cos + 2b cos

or T(a sin + b sin - Wa sin - W(2a sin + b sin

But and are connected by the relation

a sin = b sin

so that

a cos = b cos

hence

T(tan + tan = W tan + W(2 tan + tan

Now put cot instead of tan and we find that

T = (W+W )sin2 + W.

Ex.3 A regular hexagon ABCDEF consists of six equal rods, which are each of weight W

and are freely jointed together. The hexagon rests in a vertical plane and AB is in contact

with a horizontal table. If C and F be connected by a light string. Prove that its tension is

W3. E D

Sol. W

Let ABCDEF be the regular hexagon

formed of the six rods AB, BC, CD, DE, EF W W

and FA, each of weight W. The rod AB is in F C

contact with a horizontal table LABM, and C T T

and F are connected by a string. Suppose that

the angle FAL or CBM is . W W

Give a small displacement such that

changes to + This displacement changes L A B M

W

28

the positions of the middle points of the rods BC, CD, DE, EF and FA (AB remaining in

contact with the table), but the lengths of the rods are not altered. Since AB is fixed, we

find the heights of the middle points of these five rods above AB. The height of the

middle point of AF or BC is a sin, where 2a is the side of the hexagon and hence work

done by the weight of AF or BC during the change of to + is - W(a sin). Minus

sign is prefixed, since the point of application moves upwards.

Again the height of the middle point of CD or EF is 3a sin and the work done by

weight CD or EF is - W(3a sin). Similarly we see that the work done by the weight of

DE is - W(4a sin). Since the middle point of AB is not shifted, no work is done by the

weight of AB.

Lastly the length of the string FC is 2a + 4a cosand hence, the work done by the

tension T in the string is

- T(2a + 4a cos).

The equation of virtual work is then

- 2W(a sin) - 2W(3a sin) - W(4a sin ) - T(2a + 4a cos) = 0,

or 4a(T sin - 3W cos) = 0.

Since

0,

T = 3 cotW

In the position of equilibrium, FAB = 1200 and accordingly = 60

0. It follows

that T = W3.

Ex.4 A frame ABC consists of three light rods, of which AB, AC are each of length a,

BC of length 3a/2 freely jointed together. It rests with BC horizontal, A below BC and

the rods AB, AC over two smooth pegs E and F, in the same horizontal line, distant 2b

apart. A weight W is suspended from A, find the thrust in the rod BC.

Sol.

Let AB, BC, AC be three rods forming

a framework, such that AB = AC = a and B T M T C

BC = 3a/2. Let BAM be equal to , where

AM is the perpendicular from A on BC.

Let us imagine a displacement which K

alters to . Since BC = 2a sin, increase E F

in its length is (2a sin) and the work done by

the thrust in BC

= T.(2a sin).

Since the pegs are fixed, we consider the A

depth of A below EKF. Now AK = bcot , hence

the work done by W is W

= W.(b cot).

The reactions at the pegs do not work, accordingly the equation of virtual work is

T.(2a sin) + W.(b cot) = 0,

or, since 0,

T.2a cos - W.b cosec2 = 0.

Therefore

29

Wb cosec2 sec

2a

In the position of equilibrium,

sin = ¾, cos = 7/4

hence

Wb 16 4

2a 9 7

32Wb

97 a

Ex.5 A smooth parabolic wire is fixed with its axis vertical and vertex downwards, and in

it is placed a uniform rod of length 2l with its ends resting on this wire. Show that, for

equilibrium, the rod is either horizontal, or makes with the horizontal an angle given by

cos2 = 2a/l,

4a being the latus rectum of the parabola.

Sol.

Let AB be the rod of length l. Choose Y

OX and OY as coordinate axes, thus the

equation of the parabola may be written as B

x2 = 4ay.

If the coordinates of the point A be G

(2at, at2) and if BAC be equal to , then A

C

coordinates of B will be (2at + 2l cos, at2 + 2l sin).

Since B lies on the parabola, A O M B X

(2at + 2l cos)2 = 4a(at

2 + 2l sin), W

giving

t = tan – (l cos)/(2a).

If y be the height of the centre of gravity G of the rod,

y = ½ [at2 + (at

2 + 2l sin)] = at

2 + l sin

y = a[tan – (l cos)/(2a)]2

+ l sin

= (l2 cos

2)/(4a) + a tan

2.

Now displace the rod so that increases to , the ends of the rod remaining

in contact with the wire. By the principle of virtual work,

- W y = 0, where W is the weight of the rod,

or,

W(l2 cos

2)/(4a) + a tan

2] = 0,

or

W[- (l2 2cossin)/(4a) + 2a tan

2sec

2] = 0.

Since W 0 and 0, we must have

sin (cos4 – 4a

2/l

2) = 0.

Therefore either

sin = 0, i.e., = 0,

or,

cos4 – 4a

2/l

2 = 0,

i.e., cos2 = 2a/l.

T =

T =

T =

30

It is clear that = 0 corresponds to the horizontal position of the rod.

Ex.6 Four equal jointed rods, each of length a, are hung from an angular point, which is

connected by an elastic string with the opposite point. If the rods hang in the form of a

square, and if the modulus of elasticity of the string be equal to the weight of a rod, shew

that the unstretched length of the string is (a2)/3.

Sol.

Let AB, BC, CD and DA be four equal A

rods of length 2a and let the framework hang

from A, the corners A and C being connected

by an elastic string. Let T be the tension of the

elastic string. If we suppose that the depth of B W W D

the centres of gravity of the two upper rods

below A is x, then the depth of the centres of

gravity of the two lower rods would be 3x and W W

the length of the string 4x. C

Now give a displacement to the figure such that AC is stretched. By the principal

of virtual work,

2W(x) + 2W(3x) – T(4x)= 0,

or,

4(2W – T)x = 0.

Since x 0, T = 2W.

But, by Hooke‟s law,

T = W.[(a2 – l)/l],

where l is the unstretched length of the string, hence

(a2 – l)/l = 2.

Therefore

l = (a2)/3.

Ex.7 One end of a beam rests against a smooth vertical wall and the other on a smooth

curve in a vertical plane perpendicular to the wall; if the beam rests in all positions, show

that the curve is an ellipse whose major axis lies along the horizontal line described by

the center of gravity of the beam.

Sol.

Let OY be the wall and CBD be the

curve on which the end B of the rod AB rests. Y

During a small displacement of the rod at A

and B do no work, hence by the principle of A

virtual work, if h be the height of the centre of G

gravity G of the rod above a fixed horizontal D

line OX, Wh = 0, which gives that h = constant. M B

Hence the centre of gravity of the rod

lies always on a horizontal line for all positions of the

rod. Now take OX and OY as the coordinate

axes. If AB is inclined at an angle to the O C X

horizontal and (x, y) are the coordinates of B,

31

we have

x = 2a cos, where AB = 2a. (1)

The height of G above OX = y + a sin.

Hence

h = y + a sin. (2)

Eliminating between (1) and (2), we get

x2 (y – h)

2

4a2 a

2

which represents an ellipse with its centre at (0, h). Its major axis is y = h, the horizontal

line described by the centre of gravity.

Ex.8 An endless chain of weight w rests in the form of a circular bond round a smooth

vertical cone, which has its vertex upwards. Find the tension in the chain due to its

weight, assuming the vertical angle of the cone to be 2.

Sol.

Let AB be the endless chain resting O

under its own weight on the cone. If its length

be 2x, the radius AC of the circle which it

form will be x. A B

Suppose that a small displacement is

given to the chain such that x becomes x + x.

Then the work done by the tension is A B

T(2x). C

We may take the vertex O to be the

fixed position from which the level of the

centre of gravity is to be measured. The centre of gravity of the chain is at depth x cot

from O in the position of equilibrium. During the displacement the centre of gravity goes

lower and accordingly the work done by the weight of the chain is

w(x cot).

Since the reactions at the various points of contact do no work, we have, by the

principle of virtual work,

- T.(2x) + w(x cot) = 0,

i.e., x(- 2T + w cot) = 0.

Therefore,

since x 0,

T = (w cot)/(2).

Ex.9 Two small smooth rings of equal weight slide on a fixed elliptical wire, whose

major axis is vertical and they are connected by a string, which passes over a small

smooth peg at the upper focus; show that the weights will be in equilibrium wherever

they are placed.

+ = l,

C

32

Sol.

Let APAQ be the elliptical wire with A

its upper focus at S, the major axis AA being

vertical. If P and Q be the positions of the S

smooth rings, each of mass m, then principle

of virtual work gives M P

mg (y) + mg (y) = 0,

i.e., y + y = 0, (1) Q N

where y and y represent the lengths SM and

SN respectively.

Taking S as the pole and SA as the A

initial line, the equation of the elliptical wire

can be written

l .

1 + e cos (2)

Let SP be r, then

SQ = a – r ( = r, say), (3)

a being the length of the string.

From (2)

y = SM = SP cos (1800 ) = (r – l)/e, where we have used (2)

and y = SN = SQ cos ( ), where is the vectorial angle of Q,

= (r – l)/e = (a – r –l)/e, on using (3).

Hence the left hand side of (1) is

r – l a – r – l

e e

which is identically zero. Hence the weight will be in equilibrium wherever they are

placed.

Ex.10 A frame consists of the bare forming the sides of a rhombus ABCD with the

diagonal AC. If four equal forces P act inwards at the middle points of the sides, and at

right angles to the respective sides, prove that the tension in AC is

P cos2

sin

where denotes the angle BAC.

Sol.

Let ABCD be the rhombus, connected Y

by the rod AC forming its diagonal. Let BAC A

be the angle . For the sake of simplicity we P P

can suppose that we give displacement such

that becomes , the point of intersection H E

O of the diagonals and the directions of the lines D X

OB and OA remaining fixed. This means that O B

there is alteration in the length of AC and also in G F

the position of the middle points E, F, G, and H of P P

the sides. C

r =

( ) + ( ),

33

Now the length of the diagonal AC is 4a cos, where 2a is the side of the

rhombus. The work done by the tension in the rod AC is

T(4a cos),

where T is the tension.

To find the work done by the equal forces P acting at the middle points E, F, G

and H, take an axis OX along OB and a perpendicular axis OY along OA as shown in the

figure. By symmetry the force P at the four points will do equal work, so that it will

suffice to find the work done by the force at E. The coordinates of E are

x = a sin , y = a cos

and the components of P at E along the positive directions of the axes are

X = - P cos , Y = - P sin

The work done by P in a small displacement is therefore

X x + Y y = - P cos(a sin) - P sin(a cos)

= - aP (cos2 – sin

2)

= - aP cos2

Hence the equation of virtual work gives

- T (4a cos) – 4aP cos2= 0,

i.e., 4a [T sin – P cos2= 0,

so that P cos2

sin

Ex.11 A uniform square lamina rests equilibrium in a vertical plane under gravity with

two of its sides in contact with smooth pegs in the same horizontal line at a distance „c‟

apart. Show that the angle made by a side of the square with the horizontal in a non-

symmetrical position of equilibrium is given by

c(sin + cos) = a,

2a being the length of a side of the square.

Sol.

Let the reactions at the pegs P and Q

meet in the point E, then the vertical through C

G must pass through E.

Let the body be given a small vertical G

displacement. If z be the height of C.G. above D S

the horizontal line PQ, the equation of virtual E R B

work is Q M

Wdz = 0 or dz = 0 45 P

But z = GM = GN – NM = GN – LP

= AG sin( + 45) – AP sin A N L

= 2 a sin( + 45) – c cos sin

dz = +2 a cos( + 45) – c (cos2 - sin

2)

So 0 = +2 a (cos cos45 – sin sin45) – c (cos2 - sin

2)

Hence a = c (cos + sin) which gives the position of equilibrium.

T =

34

Ex.12 Two equal light rods AOB, COD freely jointed at O, their middle points, are rest in

a vertical plane with their ends B, C on smooth horizontal table. A string to the ends of

which equal weight are attached, passes over A and D. Show that, in the position of

equilibrium, in the angle between the rods is tan1

(4/3).

Sol.

Since the equal rods are joined at their

middle point and the weights are equal. It is D A

clear that the string must be placed symmetrical

over the rods. Let 2 be the supposed angle

between the rods. a

Let the systems be given a small vertical W O W

displacement so that angle changes. Let 2a be

the length of each rod.

The equation of virtual work is y y

- 2W dy = 0, i.e., dy = 0.

Also the length of the string = l.

So 2AC – 2y + AD = l. B C

Hence 4a cos – 2y + 2a sin = l,

i.e. (- 2a sin + a cos) d – dy =

Therefore 2a sin = a cos

i.e. tan = ½

So 2 tan 2½ 4

1 – tan2 1 – ¼ 3

Hence angle between the rods = tan-1

(4/3).

Ex.13 A solid hemisphere is supported by a string fixed to a point on its rim and to a

point on a smooth vertical wall with which the curved surface is in contact. If are the

inclinations of the string and the plane base of the hemisphere to the vertical show that.

tan = 3/8 + tan

Sol.

Let O be the point of suspension and

C the centre of the hemisphere. The normal O

reaction at F will be perpendicular to the l

wall and also pass through C. Hence FC is D

horizontal.

If G is the C.G. of the hemisphere, K A

CG = 3/8.

Let OA = l.

Let y be the depth of G below the F E H C

fixed point O.

y = OK + AE + HG B

= l cos + a cos + (3/8) a sin (1) G

The only force doing work will be the

weight of the hemisphere.

The equation of virtual work is

tan2

35

wdy = 0; so dy = 0.

But from (1),

dy = - l sin d – a sin d + (3/8) a cos d

Hence l sin d = [(3a/8) cos - a sin] d

Also and are connected by the relation

l sin + a sin = a [since CF = CE + AK].

Hence l cos d + a cos d= 0.

Therefore l sin d = a[(3/8) cos - sin] [- (l cos)/(a cos)]d

So sin cos = - (3/8) cos cos + cos sin

Dividing by cos cos,

we get tan = - 3/8 + tan

or tan = 3/8 + tan

Ex.14 Six equal heavy beams are freely jointed at their ends to form a hexagon are placed

in a vertical plane with one beam resting on a horizontal plane; the middle points of the

two upper slant beams, which are inclined at an angle to the horizon, are connected by a

light cord. Show that its tension is 6W cot, where W is the weight of each beam.

Sol.

Let a be the length of each beam and

G1, G2, G3, G4, G5, G6 be their centres of gravity. G1

The points G2 and G6 are connected by E D

a cord of tension T. T W

DCH = EFK = G2 G6

Let the system be given a slight downward W T

vertical displacement. If the heights of G3, G4, F

C

above AB be each equal to x, that of G2, G6 will be K H

3x and that of G1 will be 4x. G3 G5

The work done by the weights W G4 W

= - 2Wdx – 2Wd(3x) – Wd(4x) = - 12Wdx

[- ve, since weights act downward A B

and x increases upward].

Work done by tension = - Tdp, where p = G2G6

= - T(- 2a sin d) = 2a + 2a cos

and so dp = - 2a sin d = T.2a sin d.

The equation of virtual work is

- 12Wdx + T.2a sin d= 0

Again, since x = a sin , so dx = a cos d

- 12W a cos d+ T.2a sin d= 0.

Hence T = 6W cot.

Ex.15 Four rods are jointed together to form to form a parallelogram, the opposite joints

are joined by strings forming the diagonals, and the whole system is placed on a smooth

horizontal table. Show that their tensions are proportional to their lengths.

Sol.

Let the string AC = x and string BD = y.

W

36

Let the system be supposed given a slight displacement so that only the lengths of

the strings change.

The equation of virtual work is D C

T x – T y = 0. (1)

If AB = a and AD = b, then T T

x2 + y

2 = 2(a

2 + b

2) (by geometry)

Hence 2x x + 2y y = 0 T T

From (1) and (2) A B

T/x = T/y.

Hence the tensions are proportional to the lengths of the strings.

Ex.16 A uniform square lamina of side a and weight w is suspended from four points in a

horizontal plane by equal inextensible vertical strings of length l attached to the corners

of a square. Show that the couple that would be necessary to hold the lamina in this

position in which it has been turned through an angle from the former position is

wa2 sin

2l2 – 2a

2 sin

2½)

Sol.

ABCD is the square and AL, BM, CN

and DP are the strings. If O is the centre of the P

square, let O be the displaced position of O

when a couple of moment G is applied to it. L N

Draw OK CN

So KOC = , CN = CN = l M

and OK = OC = a/2 O

D K

If OO = y, the equation of virtual work is

wdy – Gd = 0 (1) A O C

also CN2 = CK

2 + KN

2 B

So l2 = 2a

2 sin

2½ + (l – y)

2 (2)

Hence 0 = 2a2.2 sin½ cos½ ½d + 2(l – y)(- dy)

So (l – y)dy = ½ a2 sin d

Comparing with (1);

w 2G

l – y a2 sin

wa2 sin

2l2 – 2a

2 sin

2½)

Ex.17 Six equal rods AB, BC, CD, DE, EF, and FA are each of weight W and are freely

jointed at their extremities so as to form a hexagon; the rod AB is fixed in a horizontal

position and the middle point of AB and AD are jointed by a string; prove that its tension

is 3W.

Sol.

Let G1, G2, G3, G4, G5 and G6 be the middle points of the rods. Since, by

symmetry, BC and CD are equally inclined to the vertical the depths of the points C, G3

and D below AB are respectively 2, 3 and 4 times as great as that of G2.

=

G =

C

37

Let the system undergo a displacement in the vertical plane of such a character

that D and E are always in the vertical lines through B and A and DE is always

horizontal. If G2 descend a vertical distance x, then G3 will descend 3x, G4 will descend

4x, whilst G5 and G6 will descend 3x and x respectively.

The sum of the virtual works done by

the weights G1

= W.x + W.3x + W.4x + W.3x + W.x A B

= 12 W.x. G6 T G2

If T be the tension of the string, the W W

virtual work done by it will be F C

T (- 4x). G5 G3

For the displacement of G4 is in a W G4 T W

direction opposite to that in which T acts E D

and hence the virtual work done by it is negative. W

The principle of virtual work then gives

12W.x + T(- 4x) = 0, i.e. T = 3W.

Ex.18 Four equal uniform rods are jointed to form a rhombus ABCD, which is placed in

a vertical plane with AC vertical and A resting on a horizontal plane. The rhombus is

kept in the position which BAC = by a light string joining B and D. Shew that its

tension is 2W tan, where W is the weight of rod.

Sol.

Let x be the height above A of the middle points of AB and AD, so that 3x is

clearly the height of the middle points of BC and CD.

Let BO = y = OD, where O is the centre C

of the rhombus. Choose as our displacement one

in which becomes and hence x becomes G3 G2

x + x and y becomes y + y. W W

Then, T being the tension of BD, the

equation of virtual work is D T O T B

2T(-y) + W(-x) + W(-x) + W[-(3x)] + W[-(3x)] = 0. G4 G1

So T = - 4W(x/y), W W

Now, if AB = 2a, we have x = a cos and y = 2a sin. A

Hence x - a sin

y 2a cos

Therefore T = 2W tan

Ex.19 A uniform beams rests tangentially upon a smooth curve in a vertical plane and

one end of the beam rests against a smooth vertical wall; if the beam is in equilibrium in

any position, find the equation to the curve.

Sol.

Take the wall as the axis of y and any point O on it as the origin. If y be the height

of the centre of gravity of the beam above Ox, the equation of virtual work becomes

W.y = 0. So y = constant = h.

Hence G is the point (a cos, h), where 2a is the length of the rod and is its

inclination to the horizontal.

= = - ½ tan

38

Hence the equation to AG is B

y – h = tan (x – a cos) y G

= x tan – a sin. P

For its envelope, differentiating with

respect to , we have x = a cos3 and A

y – h = - a sin3.

Hence x2/3

+ (y – h)2/3

= a2/3

, so that

the required curve is a portion of a four-cusped O M x

hypocycloid.

1.5 Catenary

Definition:

A string is said to be perfectly flexible if the action across any normal section of it

is a single force acting along the tangent to the string. Such a string, therefore, offers no

resistance to any action tending to bend it at a point i.e., it possesses no rigidity of shape.

The section of the string is considered to be so small that it may be treated as a

curved line.

A chain, whose links are very small and perfectly smooth, may be treated as a

flexible string.

A string or a chain is taken as uniform if its weight per unit length is the same

throughout.

The curve in which a uniform inextensible heavy string hangs freely under gravity

is called a Catenary.

Equation of Catenary:

Theorem:

A uniform heavy inextensible string hangs freely under the action of gravity. Find

the equation of the curve, which it forms.

Proof:

Let A be the lowest point of the string where the tangent is horizontal. Let P be

any point on it. If arc s be measured from A, let arc AP = s.

Consider the equilibrium of the portion AP of the string. The string is under

tension due to rest of the part.

The three forces acting on it are:

(i)Its weight acting vertically down through its C.G.

(ii)Tension T0 at A along the tangent in the sense PA.

(iii)Tension T at P along the tangent in the sense AP.

Since these three forces are in equilibrium, the line of action of the weight must

pass through the point of intersection of tangents at A and P.

39

Resolving horizontally and vertically,

T cos = T0

T sin = ws

where w is the weight per unit length of the string.

tan = ws/T0

If T0 be taken equal to ws i.e., weight of T

length c of the string, then P

tan = (ws)/wc S

s = c tan T0 A Q

which is the required intrinsic equation of the catenary.

Note: ws

The point A is called the vertex of the catenary and c is called the parameter.

The horizontal line below A at a distance c from it is called the directrix.

The vertical line through the lowest point A is called the axis of the catenary.

If the catenary is formed by suspending the string from two points R and S at the

same horizontal level the distance RS is called the Span.

Relation between x, y and :

The intrinsic equation is

s = c tan

So ds/d = c sec2

Now dy/d = (dy/ds).(ds/d)

= sin.c sec2

So dy/d = c tan sec

Integrate

y = c sec + A

Let y = c when = 0

i.e., the origin is taken at depth c below A.

So c = c.1 + A, hence A = 0

Therefore y = c sec.

Again dx/d = (dx/ds).(ds/d) = cos c sec2 = c sec

Integrate

x = c log (sec + tan) + B

Let us take the y-axis passing through the lowest point,

therefore = 0 when x = 0

i.e., 0 = 0 + B, hence B = 0.

Therefore x = c log (sec + tan).

Cartesian Equation of the catenary:

The intrinsic equation is

s = c tan = c (dy/dx)

40

So ds d2y

dx dx2

dy d2y

dx dx2

So 1 d2y/dx

2 .

c [1 + (dy/dx)2]

Integrating

x/c = sinh– 1

(dy/dx) + A

where A is an arbitrary constant.

Let the y-axis be taken as a vertical line through A

dy/dx = 0 when x = 0, so A = 0

Therefore x/c = sinh– 1

(dy/dx),

hence dy/dx = sinh(x/c)

Integrate again,

we get y = c cosh(x/c) + B

Let the origin be taken vertically below A at a depth c from it. Hence at A, y = c

when x = 0.

Therefore c = c + B i.e., B = 0

Hence y = c cosh(x/c) is the required equation of the catenary.

Relation between x and s:

y = c cosh(x/c)

So dy/dx = sinh(x/c)

Therefore tan = sinh(x/c)

or s/c = sinh(x/c)

i.e., s = c sinh(x/c)

Tension at a point:

(a)We know that T cos = T0 = wc.

Therefore T = wc sec

Hence T = wy [since y = c sec]

Therefore tension at a point is equal to the weight of a length of the string equal to

its height above the directrix.

In other words, the tension at any point P of a catenary is equal to the weight of

the portion of the string whose length is the vertical distance between P and the directrix.

(b)Since T0 = wc, the tension at the lowest point is equal to the weight of string of length

equal to its height above the directrix.

(c )Again, since T cos = T0 = wc, the horizontal component of the tension at any point

is constant and equal to wc.

(d)Since T sin = ws, the vertical component of the tension at any point is equal to the

weight of the length of string lying between the point and the vertex.

Cor:

Since T = wy.

= c

)2] = c [1 + (

=

41

If a heavy string hangs over two smooth pegs not in the same vertical line, since

the tension at the pegs must support the vertical part of the string, the ends of the string

must rest on the directrix.

Geometrical Properties of Catenary:

1.From the equation T cos = T0, we get that the horizontal component of the tension at

every point of the curve is the same and is equal to wc or T0.

2.From the equation T sin = ws, we deduce that the vertical component of the tension at

any point is equal to ws, i.e. equal to the weight of the portion of the string lying between

the vertex and the point.

3.The equation T0 = wc, shows that the tension at the lowest point is equal to the weight

of the string whose length is the same as the distance between the origin and the vertex.

4.On squaring each side of

y = c sec and s = c tan,

and substituting we have

y2 = s

2 + c

2.

5.If denotes the radius of curvature at any point, then

= ds/d = c sec2

Approximations to the common catenary:

1.The equation of the catenary has been shown to be

y = c cosh(x/c) = ½ c(ex/c

+ e– x/c

)

= c[1 + x2/(2! c

2) + x

4/(4! c

4) + … ],

on expending the exponentials.

If x/c is small, the series on the right can be limited to the first few terms and the

Cartesian equation approximates to

y = c + x2/(2c),

where we have retained the first two terms.

This shows that so long as x is small and c large, the curve coincides very nearly

with a parabola of latus rectum

2c or 2T0/w.

2.We now obtain an approximation to the shape of catenary when x is large, i.e. at points

far removed from the lowest point. When x is large, e– x/c

becomes very small, hence

y = ½ c(ex/c

+ e– x/c

) behave as y = ½ c ex/c

.

Thus for very large values of x, catenary behaves as an exponential curve.

Sag of a tightly stretched wire:

We shall now consider a wire stretched nearly horizontal – as for instance a

telegraph wire. Thus let B, C be two points in a horizontal line between which a wire is

stretched. Let l be the length, W the total weight, T0 the horizontal tension, k the sag AN

and h the span BC.

42

The tension T0 may be found in terms

of W from the first principles. For instance, B N C

taking moments about C for the portion AC of

the wire, we have

T0.k = ½ W.¼l

approximately, or T0

T0 = Wl/(8k). A

We can calculate the increase in the W/2

length of the wire on account of the sag by

considering the equation s = c sinh(x/c).

Since cos2 = c, c will be large if the curve is flat near its vertex. It follows that

x/c will be small in the case of a tightly stretched string. Retaining only the two terms in

the expansion of sinh(x/c) we get from (1), s = x + x3/(6c

2).

Hence

s – x = AC – NC = x3/(6c

2) approximately

w2x

3

6 T02

If we put x = ½ h, we obtain that the total increase, due to sagging, in a span of

length h, is

w2h

3

24T02

Ex.1Show that the length of an endless chain which will hang over a circular pulley of

radius a so as to be in contact with two-thirds of the circumference of the pulley is

3 . 4

log(2 + 3) 3

Sol.

Let ABLCA be the endless chain Y

hanging over the circular pulley MBLC of L

radius a. Since the chain is in contact with

two-thirds of the pulley, this portion

CLB = (2/3).(2a) = 4a/3. O

The remaining portion CAB will

hang in the form of a catenary with its N

vertex at the lowest point A. Its length is C B

therefore twice the length of AB.

The tangent at B is perpendicular to

OB and since A

COB = 2/3, NOB = /3, 60

and hence the tangent at B is inclined at /3 O X

to the horizontal.

And NB = a cos(/6) = (a3)/2.

Applying the formula x = c log(tan + sec) for the point B, we have

a3 .

2 log(2 + 3) [since = /3].

Hence the length

AB = c tan

=

2s – h =

a + .

c =

0

M

43

3a .

2 log(2 + 3)

The length CAB is accordingly

3a .

log(2 + 3)

It follows therefore that the total length of the chain is

3 . 4

log(2 + 3) 3

Ex.2 A weight W is suspended from a fixed point by a uniform string of length l and

weight w per unit of length. It is drawn aside by a horizontal force p. Shew that in the

position of equilibrium, the distance of W from the vertical though the fixed point is

P W + lw W

w P P

Sol.

Let BC be the string of length l, suspended from the fixed point B, with weight W

hanging at C and drawn aside by a horizontal force P acting at C. The tension at C

balances the resultant of P and W. Hence the direction of the tangent at C is given by

tanc = W/P.

We shall next find the direction of the tangent at B. If X and Y be the horizontal

and vertical components of the tension at B, we have, for the equilibrium of BC,

X = P,

Y = W + lw,

where lw is the weight of the string BC. Thus Y

the direction of the tangent at B is given by

tanb = (W + lw)/P. B

Let the horizontal distance of C from A, X

the lowest point of the catenary formed by BC,

be x. Then C

P

s = c tanc = c sinh(x/c), where AC = s, A

whence

x = c sinh–1

(tanc) = c sinh–1

(W/P). W

Also the horizontal component of the

tension at any point is P. Hence

P = wc,

therefore

c = P/w

Accordingly

x = (P/w) sinh–1

(W/P).

Similarly we can show that the horizontal distance of B from A is given by

x = (P/w) sinh–1

[(W + lw)/P].

Hence the required horizontal distance

= x – x

P W + lw W

w P P

=

=

a + .

sinh–1

- sinh–1

sinh–1

- sinh–1

=

44

Ex.3 The end links of a uniform chain slide along a fixed rough horizontal rod. Prove that

the ratio of the maximum span to the length of the chain is

1 + (1 + )2

where is the coefficient of friction.

Sol.

Let AB be the maximum span. In this

position, the end links A and B are in a state

of limiting equilibrium. If the reaction at A be R R

R, then (where tan = ) is the angle that

the resultant of R and the frictional force R

makes with the direction of R. For the R R

equilibrium of A, the direction of this resultant B

o A

should be opposite to the direction of the

tension at A, hence the tangent at A makes an angle 900 – with the horizontal.

The length of the chain

= 2s = 2c tan = 2c cot= 2c/,

and the maximum span

AB = 2x = 2c log(tan + sec) = 2c log (cot + cosec)

1 + (1 + )2

Hence the required ratio

2x

2s

1 + (1 + )2

Ex.4 A kite is flown with 600 ft. of string from the hand to the kite, and a spring balance

held in a hand shows a pull equal to the weight of 100 ft. of the string, inclined at 300 to

the horizon. Find the vertical height of the kite above the hand.

Sol.

Let PB be the string with the kite at B.

The hand at P experiences a pull of 100w,

where w is the weight per unit length of the Y

string. PB forms part of the catenary APB, B

tangent at P giving the direction of the pull.

Since the tension T at any point is equal to

wy, where y is its ordinate, we have at P,

T = wy = 100w,

therefore y = 100.

Also y = c sec, A P

hence c = y cos = 100 cos300

= 503 ft. 30

L O M X

log

= 2c log

=

= log

0

45

Further

AP = s = c tan = 503.tan300 = 50 ft.

We have to find the vertical height of B above P. Let this be k. Then the ordinate

of B will be (100 + k) ft. Applying the formula

y2 = c

2 + s

2,

for the point B, we get that

(100 + k)2 = (503)

2 + (650)

2, since APB = AP + PB

= 430000 = 431002.

Therefore

k = 100(43 – 1) = 555.7 ft. nearly.

Ex.5 A uniform chain is hung up from two points at the same level and distance 2a apart.

If z is the sag at the middle, shew that z = c [cosh(a/c) – 1].

If z be small compared with a, shew that 2cz = a2 nearly.

A telegraph wire is supported y two poles distant 40 yards apart. If the sag be one

foot and the weight of the wire half an ounce per foot, show that the horizontal pull on

each pole is ½ cwt. nearly.

Sol.

Let CAB be the uniform chain, its equation, referred to OX and OY as axes, is

y = c cosh(x/c),

where OA = c. Y

The sag is, therefore C B

= OD – OA = c cosh(a/c) – c D

= c [cosh(a/c) – 1],

for the „y‟ of D is the same as the „y‟ of B A

whose x-coordinate is given to be a.

Now

a 1 a2 1 a

4 O X

c 2! c2 4! c

4

Hence substituting in (1), we get

1 a2 1 a

4

2! c 4! c3

When z is small compared with a, c must be large and hence neglecting the

second subsequent terms in the above expansion, we get

2cz = a2 nearly.

In the numerical part,

z = 1 ft. and a = 60 ft., hence

c = a2/(2z) = 6060/2 = 1800 ft.

The required horizontal pull at A

= wc = (1/32).1800 lbs. = ½ cwt. nearly.

Ex.6 A telegraph wire, stretched between two points at a distance „a‟ feet apart, sags n ft.

in the middle; prove that the tension at the ends is approximately w[a2/(8n) + 7n/6].

Sol.

If y be the ordinate at the extreme

y = n + c, so y – c = n

cosh = 1 + + + … ,

z = + + … ,

46

Also y = c cosh(x/c) = c cosh(a/2c), since x = a/2

1 a2 1 a

4 Y

2! 4c2 4! 16c

4

Therefore

a2 a

4

8c 24c3 B a/2 A

a2 a

4 n

8 24c2 y

But to the first approximation c

C = a2/(8n)

O X

Hence

a2 a

464n

2

8 24a4

= a2/8 + n

2/6

or c = a2/(8n) + n/6

Now T = wy = w(c + n)

a2 7n

8n approx.

Ex.7 A chain ABC is fixed at A and passes over a smooth peg at B and the portion BC is

vertical. The length of each of the portions AB, and BC is 15 ft. and depth of B below the

horizontal through A is 5 ft. Find the horizontal pull at A, having given that wt. of whole

string is 45 lbs.

Sol.

Since BC = 15 ft. and the string passing over a smooth peg retains the same

tension, this tension balances the weight of the string BC. So, if B is the point (x, y) then

wy = w.15 or y = 15 ft.

Let arc DB = s1 and arc DA = s2.

Therefore, at A(x, y), y = 15 + 5 = 20 ft.

Now from the relation,

y2 = c

2 + s

2

At B; (15)2 = c

2 + s1

2 (1) A Y

At A; (20)2 = c

2 + s2

2 (2)

Subtracting them, we get s2 s1

(20)2 – (15)

2 = s2

2 – s1

2

175 = (s2 – s1)(s2 + s1) D 15 ft.

But s2 + s1 = 15, therefore

s2 – s1 = 175/15 = 35/3 ft.

Hence s1 = 5/3 ft. O C X

Now from (1)

225 = c2

+ 25/9

So c = 205/3 ft.

Hence w = wt. per unit foot of length = 45/30 = 3/2 lbs.wt.

Therefore, tension at the lowest point = wc

= (3/2)205/3 lbs.wt. = 105 lbs.wt.

= c[1 + + + … ]

n = y – c = + approx.

cn = +

cn = +

= w +

47

Ex.8 A uniform chain of length 2l and weight W is suspended from two points, A and B,

in the same horizontal line. A load P is now suspended from the middle point D of the

chain and the depth of this point below AB is found to be h. Show that each terminal

tension is

Pl W(h2 + l

2)

h 2hl

Sol.

Let the weight P be suspended from the middle point D of the string. Let C be the

lowest point of the catenary of which DA is part. (DA and DB are not parts of the same

catenary)

Let arc CD = s and ordinate of D be y. Let T, T be the tension at D towards the

two parts. Then the resultant of T, T at angle (180 – 2) = 2T cos(90 - ) = 2T sin.

For equilibrium at the point D,

2T sin = P

2wy sin = P

or 2Wys B A

2ly

or s = lP/W h l

Also at D, y2 = c

2 + s

2

at A, (y + h)2 = c

2 + (s + l)

2 T T

Subtract, 180 – 2

2hy + h2 = 2ls + l

2

So l2 – h

2 2l lP

D

2h 2h W

or l2 – h

2 l

2P C P lbs C

2h hW

Therefore, tension at A

= w(h + y)

W l2 – h

2 l

2P

2l 2h hW

Pl W(h2 + l

2)

h 2hl

Ex.9 A uniform chain, of length 2l, has its ends attached to two points in the same

horizontal line at a distance 2a apart. If l is only a little greater than a, shew that the

tension of the chain is approximately equal to the weight of a length

a3

. 1/ 2

6(l – a)

of the chain, and that “sag”, or depression of the lowest point of the chain below its ends

is ½[6a(l – a)]nearly.

Sol.

Since l is very little greater than a, the tension of chain must be very great and

hence c must be large.

Now from

½ +

= P

y = +

y = +

h + + =

= ½ +

48

s = c tan = ½ c[ex/c

– e–x/c

], we get

l = ½ c[ea/c

– e–a/c

], we get

or 2a 2a3

c 6c3

or a3 a

5

6c2 120c

4

A first approximation is then l – a = a3/(6c

2), and hence

a3

. 1/ 2

6(l – a)

so that the tension at the lowest point is equal to a weight of this length of the chain.

The ordinate of the end of the chain is,

y = ½ c[ea/c

+ e–a/c

], we get

or 2a2 2a

4

2c2 24c

4

or a2 a

4

2c 24c3

Hence the sag of the lowest point

= y – c = a2/(2c) approx.

6(l – a) 1/ 2

a3

= ½[6a(l – a)]

It follows easily that if d be the sag of the lowest point, then the tension there is

approximately equal to a length a2/(2d) of the chain.

Ex.10 A uniform chain of length l hang between two points A and B which are at a

horizontal distance „a‟ from one another, with B at a vertical distance b above A. Prove

that the parameter of the catenary is given by 2c sinh(a/2c) = (l2 – b

2).

Sol.

Let C be the vertex of the catenary of which AB is a part. Let A be the point (x, y)

and arc AC = s. So

y = c cosh(x/c) = ½ c[ex/c

+ e–x/ c

] (1)

At B; y + b = c cosh[(x + a)/c] Y

= ½ c[e(x + a)/c

+ e–(x + a)/ c

] (2)

s = c sinh(x/c) = ½ c[ex/c

– e–x/ c

] (3) a B

s + l = c sinh[(x + a)/c] b

= ½ c[e(x + a)/c

– e–(x + a)/ c

] (4) C l

From (1) and (2) subtracting: A

b = ½ c[e(x + a)/c

+ e–(x + a)/ c

] – ½ c[ex/c

+ e–x/ c

]

From (3) and (4) subtracting:

l = ½ c[e(x + a)/c

– e–(x + a)/ c

] – ½ c[ex/c

– e–x/ c

] O X

Therefore,

l + b = c[e(x + a)/c

– ex/ c

] = c ex/c

[ea /c

– 1]

l – b = – c[e–(x + a)/ c

– e–x/ c

] = – c e–x/ c

[e–a / c

– 1]

Multiplying them, we get

l2 – b

2 = – c

2[– e

a / c – e

– a / c + 2] = c

2[e

a / c + e

– a / c – 2]

= c2[e

a / 2c e

– a / 2c]2 = c

2 4 sinh

2(a/2c), i.e. (l

2 – b

2) = 2c sinh(a/2c)

l = ½ c[ + + … ]

l = a + + +

…

c =

y = ½ c[2 + + + … ]

y = c + + + …

= ½ a2

49

Ex.11 A uniform string, of length 18a ft. and weight 3W lbs., lies on a horizontal smooth

table along with a rod of weight 2w 1bs., to the ends of which its ends are attached. When

the middle point of the string is raised to a height 7a, the pressure on the table just

vanishes. Show that the length of the rod is 16a log 2 ft.

Sol.

Let C be the vertex of the catenary of which AD is a part. When the pressure of

the rod on the table just vanishes, the tension at A and B just balance the weight of the

rod.

Therefore, resolving the forces vertically;

2T sin = 2W,

wy(s/y) = W

where A is the point (x, y) and arc AC = s D

But 3W W

18a 6a 9a 9a

So 3Ws 7a

18a

Therefore s = 6a T T

Now at A; y2 = c

2 + (6a)

2

and at D; (y + 7a)2 = c

2 + (15a)

2 C

Subtract; 14ay + 49 a2 = 189a

2

or y = 10a, so c = 8a

Again since s = c sinh(x/c)

At A; 6a = 8a sinh(x/c)

Therefore

x/c = sinh–1

(3/4) = log[3/4 + (1 + 9/16)]

So x = c log2

Again at D; 15a = 8a sinh[(x + b)/c]

So (x + b)/c = sinh–1

(15/8) = log[15/8 + (1 + 225/64)]

= log4 = 2 log2

Therefore x + b = 2c log2, so b = c log2

Hence length of the rod = 2c log2 = 16a log2

Ex.12 A heavy uniform string 90 inches long hangs over two smooth pegs at different

heights. The parts, which hang vertically, are of lengths 30 and 33 inches. Prove that the

vertex of catenary divides the whole string in the ratio 4:5, and find the distance between

the pegs.

Sol.

Let s1 and s2 be the actual distances of the two pegs from the vertex C of the

catenary, so that s1 + s2 = 27. If c be the parameter of the catenary then by y2 = c

2 + s

2, we

get s12 + c

2 = 30

2 and s2

2 + c

2 = 33

2; for the ends of the strings must lie on the directrix of

the catenary by the property (a) of Art 4.6, since the tension of the string is unaltered by

its passing round a smooth peg.

Hence, easily, s1 = 10, s2 = 17 and c = 202, so that

s1 + 30 4

s2 + 33 5

w = =

= W

A B

=

50

Also from y = c cosh(x/c), we have

y + (y2 – c

2)

c

Hence, when y = 30 or 33, x = 102 loge2 or 102 loge(25/8).

Therefore,

x1 + x2 = 202 loge2.5 = 28.2840.9163 = 25.92,

so that the horizontal and vertical distance between the pegs are 25.92 and 3 inches.

Ex.13 A box kite is flying at a height h with a length l of wire paid out, and with the

vertex of the catenary on the ground; shew that at the kite the inclination of the wire to

the ground is 2 tan–1

(h/l), and that its tensions there and at the ground are

w(l2 + h

2)

2h

and

w(l2 – h

2)

2h

where w is the weight of the wire per unit of length.

Sol.

Let P is the kite and C is on the ground, so that NP = h + c and hence, from the

triangle NYP, Y

(h + c)2 = NY

2 + YP

2 = c

2 + l

2

So (l2 – h

2) A B

2h

and NP = c + h T

(l2 + h

2) ws P

2h

Also cos = c/(c + h) T0 C Y

(l2 – h

2) c

l2 + h

2

so that tan(/2) = h/l O T N X

Also the required tensions at P and C are w.PN and w.c.

Ex.14 A uniform chain, of length 2l and weight W, is suspended from two points, A and

B, in the same horizontal line. A load P is now suspended from middle point D of the

string; if AB=2a, find the depth below AB of the position of D.

Sol.

Let C be the lowest point of the catenary of which DA is a part; let its parameter

be c, let the arc CD = s and let the ordinate of D be y. Then

P = vertical component of the tensions at D

= 2T sin

2Wys Ws

2ly l

Also

y2 = c

2 + s

2

and (y + h)2 = c

2 + (s + l)

2 (1)

if h be the required depth.

c =

=

=

= =

x = c loge

51

Theses equations give s = lP/W, and

l2(W + 2P) h

2hW 2 (2)

Also, if x be the abscissa of D, then by

y = c cosh(x/c) and s = c sinh(x/c), we have

y + s = c ex/c

and y + h + s+ l = c e(x + a)/c

So h + l

y + s

On substituting from (1) and (2), we have an equation to give h.

Ex.15 A chain of length 2l, is hung over two small smooth pulleys which are in the same

horizontal line at a distance 2a apart; to find the positions of equilibrium and to determine

whether they are stable.

Sol

Since the tension of the chain is A a a A

unaltered by passing over the pulley and since

on one side it is equal to the weight of the

free part AN and on the other it is equal to C

the weight of the chain that would stretch

vertically down to the directrix, it follows c

that N and also N lie on the directrix of the

catenary. x

Hence N O N

l = arc CA + line AN

= ½ c[ea/c

– e–a/c

] + ½ c[ea/c

+ e–a/c

]

= c ea/c

(1) R

where c is the parameter of the catenary.

Equation (1) cannot be solved

algebraically, but a graphic solution may be Y P

obtained as follows when a and l are given

numerically. Q

Put a/c = X; then eX = (l/a)X (2)

Draw the curve Y = eX and

the straight line Y = (l/a)X.

The points Q and R in which they cut give,

on measurement of their abscissa, approximate O X

solutions for X and hence for c. It is clear

that there will be two real, coincident or

imaginary solutions according as A a a A

(l/a) = tan POX,

where OP is the tangent from O to the curve.

Now P is given by C

Y/X = tan POX = dY/dX = eX =Y,

and is therefore the point (1, e), so that c

tan POX = e.

– y =

ea/c

= 1 +

> <

52

There are therefore two, one or no possible catenaries according as

l = ae.

One catenary will be somewhat as drawn in the first figure and the other as in the

annexed one. The parameter c of the first case is clearly greater than in the second.

Stability or instability:

The height of the centroid of the catenary above its directrix = (cx + ys)/(2s).

So its depth below AA

cx + ys ys – xc

2s 2s

Hence the depth below AA of the C.G. of the whole chain

2s.(ys – xc) 2y.y

2s 2

2s + 2y

y2

+ ys – xc

2(y + s)

Now in the above case x = a; y = ½ c[ea/c

+ e–a/c

], s = ½ c[ea/c

– e–a/c

] and

l = y + s = c ea/c

.

Therefore depth below AA of the centre of gravity of the whole chain

yea/c

– a

2ea/c

(l/c).(c/2)[l/c + c/l] – a

2l/c

l2

+ c2 – 2ac

4l

(l2

– a2) + (c – a)

2

4l

Hence, the greater c, the greater is the depth of the centre of gravity below AA‟.

Hence the first form of the possible curves the stable one, and the second is the unstable.

Ex.16 A uniform heavy string of given length l is attached to two points P and Q, the

latter point being horizontal and vertical distances, h and k, from P; to find the parameter

c of catenary in which it rests.

Sol.

Let P be the point (x, y) referred to the directrix Ox and the vertical line through

the lowest point of the Catenary. Then,

y = ½ c[ex/c

+ e–x/c

] (1)

y + k = ½ c[e(x + h)/ c

+ e–(x + h)/ c

] (2)

and l = SQ – SP = ½ c[e(x + h)/ c

– e–(x + h)/ c

] – ½ c[ex/c

– e–x/c

] (3)

Therefore

l + k = ce(x + h)/c

– cex/c

= cex/c

(eh/c

– 1) (4)

and

l – k = – ce–(x + h)/c

+ ce–x/c

= ce–x/c

(1 – e–h/c

) (5)

Hence

l2 – k

2 = c

2 (e

h/c – 2 + e

–h/c)

giving

< >

= y – =

= +

=

=

=

=

=

53

(l2 – k

2)= c (e

h / 2c – e

–h / 2c) (6)

This equation can not be solve algebraically. A graphic solution may be obtained

by putting h/2c = X, so that (6) gives

(l2 – k

2) X

h (7)

and hence X is given as the point in which the straight lines

(l2 – k

2) X

h

meet the curve Y = sinhX.

On drawing the curve we obtain two equal and opposite values for X, and hence

two equal and opposite values for c, provided that

(l2 – k

2)

h

i.e. provided that l is greater than length PQ.

The value of c being now know to any degree of approximation, equation (4)

gives x and then equation (1) gives y. The solution is therefore complete.

It is clear that only the positive value of c need be taken. For a negative value of c

would make y negative.

Supposing that a value of X1 as an approximate solution of (7) has been obtained

by graphical methods, a further approximation may be obtained by analysis. For, taking

the upper sign in (6) and putting

(l2 – k

2)

h

we have to solve sinh X = X, where X1 is an approximate solution.

Putting X = X1 + , where is small, we have

sinh(X1 + ) = (X1 + ),

i.e. sinhX1 + coshX1 + … = (X1 + ), by Taylor‟s theorem.

Hence, neglecting square of , we have

X1 – sinhX1 (l2 – k

2) X1 – h sinhX1

coshX1 – h coshX1 – (l2 – k

2)

so that X1 + is a second approximation.

1.6 Unit Summary:

1. A system of forces acting in one plane at different points of a rigid body can always be

reduced to a single force through any given point and a couple.

2. A system of forces acting in one plane at different points of a rigid body can be

reduced to a single force, or a couple.

3. If G = 0, the given forces reduce to the single force R. Hence in every case the forces

can be reduced either to a single force, or a couple.

4. If three forces, acting in one plane upon a rigid body, keep it in equilibrium, they must

either meet in a point or be parallel.

5. If a straight line CD drawn from the vertex C of a triangle ABC to the opposite side

AB divides it into two segments in the ratio of m: n, then

(m + n) cot = m cot – n cot and (m + n) cot = n cotA – m cotB,

sinhX =

Y =

> 1

=

= =

54

where and are the angles which CD makes with CA and CB and the angle which

CD makes with the base AB.

6. The resultant R and the couple G must separately vanish.

7. A system of forces in a plane will be in equilibrium if the algebraic sum of the

moments of all the forces with respect to each of three non-collinear points is zero.

8. A system of forces in a plane will be in equilibrium if the algebraic sum of the

moments about each of any two different points is zero and the algebraic sum of the

resolved parts of the forces in any given direction not perpendicular to the line joining the

given points is zero.

9. Resultant forces and couple corresponding to any base point O of a system of coplanar

forces.

10. Equation to the resultant of a system of forces in one plane.

11. In any given problem, we imagine the body to be displaced a little and find out the

work done during the displacement. The condition of equilibrium is obtained by equating

to zero the sum of the work done. Since the body is not actually displaced, the work done

is called virtual work. The virtual work that is calculated is the amount of work that

would have been done if the displacements had actually been made.

12. The necessary and sufficient condition that a rigid body acted upon by a number of

coplanar forces be in equilibrium is that the algebraic sum of the virtual works done by

the forces in any small displacement consistent with the geometrical conditions of the

system is zero.

13. If the distance between two particles of a system is invariable, the work done by the

mutual action and reaction between the two particles is zero.

14. The reaction R of any smooth surface with which the body is in contact does no work.

15. If the force between two bodies of a system is a mutual pressure at a point of contact,

then no work is done in any virtual displacement of the system by the action and reaction

if the same points of the bodies remain in contact during the displacement.

16. When a body rolls without sliding on any fixed surface the work done in a small

displacement by the reaction of the surface on the rolling body is zero.

17. If any body is constrained to run round a point or on an axis fixed in space, the virtual

work of the reaction at the point or on the axis is zero.

18. The curve in which a uniform inextensible heavy string hangs freely under gravity is

called a Catenary.

19. We know that T cos = T0 = wc.

Therefore T = wc sec

Hence T = wy [since y = c sec]

Therefore tension at a point is equal to the weight of a length of the string equal to

its height above the directrix.

20. Since T0 = wc, the tension at the lowest point is equal to the weight of string of length

equal to its height above the directrix.

21. Again, since T cos = T0 = wc, the horizontal component of the tension at any point

is constant and equal to wc.

22. Since T sin = ws, the vertical component of the tension at any point is equal to the

weight of the length of string lying between the point and the vertex.

23. From the equation T cos = T0, we get that the horizontal component of the tension

at every point of the curve is the same and is equal to wc or T0.

55

24. From the equation T sin = ws, we deduce that the vertical component of the tension

at any point is equal to ws, i.e. equal to the weight of the portion of the string lying

between the vertex and the point.

25. The equation T0 = wc, shows that the tension at the lowest point is equal to the weight

of the string whose length is the same as the distance between the origin and the vertex.

26. On squaring each side of

y = c sec and s = c tan,

and substituting we have

y2 = s

2 + c

2.

27. If denotes the radius of curvature at any point, then

= ds/d = c sec2

1.7 Assignments:

1. The wire passing round a telegraph pole is horizontal and the two portions attached to

the pole are inclined at an angle of 600 to one another. The pole is supported by a wire

attached to the middle point of the pole and inclined at 600 to the horizon; show that the

tension of this wire is 43 times that of the telegraph wire.

2. AB is a diameter of a circle and BP and BQ are chords at right angles to one another;

show that the moments of forces represented by BP and BQ about A are equal.

3. Three forces act along the sides of a triangle; show that, if the sum of two of the forces

be equal in magnitude but opposite in sense to the third force, then the resultant of the

three forces passes through the center of the inscribed circle of the triangle.

4. At what height from the base of a pillar must the end of a rope of given length be fixed

so that a man standing on the ground and pulling at its other end with a given force may

have the greatest tendency to make the pillar overturn.

5. Forces proportional to AB, BC and 2CA act along the sides of a triangle ABC taken in

order; show that the resultant is represented in magnitude and direction by CA and that its

line of action meets BC at a point X where CX is equal to BC.

6. A smooth rod, of length 2a, has one end resting on a plane of inclination to the

horizon, and is supported by a horizontal rail, which is parallel to the plane and at a

distance c from it. Show that the inclination of the rod to the inclined plane is given by

the equation

c sin = a sin2 cos().

7. Two equal circular discs of radius r, with smooth edges, are placed on their flat sides in

the corner between two smooth vertical planes inclined at an angle 2, and touch each

other in the line bisecting the angle. Show that the radius of the smallest disc that can be

pressed between them, without causing them to separate, is r (sec – 1).

8. The altitude of a cone is h and the radius of its base is r; a string is fastened to the

vertex and to a point on the circumference of the circular base, and is then put over a

smooth peg; show that, if the cone rest with its horizontal, the length of the string must be

(h2 + 4r

2).

9. A cylinder, of radius r, whose axis is fixed horizontally, touches a vertical wall along a

generating line. A flat beam of uniform material, of length 2l and weight W, rests with its

56

extremities in contact with the wall and the cylinder, making an angle of 450 with the

vertical. Show that, in the absence of friction, l/r = (5 – 1)/10, that the pressure on the

wall is ½ W, and that the reaction of the cylinder is ½5W.

10. A solid cone of height h and semi-vertical angle , is placed with its base against a

smooth vertical wall and is supported by a string attached to its vertex and to a point in

the wall; Show that the greatest possible length of the string is h[1 + (16/9) tan2].

11. A regular hexagon is composed of six equal heavy rods freely jointed together, and

two opposite angles are connected by a string, which is horizontal, one rod being in

contact with a horizontal plane; at the middle point of the opposite rod is placed a weight

W1; if W be the weight of each rod, shew that the tension of the string is

3W + W1

3

12. Six equal heavy rods, freely hinged at their ends, form a regular hexagon ABCDEF

which when hung up by the point A is kept form altering its shape by two light rods BF

and CE. Prove that the thrusts of these rods are (53W)/2 and (3W)/2, where W is the

weight of either rod.

13. A prism whose cross section is an equilateral triangle rests with two edges on smooth

planes inclined at angles to the horizon. If be the angle which the plane containing

these edge makes with the vertical shew that

23 sin sin + sin ()

3 sin()

14. A smooth rod passes through a smooth ring at the focus of an ellipse whose major

axis is horizontal, and rests with its lower end on the quadrant of the curve, which is

furthest, removed from the forces. Find its position of equilibrium, and shew that its

length must at least be (3a/4) + (a/4)(1 + 8e2), where 2a is the major axis and e is the

eccentricity.

15. A heavy rod AB of length 2l rests upon a fixed smooth peg at C and with its end B

upon a smooth curve. If it rests in all positions, shew that the curve is a conchoids whose

polar equation, with C as origin is

r = l + a/sin

16. A heavy uniform string, of length l, is suspended from a fixed point A, and its other

end B is pulled horizontally by a force equal to the weight of a length a of the string.

Shew that the horizontal and vertical distances between A and B are

a sinh–1

(l/a) and (l2 + a

2) – a .

17. A string, of length l, hangs between two points (not in the same vertical) and makes

angles with the vertical at the points of support. Shew that, if k is the height of one

point above the other and the vertex of the catenary does not lie between them, then.

k cos[()/2] = l cos[()/2].

18. A uniform chain, of length l is suspended from two points A and B, in the same

horizontal line so that either terminal tension is n times that at the lowest point. Shew that

the span AB must be

l .

(n2 – 1)

If l = 100 feet and n = 3 show, from the tables, that the length is about 62.3 feet.

tan =

loge[n + (n2 – 1)]

57

19. A heavy chain, of two length 21 , has one end tied at A and the other is attached to a

small heavy ring which can slide on a rough horizontal rod which passes through A. If

the weight of the ring be n times the weight of the chain, shew that its greatest possible

distance from A is.

2l .

where 1/(2n + 1) and is the coefficient of friction.

1.8 References:

1. Loney, S. L.: An Elementary Treatise on Statics, Cambridge University Press, 1956

2. Varma, R. S.: Text-Book on Statics, Pothishala (Private) Limited Lajpat Road,

Allahabad – 2, 1962.

UNIT –II Forces in Three Dimensions, Stable and

Unstable Equilibrium

STRUCTURE

2.1 Introduction

2.2 Objectives

2.3 Forces in three dimensions

2.4 Poinsot‟s central axis

2.5 Null lines and planes

2.6 Stable and unstable equilibrium

2.7 Unit Summary

2.8 Assignments

2.9 References

loge[ + (1 + 2)]

58

2.1 Introduction:

This unit provide us to understanding the basic concepts of Forces in

three dimensions, Poinsot‟s central axis, Null lines and planes, Stable and

unstable equilibrium. This unit gives a brief idea to understand the above

topics. In this unit we shall study the basic ideas of Forces in three dimensions, Poinsot‟s

central axis, Null lines and planes, Stable and unstable equilibrium. It is hopped the unit

help students in studying.

2.2 Objectives:

At the end of the unit the students would be able to understand the concept of:

Forces in three dimensions

Parallelopiped of Forces

Moment of a force about a point

Couples

General Conditions of Equilibrium

Poinsot‟s Central Axis

Null Lines and Planes

Stable and Unstable Equilibrium

Conditions of stability for a body with one degree of freedom

59

2.3 Forces in three dimensions:

Parallelopiped of Forces:

Forces in three dimensions may be compounded by the law of parallelopiped of

forces, which is an extension of law of parallelogram of forces in two dimensions. It may

be stated as follows:

If three forces acting at a point O are

represented, in magnitude and direction, by

straight lines OA, OB, OC their resultant is

represented in magnitude and direction, by

the diagonal OD of the parallelopiped whose

edges are OA, OB and OC.

For the two forces represented by OA Z

and OB are equivalent to force represented

by OE, the diagonal of the parallelogram C

OAEB. Again the resultant of the forces D

represented by OE and OD, the diagonal of

the parallelogram OEDC, represents OC. O A X

Hence the resultant of forces represented by

OA, OB, OD represents OC. B E

60

If the parallelopiped is rectangular, so Y

that OA, OB, OC are taken along the

rectangular axes of OX, OY, OZ respectively

and the forces. OA, OB, OC be respectively

X, Y and Z and their resultant OD be R, then R2 = X

2 + Y

2 + Z

2

and acts along OD whose direction cosines are cos AOD, cos BOD, cos COD. i.e. OA X

OD R

OB Y

OD R

OA Z

OD R

i.e. X/R, Y/R, Z/R

Conversely a force R acting along a line with direction cosine l, m, n has as

components along axes of co-ordinates

X = lR, Y = mR. Z = nR. Forces in three dimensions:

When forces in different direction in space are acting upon a rigid body, we say that the

body is under the action of forces in three dimensions.

Moment of a force about a point:

The moment of a forces F about any point is equal to F.p, where p is perpendicular

distance of the point from the line of action of force. It is said to be +ive or –ive according as the

force has a tendency to rotate the body about the point in anticlockwise or clockwise direction.

However, in three dimension those forces which try to rotate the body from axis

of x to axis of y, from axis of y to axis of z and from axis of z to axis of x will have + ive

moments about the axis, similarly the moments of couples.

Couples:

We already know that two equal and unlike parallel forces form a couple whose

moment is equal to the product of the force and the arm of the couple. (By arm we mean

the perpendicular distance between the lines of action of the forces of the couple.)

A couple is represented by a line perpendicular to the plane of the couple and

whose length is proportional to the magnitude of the moment. If G be the moment of the

couple and L, M, N be its components on the axis of co-ordinates then G2 = L

2 + M

2 + N

2

and d.c.‟s of its axis are

L/G, M/G, N/G

Conversely a couple G about a line whose d.c.‟s are l, m, n is equivalent to three

couples about the axis whose moments are lG, mG, and nG.

Theorem: Any given system of forces acting at any given point of a rigid body can be

reduced to a single force acting through an arbitrarily chosen point and a couple whose

axis passes through that point.

In other words, a system of forces acting on a rigid body can be reduced in

general to a force acting at an arbitrarily chosen point of the body and a couple.

Or

The resultant of any given system of forces acting at points of a rigid body.

Proof:

Let any arbitrarily chosen point O be taken as origin and three mutually

perpendicular lines through it as axes of co-ordinates.

l = =

m = =

n = =

61

Again suppose that at any point

P1 (x1, y1, z1) acts one of the given forces z Z1

whose components parallel to axes are

X1, Y1, Z1.

From P1, draw P1M1 perpendicular P1 Q1 X1

to x-y-plane and from foot of the

perpendicular M1 draw M1N1 parallel to Z1

axis of y meeting the axis of x in N1, so that Y1 z1 Z1

ON1 = x1, M1N1 = y1 and P1M1 = z1. Now

draw a line Q1N1S1 through N1 parallel to O x

axis of z. x1 N1

Introduce force each equal to Z1 along y y1

N1Q1, N1S1, OZ and OZ. These being two Z1 M1 Z1

sets of equal and opposite forces do not alter

the effect of given forces. Thus we have five z S1

equal forces parallel to axis of z.

(i) Z1 along P1Z1 and Z1 along N1S1 from a couple whose moment is Z1.M1N1 = Z1. y1

in a plane perpendicular to Ox. Also it has tendency to rotate the body from axis of y to

axis of z and as such its moment is +ive and its axis is Ox.

(ii) Z1 along N1Q1 and Z1 along OZ form a couple whose moment is Z1.ON1 = Z1x1 in a

plane at right angles to Oy. Also it has a tendency to rotate the body from axis of x to axis

of z and as such its moment is – ive, i.e. its moment is – Z1x1 about Oy and its axis is Oy.

(iii) A single force Z1 along OZ

Thus the component Z1 of a forces at P1 is equivalent to

A couple of moment +y1Z1 about Ox

A couple of moment –x1Z1 about Oy (1)

A single force Z1 along Oz.

Similarly the component X1 of a force is equivalent to

A couple of moment +z1X1 about Oy

A couple of moment –y1X1 about Oz (2)

A single force X1 along Ox

Again the component Y1 of a force is equivalent to

A couple of moment +x1Y1 about Oz

A couple of moment –z1X1 about Ox (3)

A single force Y1 along Oy

Combining (1), (2) and (3), we can say that the components X1, Y1, Z1 of a force

at P are together equivalent to

A couple of moment (y1Z1- z1Y1) about Ox

A couple of moment (z1x1 –x1Z1) about Oy (4)

A couple of moment (x1Y1- y1X1) about Oz

and forces X1, Y1, Z1 along Ox, Oy, and Oz respectively.

The result D is quite symmetrical and there should be no difficulty in

remembering it.

In a similar manner we can replace other forces acting at other points like (x2, y2,

z2), (x3, y3, z3) etc. whose components are X2, Y2, Z2; X3, Y3, Z3 etc. by couples about the

axes of co-ordinates and forces along the axes.

62

Thus the whole system is equivalent to

A couple of moment (y1Z1- z1Y1) about Ox = L

A couple of moment (z1X1- x1Z1) about Oy = M

A couple of moment (x1Y1- y1X1) about Oz = N

A single force X1 + X2 +……= X1 along Ox = X (5)

A single force Y1 + Y2 +……= Y1 along Oy = Y

A single force Z1 + Z2 +…… = Z1 along Oz = Z

Three forces X, Y, Z, acting at O, are together equivalent to a single force R

through O where R2 = X

2+Y

2+Z

2 and the direction – cosines of its line of action are X/R,

Y/R, Z/R.

Similarly the three couples of moment L, M, N are together equivalent to a single

couple of moment G, where

G2 = L

2+M

2+N

2

and the direction-cosines of its axes are L/G, M/G, N/G.

Thus the entire system has been reduced to a single force R acting at arbitrarily

chosen point O together with a couple of moment G whose axis passes through O.

General Conditions of Equilibrium:

The system will be in equilibrium when R = 0 and G = 0, which in other words

means that X = 0, Y = 0, Z = 0. Also L = 0, M = 0, N = 0.

The sum of the resolved parts of the system of forces parallel to any three axes of

co-ordinates must vanish separately, and also the sums of their moments about the three

axes must vanish separately.

Elements of System:

The six quantities X, Y, Z, L, M, N. are called the elements of the system. It is

very easy to write down their values. If the forces are acting at (x1, y1, z1) and its

components along the axes be given, then write

x1 y1 z1

X1 Y1 Z1

Then X = X1, Y = Y1, Z = Z1

and y1 z1

L1 = i.e. (y1Z1 – z1Y1); so L = L1

Y1 Z1

z1 x1

M1 = i.e. (z1X1 – x1Z1); so M = M1

Z1 X1

x1 y1

N1 = i.e. (x1Y1 – y1X1); so N = N1

X1 Y1

Another form:

If the force P acts along a line

x – x1 y – y1 z – z1

l m n

then its components X1, Y1, Z1 along the axes will be Pl, Pm, Pn respectively and we

should in this case write

= =

63

x1 y1 z1

Pl Pm Pn

then X = X1 = Pl, Y = Y1 = Pm, Z = Z1 = Pn

and y1 z1

L1 = i.e. L = L1

Pm Pn

and z1 x1

M1 = i.e. M = M1

Pn Pl

and x1 y1

N1 = i.e. N = N1

Pl Pm

Rule:

Write down in a line the co-ordinates of the point at which the forces is acting. In

the second line write down the components of the force parallel to axes of co-ordinates

and then form determinates of 2nd

order to get the values of L1, M1, N1 etc. Similarly treat

other force.

2.4 Poinsot’s Central Axis:

Theorem: Any system of forces acting on a rigid body can be reduced to a single force

together with a couple whose axis is along the direction of the force.

Proof:

We know that a system of forces acting on a rigid body can be reduced to a single

forces R acting at O along OA together with a couple G whose axis OD is a line passing

through O.

A D A

G G cos

R R

G sin

O B O B

C C

Fig. (1) Fig. (2) A A

G cos

R

A G cos R

R

O B O B

64

O

C OR C

Fig. (3) Fig. (4)

Let the axis OD of the couple be inclined to the direction of R at an angle Draw

OB perpendicular to OA so that OA, OD and OB all lie in one plane AOB. Now draw

OC perpendicular to plane AOB.

The couple G about OD as axis is equivalent to a couple G cos about OA as axis

and a couple G sin about OB as axis (Fig. 2). The latter couple G sin about OB as axis

acts in the plane AOC because its axis OB is perpendicular to plane AOC. Hence the

couple G sin can be replaced by two equal and unlike parallel forces of moment G sin.

Let one of the forces be R acting at O in a direction opposite to R along OA and

the other will also be R acting parallel to OA at some point O in OC such that its

moment R.OO = G sin, so OO = (G sin)/R (Fig. 3)

The two equal forces R at O in opposite directions each other and we are now to

deal with force R at O and a couple of moment G cos about OA as axis. The axis of a

couple can always be transferred to a parallel axis; hence we can choose OA as the axis

of couple G cos. Thus we are left with a force R along OA and a couple of moment G

cos whose OA is along the direction of force R.

Note:

1.Poinsot’s Central axis:

The line OA, the axis of the single couple and also the line of action of single

forces R to which a system of forces is reduced is called Poinsot‟s Central Axis.

The moment of the resultant of couple about the central axis i.e. G cos is less

than the moment of the resultant couple corresponding to any other point O which is not

on the central axis because G cos <G.

Also central axis for a system of forces is unique.

2.Wrench:

A single force R together with a couple K whose axis coincides with the direction

of the force when taken together is called a wrench of the system. The single force R is

called the intensity of the wrench.

3.Pitch:

The ratio K/R i.e. moments of the couple about central axis divided by forces is

called the pitch and is of a linear magnitude. Also it is clear that when the pitch is zero

the wrench reduces to a single force. When the pitch is infinite then R = 0 and hence the

wrenches reduce to a couple K only.

4.Screw:

The straight line, along which the single forces acts, when considered together

with the pitch, is called a screw. Hence screw is a definite straight line associated with a

definite pitch.

Theorem: Find the condition that a given system of forces should compound into a

single force.

Proof:

65

We know that the forces are equivalent to a single force R acting at an arbitrary

origin O and a single couple G. If be the angle between R and the axis of G, then R is

equivalent to a force R cos along the axis OB of the couple, and a force R sin in the

plane of the couple. This force R sin, together with the parallel forces of the couple, are

equivalent to a parallel force R sin that does not pass through O and therefore cannot in

general, compound with R cos into a single force.

But, if R cos = 0, i.e. if cos = 0, then we are left with a single force R sin.

Hence must be 900, i.e. the angle between the straight lines whose direction cosines are

(X/R, Y/R, Z/R) and (L/G, M/G, N/G) must be a right angle.

So X L Y M Z N

R G R G R G

Therefore XL + YM + ZN = 0 is the required condition.

Theorem: Whatever origin or base point and axes are chosen for any system of forces

the quantities

X2 + Y

2 + Z

2 and LX + MY + NZ

are invariable, where X = X1, etc. and L = (y1Z1 – z1Y1) etc.

Proof:

We know that X2 + Y

2 + Z

2 is the square of the resultant force R corresponding to

the Central Axis and is therefore invariable.

Again, if (l, m, n) are the direction cosines of the resultant force and (l1, m1, n1)

those of the axis of the resultant couple, then

X L Y M Z N

R G R G R G

= the cosine of the angle between the resultant force and

the axis of the resultant couple = cos

Therefore

LX + MY + NZ = R.G cos = R.K,

where K is the moment of the couple about the Central Axis.

Hence I = LX + MY + NZ is an invariant.

It follows that if K be zero, that is, if the given system reduces to a single force,

then LX + MY + NZ = 0.

This second invariant will be zero also when the resultant force R is zero. In this

case the first invariant is zero also.

The pitch, p, of the resultant Wrench of the system = K/R

= the invariant I of the system divided by the square of the invariant R.

Theorem: Equation of Central Axis of any given system of forces.

Proof:

Let (f, g, h) be the coordinates referred to the axes Ox, Oy, Oz of any point Q.

The moment about a line through Q parallel to Ox is clearly obtained by putting x1 – f, y1

– g, z1 – h instead of x1, y1, z1.

Hence the moment

= [(y1 – g)Z1 – (z1 – h)Y1]

= (y1Z1 – z1Y1) – gZ1 + hY1 = L – g Z + h Y

So the moments about lines through Q parallel to the other axes are

+ + = cos900 = 0

+ + = ll1 + mm1 + nn1

66

M – h X + f Z and N – f Y + g X.

Also the components of the resultant force are the same for all points such as Q

and are thus X, Y and Z.

If Q be a point on the central axis, the direction cosines of the axis of the couple

corresponding to it are proportional to those of the resultant force. Hence

L – gZ + hY M- hX + fZ N – fY + gX

X Y Z

LX + MY + NZ K

X2 + Y

2 + Z

2 R

Hence the equation of the locus of the point (f, g, h), i.e. the required equation of

the central axis is

L – yZ + zY M- zX + xZ N – xY + yX

X Y Z

= K/R = the pitch p of the wrench.

Ex.1 A single force is equivalent to component forces X, Y, Z along the axes of

coordinates and to couples L, M, N about three axes. Prove that the magnitude of the

single force is (X2 + Y

2 + Z

2) and the equations to its lines of action are

yZ – zY zX – xZ xY – yX

L M N

Sol.

If the system reduces to a single force, then LX + MY + NZ = 0 and the equation

of central axis i.e. line of action of single force are

L – yZ + zY M- zX + xZ N – xY + yX

X Y Z

or L = yZ – zY , M = zX – xZ , N = xY – yX

or yZ – zY zX – xZ xY – yX

L M N

Since the couples do not affect the magnitude of the single force and hence in

magnitude R = (X2 + Y

2 + Z

2).

Ex.2 Forces P, Q, R act along three non-intersecting edges of a cube; find the central

axis.

Sol.

Let the three edges be OA, CA, CO, whose equations are

x y z

1 0 0

x y z – a

0 1 0

x – a y – a z

0 0 1

z

C (0, 0, a)

B

Q

Y Z

= =

= =

= =

= = = l

= = = 0

= = = l

= =

= =

= =

67

A O P

O R

x

A (a, 0, 0)

B (0, a, 0) R

y X C (a, a, 0)

So X = P, Y = Q, Z = R.

Also moments of couples are

0, 0, 0; – aQ, 0, 0; aR, – aR, 0.

So L = a(R – Q), M = – aR, N = 0.

For equation of central axis write

x y z x y z

X Y Z P Q R

L – yR + zQ M – zP + xR N – xQ + yP

P Q R

a(R – Q) – yR + zQ – aR – zP + xR – xQ + yP

P Q R

Ex.3 OA, OB, OC are the three coterminous edges of a cube and AA, BB, CC and OO

are diagonal. Along BC, CA, AB and OO act forces equal to X, Y, Z and R

respectively. Show that they are equivalent to a single resultant if

(YZ + ZX + XY) 3 + R (X + Y + Z) = 0.

Sol.

Take O as origin and the axes along OA, OB, OC. If a be the edge of a cube, then

coordinates of A, B, C are as marked and those of O are (a, a, a).

z

C (0, 0, a)

B

Q

Y Z

A O

O R

x

P A (a, 0, 0)

B (0, a, 0) R

y X C (a, a, 0)

The lines of action of the three forces are parallel to axes and hence their

equations are

x y – a z

1 0 0

x y z – a

0 1 0

x – a y z

= =

= =

i.e.

= =

= =

= =

for X,

for Y,

for Z.

68

0 0 1

x y z

1/3 1/3 1/3

Since d.c.‟s are

a a a

a/3 a/3 a/3

The components of the forces parallel to axes are

X, 0, 0; 0, Y, 0; 0, 0, Z; R/3, R/3, R/3.

So X = X + R/3, Y = Y + R/3, Z = Z + R/3

It is easy to calculate L1, M1, N1 etc. and consequently

L = – aY, M = – aZ, N = – aX

The system will reduce to a single force if

LX + MY + NZ = 0

or – aY (X + R/3) – aZ (Y + R/3) – aX (Z + R/3) = 0

(YZ + ZX + XY) 3 + R (X + Y + Z) = 0.

Ex.4 Equal forces act along two perpendicular diagonals of opposite faces of a cube of

side a. Show that they are equivalent to a single force R acting a line through the centre

of the cube and a couple ½ aR with the same line for axis.

Sol.

Let the axes be along the edges OA, OB and OC. Let ON be one diagonal and LM

be the perpendicular diagonal of the opposite face. Suppose equal forces P act along these

diagonals whose equations in terms of d.c.‟s are

x y z

a/ a/2 0

x y – a z – a

a/2 –a /2 0

z

C (0, 0, a)

M (a, 0, a)

P

L P (a, a, a)

(0, a, a) O o

x

A (a, 0, 0)

B (0, a, 0) P

y N (a, a, 0)

The components of forces parallel to axes are

Pa Pa

2 2

and Pa –Pa

2 2

Therefore X = X1 = Pa2, Y = 0, Z = 0 (1)

= = for R,

, ,

= =

= =

, , 0

, 0 ,

69

Clearly L, M, N for forces P along first line are zero and for the second line we

write

0 a a

Pa –Pa 0

2 2

So Pa2 Pa

2 – Pa

2

2 2 2 (2)

R = (X2 + Y

2 + Z

2) = Pa2. (3)

If the axis of the couple coincides with the line of action of R, then it is central

axis and the moment of the couple about axis is K where

LX + MY + NZ = KR.

Pa2

2

Therefore, moment of couple

Pa2 a aR

2 2 2 [by (3)].

To write down the equation of central axis, we have

x y z x y z

X Y Z Pa2 0 0

L – yZ + zY M- zX + xZ N – xY + yX

X Y Z

and Pa2 Pa

2 – Pa

2

2 2 2

Pa2 0 0

Above gives y = a/2 and z = a/2 which is a line through the centre O of the cube

whose co-ordinates are (a/2, a/2, a/2).

Ex.5 Two forces act, one along the line y = 0, z = 0 and the other along the line x = 0, z =

c. As the forces vary, show that the surface generated by the axis of their equivalent

wrench is (x2 + y

2)z = cy

2.

Sol.

A single force R together with a couple K whose axis is along the line of action of

R is called a wrench of the system. The axis is central axis.

Let the forces be P and Q acting along the lines

x y z

1 0 0

x y z – c

0 1 0

so that their components parallel to the axis are Pl, Pm, Pn etc. i.e. P, 0, 0 and 0, Q, 0

respectively.

Therefore X = X1 = P, Y = Y1 = Q, Z = Z1 = 0. (1)

In order to find the values of L, M, N we note that force P acts at (0, 0, 0) and X1,

Y1, Z1 are P, 0, 0.

So 0 0 0

P 0 0

i.e. L1 = 0, M1 = 0, N1 = 0.

L = , M = , N =

.Pa2 + 0 + 0 = K.Pa2

K = = .Pa2 =

i.e.

= =

– (z.Pa2) – 0 – (0 – yPa2)

= =

= =

= =

70

For the force Q it acts at 0, 0, c and X2, Y2, Z2 are 0, Q, 0.

Therefore 0 0 0

0 Q 0

So L2 = 0 – cQ, M2 = 0, N2 = 0.

Hence L = L1 = –cQ, M = M1 = 0, N = N1 = 0. (2)

The equations of central axis are

L – yZ + zY M – zX + xZ N – xY + yX

X Y Z (3)

In order to write down the values of expressions within brackets, we write

x y z x y z

X Y Z P Q 0

The three brackets in order are the values of determinants

y z z x x y

Q 0 , 0 P , P Q

i.e. – zQ, Pz, xQ – yP. (4)

Hence equations of central axis by putting the values from (1), (2), (4) in (3) are

– cQ – (– zQ) – PZ – (xQ – yP)

P Q 0

From last, we get xQ – yP = 0; So P/Q = x/y. (5)

From 1st two, we get

(z – c)Q – Pz

P Q (6)

(5) and (6) together give the equation to central axis. In order to find the surface

generated by central axis, let us eliminate P and Q from (5) and (6) and we get

(z – c)y – xz

x y

or y2(z – c) = – x

2.z or z(x

2 + y

2) = cy

2

Ex.6 Forces X, Y, Z act along the three lines given by the equations y = 0, z = c; z = 0, x

= a; x = 0, y = b. Prove that the pitch of the equivalent wrench is

aYZ + bZX + cXY

X2 + Y

2 + Z

2

If the wrench reduces to a single force, show that the line of action of the force

lies on the hyperboloid (x – a)(y – b)(z – c) – xyz = 0.

Sol.

The three lines along which forces X, Y, Z act are

x y z – c

1 0 0

x – a y z

0 1 0

x y – b z

0 0 1

The components parallel to the axes are

X, 0, 0; 0, Y, 0; 0, 0, Z

So X = X1 = X, Y = Y, Z = Z (1)

Also L1, M1, N1 for force X are given by

0 0 c

= =

i.e.

= =

=

= =

= =

= =

for X,

for Y,

for Z.

71

X 0 0

i.e L1 = 0, M1 = cX, N1 = 0.

Similarly L2 = 0, M2 = 0, N2 = aY,

L3 = bZ, M3 = 0, N3 = 0.

So L = L1 = bZ, M = cX, N = aY. (2)

K LX + MY + NZ

R X2 + Y

2 + Z

2

bZX + cXY + aYZ

X2 + Y

2 + Z

2 [by (1) and (2)]

Write x y z

X Y Z

Equations of central axis are

L – yZ + zY M – zX + xZ N – xY + yX

X Y Z

LX + MY + NZ

X2 + Y

2 + Z

2

In case the system reduces to a single force, then LX + MY + NZ = 0 and hence

each of the three ratios vanish.

Hence putting the values of L, M, N we have

bZ – yZ + zY = 0, cX – zX + xZ = 0, aY – xY + yX = 0

or 0.X + zY + (b – y)Z = 0,

(c – z)X + 0Y + xZ = 0,

yX + (a – x)Y + 0Z = 0.

The locus of central axis i.e. the line of action of single force in this case is

obtained by eliminating X, Y, Z from the above three and is

0 z b – y

c – z 0 x = 0

y a – x 0

– z(0 – xy) + (b – y)[(c – z) (a – x)] = 0

or (x – a)(y – b)(z – c) – xyz = 0.

Ex.7 Forces X, Y, Z act along three straight lines y = b, z = – c; z = c, x = – a and x = a, y

= – b respectively. Show that this will have a single resultant if a/X + b/Y + c/Z = 0 and

that the equations of its line of action are two of the three

y z a

Y Z X

z x b

Z X Y

x y c

X Y Z

Sol.

The three lines are

x y – b z + c

1 0 0

= Pitch of the wrench =

=

= =

=

= 0

= 0

= 0

= = for X,

72

x + a y z – c

0 1 0

x – a y + b z

0 0 1

The components of forces parallel to axes are

X, 0, 0; 0, Y, 0; 0, 0, Z.

So X = X1 = X, Y = Y, Z = Z (1)

Also L1, M1, N1 for force X are given by

0 b – c

X 0 0

i.e L1 = 0, M1 = – cX, N1 = – bX .

Similarly L2 = – cY, M2 = 0, N2 = – aY,

L3 = – bZ, M3 = – aZ, N3 = 0.

So L = L1 = – (cY + bZ), M = – (cX + aZ), N = – (bX + aY). (2)

The system will reduces to a single forces if

LX + MY + NZ = 0

Therefore

– X(cY + bZ) – Y(cX + aZ) – Z(bX + aY) = 0 [by (1) and (2)]

or 2(aYZ + bZX + cXY) = 0 or a/X + b/Y + c/Z = 0 (3)

Now write x y z

X Y Z

Equations of central axis are

L – yZ + zY M – zX + xZ N – xY + yX

X Y Z

or – (cY + bZ) – yZ + zY = 0 (4)

– (cX + aZ) – zX + xZ = 0 (5)

– (bX + aY) – xY + yX = 0 (6)

Dividing (4) by YZ, we get

– c/Z – b/Y – y/Y + z/Z = 0 or a/X – y/Y + z/Z = 0. [by (3)]

or y z a

Y Z X (7)

Similarly dividing (5) and (6) by ZX and XY, respectively and using relation (3),

we get

z x b

Z X Y (8)

and x y c

X Y Z (9)

Thus the equations of line of action of the single force are any two of three

relations (7), (8) and (9).

Ex.8 Three forces each equal to P act on a body, one at the point (a, 0, 0) parallel to Oy,

the second at the point (0, b, 0) parallel to Oz and the third at the point (0, 0, c) parallel to

Ox, the axes being rectangular. Find the resultant wrench in magnitude and position.

Sol.

The lines of action of the equal forces are

x – a y z

= =

= =

for Y,

for Z.

= =

= 0

= 0

= 0

= =

73

0 1 0

x y – b z

0 0 1

x y z – c

1 0 0

The components parallel to the axis are

0, P, 0; 0, 0, P; P, 0, 0.

So X = X1 = P, Y = P, Z = P. (1)

For L1, M1, N1 we write

a 0 0

0 P 0

So L1 = 0, M1 = 0, N1 = aP.

Similarly L2 = Pb, M2 = 0, N2 = 0; L3 = 0, M3 = cP, N3 = 0.

Therefore

L = L1 = Pb, M = Pc, N = Pa. (2)

In order to find the magnitude of wrench we should find the values of both K and

R.

Therefore

R2 = X

2 + Y

2 + Z

2 = 3P

2, so R = P3 (3)

Also KR = LX + MY + NZ, so KP3 = P2 (a + b + c).

Hence

P(a + b + c)

3 (4)

(3) and (4) give the resultant wrench in magnitude.

Equations to the central axis are given by

x y z x y z

X Y Z or P P P

or L – yZ + zY M – zX + xZ N – xY + yX K

X Y Z R

or Pb – yP + zP Pc – zP + xP Pa – xP + yP a + b + c

P P P 3

or b – y + z c – z + x a – x + y a + b + c

1 1 1 3

or a + 2b + 3c b + 2c + 3a c + 2a + 3b

3 3 3

Above represent a line equal inclined to the axes and passing through the point

or a + 2b + 3c b + 2c + 3a ac + 2a + 3b

3 3 3

Ex.9 Equal forces act along the coordinate axes and along the straight line

x – y z

l m n

Find the equation of the central axis of the system.

Sol. The given lines are

x y z

1 0 0

= =

= =

K =

= = =

= = =

= = =

= y + x + = z +

, ,

= =

= =

74

x y z

0 1 0

x y z

0 0 1

and x – y z

l m n

and let the equal force be P, so that their components parallel to the axis are

P, 0, 0; 0, P, 0; 0, 0, P; Pl, Pm, Pn.

So X = X1 = P (1 + l), Y = P (1 + m), Z = P (1 + n) (1)

L, M, N for P along first line

0 0 0

P 0 0

So L1 = 0, M1 = 0, N1 = 0.

Similarly along other two axes their values are zero. For forth line we write

Pl Pm Pn

Therefore L = P (n – m), M = P (l – n), N = P (m – l) (2)

For writing the equation of central axis we write

x y z x y z

X Y Z or P (1 + l) P ( 1 + m) P (1 + n)

or L – yZ + zY M – zX + xZ N – xY + yX

X Y Z

The factor P will cancel throughout and so we do not write it.

(n – m) – y (1 + n) + z (l + m) (l – n) – z (1 + l) + x (1 + n)

1 + l 1 + m

(m – l) – x (1 + m) + y (1 + l)

1 + m

Ex.10 Two equal forces act one along each of the straight lines

x a cos y b sin z

a sin b cos c

show that their central axis must, for all values of , lie on the surface

y (x/z + z/x) = b (a/c + c/a).

Sol.

P being the force, then at the point (a cos, b sin, 0) we have a force whose

components are proportional to a sin.P, – b cos.P, cP and at the point (– a cos, b sin,

0) a force whose components are proportional to a sin.P, b cos.P, cP.

Hence X = X1 2a sin.P,

Y = 0 and Z 2cP,

L = (y1Z1 – z1Y1) 2bc sin.P,

M = 0 and N – 2abP.

The equation of central axis then become

bc sinyc – ab + ya sin

a sin c

and za sin = x.0.

= =

= =

= =

= =

=

=

= =

=

75

Substituting the value of sin from the second of these equations in the first we

have, as the locus of the central axis,

y (x/z + z/x) = b (a/c + c/a).

Ex.11 OBDC is a rectangle such that OB = b and OC = c, also OA is perpendicular to its

plane. Along, OA, CD and BD act forces X, Y, Z respectively. Show that the component

force R and couple K of the resultant wrench are (X2 + Y

2 + Z

2) and

X (Zb – Yc)

(X2 + Y

2 + Z

2)

Show also that with OA, OB, OC as axes of x, y, z the equation to the central axis

are

x y ZK z KY

X Y XYR Z XZR

Sol.

Given forces X, Y, Z act along OA, CD and BD respectively. Now introduce two

forces each equal to Y along OB and BO and two forces each equal to Z along OC and

CO.

We have three forces X, Y, Z acting at along OA, OB and OC which can be

compound to a single force R = (X2 + Y

2 + Z

2).

The force Y along BO and Y along CD being equal and opposite form a couple of

moment – Yc about a line OA as axis, which is perpendicular to its plane. Similarly the

force Z along BD and Z along CO form a couple of moment Zb about a line OA as axis.

Thus the moment of the couple is L = (Zb – Yc) about OA as axis

Moment about OB and OC are each zero. C

So M = 0, N = 0. Y

Also LX + MY + NZ = KR. D c

Hence (Zb – Yc) X = K (X2 + Y

2 + Z

2). Z Y

Therefore Z

X (Zb – Yc) Y O X A

(X2 + Y

2 + Z

2) b

Thus K and R the two elements of the B

resultant wrench are known. The equations

of central axes are

L – yZ + zY M – zX + xZ N – xY + yX K

X Y Z R

From 2nd

and 4th

ratio, we have

– zX xZ K

Y Y R

Multiplying by Y/(XZ).

Hence

– z x KY

Z X XZR

= = +

K =

= = =

+ =

+ =

76

or x z KY

X Z XZR

Similarly taking 3rd

and 4th

ratio and multiplying by Z/(XY), we get

or x y KZ

X Y XYR

Ex.12 If a force (X, Y, Z) act along a generator of the hyperbolic paraboloid

x2 y

2 2z

a2 b

2 c

and be equivalent to an equal force (X, Y, Z) at the origin together with a couple L, M, N,

show that aL bM = 0, bX aY = 0 and cN abZ = 0.

Sol.

The generators of the given hyperboloid are given by

x y 2z x y

a b c a b (1)

and x y 2z x y

a b c a b (2)

(1) and (2) give two generators of the opposite system. Let us reduce the

generator (1) to symmetrical form. Putting z = 0, we get

x = a/2, y = – b/2, z = 0

and direction-cosines of this can be shown to be proportional to a, b, c.

Thus actual direction cosines are aA, bA, cA, where

1

(a2 + b

2 + c

2

2)

Hence in symmetrical form the equations of the generator are

a b

2 2 z – 0

aA bA cA

Now if be the force whose components are X, Y, Z, then

X = PaA, Y = PbA, Z = PcA. (1)

For L, M, N, we have

a b 0

2 2

X Y Z

L = – bZ/2, M = – aZ/2, N = aY/2 + bX/2 (2)

– abZ ab Z

2 2

bX – aY = b.PaA – a.PbA = 0.

cN – abZ = ½ c(a.PbA + b.PaA) – ab.PcA.

= ca.PbA – ab.PcA = 0.

Similarly if we take the other generator (2) and reduce to symmetrical form, then

proceeding as above we can show that aL + bM = 0, bX + aY = 0, cN + abZ = 0.

+ =

=

=

+ = , =

= , =

A =

x –

=

y +

=

+ aL – bM = = 0

77

Ex.13 Forces act along generators of the same system of a hyperboloid. Their magnitudes

are such that if they were transferred parallel to themselves to act in point, they would be

in equilibrium. Show that they are in equilibrium when acting along the generators.

Sol.

We have written the values of X1, Y1, Z1, L1, M1, N1.

So X = R1a sin1.A1, Y = – R1b cos1.A1, Z = R1c.A1.

L = R1bc sin1.A1 = bcX/a

M = R1ca cos1.A1 = caY/b

N = – R1ab.A1 = – abZ/c

X, Y, Z are the component forces acting at O along the axes. If X, Y, Z are in

equilibrium, then their resultant R = (X2 + Y

2 + Z

2), should vanish i.e. X = 0, Y = 0, Z =

0.

Clearly if X, Y, Z are zero, then L, M, N also vanish. Thus all the six elements X,

Y, Z, L, M, N vanish and hence the forces will be in equilibrium when acting along the

generators.

Ex.14 Forces act along generators of the same system of a hyperboloid and the pitch p of

the equivalent wrench is given. Prove that the central axis is that generator of the

hyperboloid

(bc/a – p) x2 + (ca/b – p) y

2 – (ab/c + p) z

2 = (bc/a – p)(ca/b – p)(ab/c + p)

which intersects xy-plane at the point

(ac – bp) Z1 cos (bc – ap) Z1 sin

c Z1 c Z1

Sol.

The values of X, Y, Z, L, M, N are

X = R1a sin1.A1, Y = – R1b cos1.A1, Z = R1c.A1.

L = R1bc sin1.A1 = bcX/a

M = R1ca cos1.A1 = caY/b, N = – R1ab.A1 = – abZ/c

The pitch of the wrench i.e. K/R = p given.

Equation of central axis is

L – yZ + zY M – zX + xZ N – xY + yX K

X Y Z R

Therefore

(bc/a – p) X + zY – yZ = 0, (1)

(ca/b – p) Y + xZ – zX = 0, (2)

(ab/c + p) Z + yX – xY = 0, (3)

Eliminating X, Y, Z the locus of the central axis

(bc/a – p) z – y

– z (ca/b – p) x = 0

y – x (ab/c + p)

Expending the above determinants we get the locus as given. In order to find the

point on xy-plane put z = 0 in (1) and (2), we get

y = (bc/a – p) (X/Z)

(bc/a – p)R1a sin1.A1

R1c.A1

(bc – ap) Z1 sin

x = , y =

= = = = p

=

y =

78

c Z1

Similarly the value of x is as given.

2.5 Null Lines and Planes:

1.Null Lines:

Corresponding to a given system of forces referred to a base or origin O, all those

lines which radiate from O, such that the moment of the given system of forces about

these lines is zero are called null lines at O.

2.Null Plane:

The plane in which all these null lines lie is called null plane of the point O.

3.Null Point:

The point O itself is called null point.

Suppose that, corresponding to any origin or base point O, the resultant force is R

and the resultant couple is G. Take any line through O perpendicular to the axis of G;

then the sum of the moments of the forces of the system about this line is zero; for the

axis of G has no component along it and R meets it.

For this reason the line is called a null line and its locus, which is the line

perpendicular to the axis of G, is called the null plane of O. Also the point O is called the

null point of the plane.

Theorem1: Find the equation of the null plane of a given point (f, g, h) referred to any

axes Ox, Oy, Oz.

Proof:

Let X, Y, Z be the component forces along Ox, Oy, Oz and L, M, N the

component couples about them.

The component couples about lines parallel to the axes through (f, g, h) are

L – gZ + hY, M – hX + fZ, N – fY + gX,

and these are proportional to the direction cosines of the axis of the resultant couple at (f,

g, h), which is the normal to the null plane there.

Hence the equation to the null plane is

(x – f)(L – gZ + hY) + (y – g)(M – hX + fZ) + (z – h)(N – fY + gX) = 0.

i.e. x (L – gZ + hY) + y (M – hX + fZ) + z (N – fY + gX) = fL + gM + hN. … (1)

The above equation of null plane can be put in determinant form as follows:

x y z

f g h = L(x – f) + M(y – g) + N(z – h)

X Y Z

Theorem2: Find the null point of a given plane lx + my + nz = 1.

Proof:

To obtain the null point of the plane

lx + my + nz = 1, (i)

compare this plane with (1) of the above theorem, we get

(L – gZ + hY) (M – hX + fZ) (N – fY + gX)

l m n = = = fL + gM + hN.

79

Since the point (f, g, h) also must lie on the plane (i), we have, on solving

f g h 1

X – nM + mN Y – lN + nL Z – mL + lM lX + mY + nZ

giving the null point of (i).

Theorem3: Find the condition that the straight line

x – f y – g z – h

l m n

may be a null line for the same system of forces.

Proof:

The component couples about lines through (f, g, h) parallel to the axes are

L – gZ + hY, M – hX + fZ, N – fY + gX,

Hence the moment of the couple about the given line

= l (L – gZ + hY) + m (M – hX + fZ) + n (N – fY + gX)

and hence is zero if

X(mh – ng) + Y(nf – lh) + Z(lg – mf) = Ll + Mm + Nn,

i.e. if X Y Z

l m n = Ll + Mm + Nn

f g h

This is therefore the condition that the given line may be a null line of the system.

Theorem4: Show that a given system of forces may be replaced by two forces, one of

which acts along a given line OA.

Proof:

With O as origin, or base point, let R and G be the resultant force and couple.

Through OA and R let a plane be drawn and let it cut the plane of the resultant couple

(i.e. the plane COD perpendicular to the axis of G) in OB. Resolve R into two forces,

one, P1 along OA, and the other, P2, along OB.

The force P2 along OB, when

compounded with the two forces in the plane A

BOC which form the resultant couple, will G R

give a force P2 in this plane which is parallel

to OB.

It follows that the given system of O C

forces is equivalent to some force P1 acting

along the given straight line OA, together with

a second force P2 which acts somewhere in the D B

null plane of O.

Note:

Such forces as P1 and P2 are called conjugate forces and their lines of action are

called conjugate lines.

Whatever point O we take on OA the force P2 will still lie in its null plane, so that,

as O moves along OA, its null plane continually turns round so that it always passes

through the line conjugate to OA. Hence the conjugate line of OA may be determined by

taking any two convenient points on it, and obtaining the equations to their null planes by

Theorem 1. The conjugate line is then the intersection of these two planes.

= = =

= =

80

Theorem5: Find the equation of a line conjugate to a given line

x – f y – g z – h

l m n

Proof:

In order to find the line conjugate to given line we have to find the equation of

null planes of any two conveniently chosen points on the given line. These two points

together will give the conjugate line.

Let one point be (f, g, h) and the other point be at infinity whose coordinates are

(l, m, n) where is infinite.

If the system reduces to X, Y, Z, L, M, N, then the null plane of (f, g, h) is

x y z

f g h = L(x – f) + M(y – g) + N(z – h) (1)

X Y Z

and the null plane of (l, m, n) is

x y z

l m n = L(x – l) + M(y – m) + N(z – n)

X Y Z

Dividing by , we get

x y z

l m n = L(x/ – l) + M(y/ – m) + N(z/ – n)

X Y Z

Now put infinite and the null plane becomes

x y z

l m n = – (Ll + Mm + Nn) (2)

X Y Z

The line of intersection of null planes (1) and (2) is a line conjugate to given line.

Ex.1 A system of forces given by (X, Y, Z, L, M, N) is replaced by two forces, one acting

along the axis of x and another force. Show that the magnitudes of the forces are

LX + MY + NZ

L

and [(MY + NZ)2 + L

2 (Y

2 + Z

2)]

1/ 2

L

and also find the equation of the line of action of the other force.

Sol.

The equations of axes of x are

x y z.

1 0 0

Let there be a force P acting along it at (0, 0, 0), so that its components parallel to

axes of co-ordinates are P, 0, 0.

The components of couples at (0, 0, 0) are all zero. Since the system is X, Y, Z; L,

M, N, therefore the components parallel to axes of the other forces are X – P, Y, Z.

If the other forces act at the point (f, g, 0), then for moments of couples we write

f g 0

X – P Y Z

= =

= =

81

So L = gZ, M = – fZ, N = fY – g(X – P)

or – MY L(X – P)

Z Z

Therefore

LX + MY + NZ

L

or LX +MY + NZ MY + NZ

L L

Y = Y, Z = Z.

Therefore resultant

= [(X – P)2 + Y

2 + Z

2]

[(MY + NZ)2 + L

2 (Y

2 + Z

2)]

1/ 2

L

In order to write down the equations of its line of action we write down the

equations of null planes of point (0, 0, 0) and a point (, 0, 0) where is infinity on the

given line.

Null plane of (0, 0, 0) is LX + MY + NZ = 0 (1)

Null plane of (, 0, 0) is

x y z

0 0 = L(x – ) + M(y – 0) + N(z – 0)

X Y Z

or – (yZ – zY) = L(x – ) + My + Nz.

Dividing by and put = .

Therefore

– (yZ – zY) = L(– 1) or yZ – zY = L. (2).

Equations (1) and (2) together give the equations of the line of action of the other

force.

Ex.2 Show that among the null lines of any system of forces four are generators of any

hyperboloid, two belonging to one system of generators and two to the other system.

Sol.

Let the hyperboloid be

x2 y

2 z

2.

a2 b

2 c

2

and referred to its centre and axes let the system be given by (X, Y, Z; L, M, N). Any

generator is

x – a cos y – b sin z.

a sin – b cos c

this is a null line of the system if

X(– bc sin) + Yca cos + Zab = La sin – Mb cos + Nc,

i.e. if sin [X/a + L/(bc)] – cos [Y/b + M/(ca)] = Z/a – N/(ab),

which clearly gives two values of , in general. Hence two generators belonging to one

system are null lines. Similarly for the other system.

Ex.3 A straight line is given by the equations

Ax + By + Cz = D, x + By + Cz = D

N =

= P

X – P = X –

=

+ = 1,

= =

82

Show that its conjugate is given by equating to zero any two of the determinants

L M N Lx + My + Nz

A B C D = 0

A B C D

where L, M, N, are the component couples at the point (x, y, z) and L, M, N those at the

origin.

Sol.

Let x, y, z be the current coordinates. If (x, y, z) be any point, then direction-

ratios of null line are x – x, y – y, z – z. Also L, M, N being the moments of couple at

the point (x, y, z), they represent the direction ratios of the axis of the couple which by

definition is perpendicular to null line.

Hence by applying the condition of perpendicularity, the equations of null plane is

L(x – x) + M (y – y) + N (z – z) = 0

or Lx + My + Nz = Lx + My + Nz (1)

But L = L – yZ + zY, M = M – zX + xZ, N = N – xY + yX, where L, M, N are

component couples at origin.

Therefore

Lx + My + Nz = (L – yZ + zY)x = Lx + My + Nz. (2)

Hence from (1) and (2), the equation of null plane is

Lx + My + Nz = Lx + My + Nz (3)

Now the conjugate line whose equations are

Ax + By + Cz = D, Ax + By + Cz = D

lie in the null plane (3).

Therefore

Ax + By + Cz = D, (4)

Ax + By + Cz = D, (5)

Hence the conjugate line i.e. locus of (x, y, z) is obtained by eliminating x, y, z

from (3), (4) and (5) or the conjugate line is given by equating to zero any two of the

following determinants:

L M N Lx + My + Nz

A B C D = 0

A B C D

2.6 Stable and Unstable Equilibrium:

Consider an object supported at one point say O. There will be equilibrium so

long as its centre of gravity G lies in the vertical line through the point of support, for, its

weight then acts through the point of support and there is no moment tending to turn the

body about that point. But the nature of equilibrium differs greatly with the position of G.

We shall distinguish three cases:

1. Suppose that the centre of gravity lies below the point of support. If the body be

slightly displaced from this position of rest, then in the displaced position there are two

equal and opposite forces acting on the body, one through O and other through G. These

83

form a couple which has a tendency to restore the body to its original position of rest. In

this case the body is said to be in stable equilibrium.

2. Next suppose that the centre of gravity lies above the point of support. If the body be

slightly displaced, a couple acts, which has a tendency to restore the body still further

from the position of rest. Such equilibrium is called unstable equilibrium.

3. If the centre of gravity is at the fixed point of support, the object will be at rest in any

position. This is called neutral equilibrium.

As another illustration consider a sphere of radius a whose centre of gravity lies at

a distance b (b < a) from its geometric centre. Suppose that the sphere rests on a

horizontal plane, the centre of gravity being either below O as at G1 or above O as at G2.

In the position of equilibrium as given by the first figure, we have two forces

acting on the body, (i) its weight acting through G1 or G2 (ii) the reaction of the

horizontal plane on the sphere acting through A. These forces act in the same vertical line

and there is equilibrium.

G2 G2

O O

G1

G1

A B

Suppose that the body, with its centre of gravity at G1, is slightly displaced so that

it assume the position shown in the second figure, the point of contact with the plane

begin B. In this position there are two equal and opposite forces acting on the sphere, the

weight of the sphere through G1 vertically downwards and the reaction, equal in

magnitude to the weight, through B vertically upwards. These forces form a couple,

which has a tendency to restore the body to the original position of rest. The sphere with

its centre of gravity vertically below O is in a position of stable equilibrium.

In case the centre of gravity of the sphere is at G2 in the first position and the

sphere is displaced slightly to assume the form as shown in the second position, there is a

couple acting on the sphere which has a tendency to turn it further away from its position

of equilibrium. Under this condition the position of the sphere is that of unstable

equilibrium.

If, however, the centre of gravity of the sphere be at O, the body will be in

equilibrium when displaced. The sphere with its centre of gravity at its geometrical centre

is in a state of neutral equilibrium.

Definitions:

If, when a body is slightly displaced from an equilibrium position, the force acting

on the body tend to make it return towards its position of equilibrium, then the

84

equilibrium is said to be STABLE. If the force acting on the body tends to move the body

away from the position of equilibrium, the equilibrium is said to be UNSTABLE. If the

forces acting on the body in the displaced position are in equilibrium, the equilibrium is

said to be NEUTRAL.

Thus the equilibrium of a pendulum is stable when it is displaced from its vertical

position of equilibrium, for its returns towards the vertical position again. Any top-heavy

thing or a stick placed vertically on a finger is an instance of unstable equilibrium. A

sphere made of wood or rubber and floating on water is an example of neutral

equilibrium. Again a cone on its base is in a state of stable equilibrium, on its vertex in an

unstable equilibrium, and when resting along a generator is in neutral equilibrium.

Conditions of stability for a body with one degree of freedom:

A body is said to have one degree of freedom when it is so constrained that only

one geometrical quantity is needed in order to fix its position.

Now consider a body with one degree of freedom. The work done on the body in

bringing it from any standard position may be written in the form

W = f(),

where is the quantity defining the position of the body. Now if F be the resultant force

on the body and s the displacement of the body in the direction of F in the position ,

then

dW = F ds,

that is, f () d = F ds.

Therefore

F = f ().[d/ds] (1)

In a position of equilibrium, F = 0, hence

f ().[d/ds] = 0.

Since there are always a number of geometrical quantities, each of which could be

used to fix the position of a body with one degree of freedom, we choose to be that one

which continually increases or continually decreases as the body is moved in the same

direction. With the assumption

d/ds 0,

hence for equilibrium

f () = 0. (2)

Suppose a root of this equation is 0. Then we have to test whether the

equilibrium is stable or unstable for = 0. Now if the force and the displacement have

opposite signs, the equilibrium will be stable.

Thus for stability dF/ds must be negative. But from (1)

dF d d2

ds 0 ds ds2

Hence for stability

d

ds

must be negative, since f () = 0, i.e., f () = 0 must be negative.

It follows therefore that, if W = f(), the conditions for stable equilibrium are

= f (0)( )2 + f (0)

f (0)( )2

85

f () = 0

and

f () = 0 is negative,

and these are the conditions for the existence of a maximum of W for = 0.

Similarly it may be prove that if W is a minimum, the equilibrium is unstable. If

W is neither a maximum nor a minimum the equilibrium is either neutral or unstable. The

equilibrium is neutral if dF is absolutely zero. But if dF has the same sign as ds for

displacements in one direction and the opposite sign for displacements in the other

direction, then, after any small displacement, the body is certain to move on the unstable

side after a short time and hence the equilibrium is really unstable.

When the forces are of the conservative type, we have seen that the potential

energy V of the body is given by

V = C – W

where C is a constant.

It follows that in the case of a conservative system, the equilibrium is stable or

unstable according as the potential energy is a true minimum or maximum. For example,

whenever gravitational energy is the only form of potential energy involved, the height of

the centre of gravity must be a minimum.

When the force of gravity is the only external force acting on a body, we arrive at

the following rule for deciding whether a given equilibrium position is stable or unstable:

Suppose that a body is tilted from its position of equilibrium, the measure of the

tilt being given by . Express the height of the centre of gravity of the body in terms of

the variable . If this height is given by

z = f(),

positions of equilibrium are given by

dz/d = f ().

We than find the sign of d2z/d

2 for the roots of

f () = 0.

If the sign is positive for a root, the height of the centre of gravity is minimum and

the equilibrium is stable for that , in case the sign is negative the centre of gravity is at a

maximum height and the equilibrium is unstable.

In some cases, it is convenient to find, instead of the height of the centre of

gravity above a fixed plane, the depth y of the centre of gravity below a fixed plane. In

such cases, if the depth x is minimum, the position of equilibrium is unstable and if the

depth y is maximum the equilibrium is stable.

Theorem: A heavy body rests on a fixed body. Find the nature of equilibrium.

Proof:

We shall suppose that (i) the upper body is free to roll, without sliding, on the

lower body, (ii) the portions of the two bodies in contact being spheres of radii r and R

respectively and (iii) the straight line joining their centres is vertical initially. Let O and

O be the centres of the spherical surfaces. The adjoining figure is a section of the bodies

through G, the centre of gravity of the upper one. Let the upper body BAC be displaced

into the position BAC, the new point of contact being A. Let G be the new position of

the centre of gravity of the upper body which now lies on OA, the new position of OA.

86

B C

A

B O G

O

A A

C

O

Further let

AOA = , AOA = .

Since we suppose the upper body displaced by rolling on the lower, we have

arc AA = arc AA;

therefore R = r. (1)

If z be the height of G above O, then

z = OO cos – OG cos (+ )

= (R + r) cos – (r – h) cos[(r + R)/r], by (1),

where AG = h.

Now for equilibrium,

dz/d = 0,

hence

– (R + r) sin + (r – h)[(r + R)/r] sin[(r + R)/r] = 0.

This is satisfied by = 0.

Now we have to find the sign of d2z/d

2 for = 0.

But

d2z r + R r + R

d2 r r

which, for = 0 reduce to

r + R

r

i.e., to r + R rR

r r + R

The equilibrium is stable or unstable

According as z is minimum or maximum,

i.e. ” ” d2z/d

2 is positive or negative,

i.e. ” ” rR/(r + R) > or < h,

i.e. ” ” (1/h) > or < 1/r + 1/R.

Particular cases:

1.If the lower surface be concave instead of convex near A, then R is negative and the

equilibrium is stable or unstable according as

(1/h) > or < 1/r + 1/R.

2.If the lower surface near A be plane, R is infinite and the equilibrium is stable or

unstable according as

h < or > r.

= (r + R) cos + (r – h)( )2 cos ( ),

(r + R) + (r – h)( )2,

( )2 [ - h].

87

3.If the upper body have a plane face in contact with the lower body, r is infinite and the

equilibrium is stable or unstable according as

h < or > R.

In the critical case, when

1/h = 1/r + 1/R, the determination of stability is difficult.

Ex.1 A uniform rod AB of length 2a is hinged at A, a string attached to the middle point

of G of the rod passes over a smooth pulley at C at a height a, vertically above A and

supports a weight P hanging freely, find the position of equilibrium and determine their

nature.

Sol.

Let AB be the rod, its weight W acting C

through the middle point G. The string of length

l attached to G passes over the pulley at C,

AC being equal to AG and carries a weight P at A

the other end. Hence the tension in the string is P.

If the angle DAB be denoted by , then P G

ACG = ½, since AC = AG. Now the depth of D

G below C = a + a cos and the depth of P below W B

C = l – CG = l – 2a cos ½.

Hence if y be the depth, below the pulley,

of the centre of gravity of the system consisting of

the weights P and W,

(P + W)y = P(l – 2a cos½ ) + Wa(1 + cos) (1)

and accordingly,

(P + W)[dy/d] = P(a sin½ ) + W(- a sin)

and

(P + W)[d2y/d

] = P[(a/2) cos½] + W(- a cos)

= P[(a/2) cos½] + Wa(1- 2a cos2½).

Positions of equilibrium are given by dy/d = 0, i.e., by

P sin½ = W sin = 2W sin½cos½

which is satisfied by

sin½0,i.e., = 0, or cos½= P/(2W).

For = 0,

(P + W)[d2y/d

] = a[½P – W].

Hence the position of equilibrium given by = 0 is stable when P < 2W and

unstable when P > 2W.

The second solution cos½= P/(2W) is possible only when P < 2W and then

(P + W)[d2y/d

] = aW[P

2/(4W

2) – P

2/(2W

2) + 1]

= aW[1 - P2/(4W

2)],

which is positive, since P < 2W. The depth y is then minimum and the position of

equilibrium given by

cos½= P/(2W)

is therefore unstable.

88

Ex.2 A homogeneous body, consisting of a cylinder and a hemisphere joined at their

bases, is placed with the hemispherical end on a horizontal table; is the equilibrium stable

or unstable.

Sol.

Let G1 and G2 be the centres of gravity of the hemisphere and cylinder, and let A

be the point of the body which is initially in contact with the table and let O be the centre

of the base of the hemisphere.

G

G1

A

If h be the height of the cylinder and r be the radius of the base, we have

OG1 = 3r/8 and OG2 = h/2

Also the weights of the hemisphere and cylinder are proportional to (2/3)r3 and

.r2h. The reaction of the plane, in the displaced position of the body, always passes

through the centre O.

The equilibrium is stable or unstable according as G, the centre of gravity of the

compound body, is below or above O, i.e., according as

OG1 wt. of hemisphere is > OG2 wt. Of cylinder,

i.e. according as 3r/8 (2/3)r3 is > h/2 .r

2h,

i.e. according as r2/2 is > h

2,

i.e. According as r is > 2h,

i.e. > h 1.42 …

Ex.3 A body consisting of a cone and a hemisphere, on the same base, rests on a rough

horizontal table, the hemisphere being in contact with the table; show that the greatest

height of the cone, so that the equilibrium may be table, is 3 times the radius of the

hemisphere.

Sol.

If O is the centre of base, then the C.G. of cone ABD and hemisphere BCD must

be below O, for stable equilibrium.

The weight of the cone, (1/3)r2h, acts at G1, where OG1 = h/4.

Weight of BCD, (2/3)r3, acts at G2 where OG2 = 3r/8. For the C.G. of the body

to be at O,

(1/3)r2h h/4 = (2/3)r

3 3r/8

Hence h < r3 (for stable equilibrium)

A

h

<

<

<

<

<

G

2

O

89

G1

B x O D

G2

C

Ex.4 A solid consists of a cylinder and a hemisphere of equal radius, fixed base-to-base,

find the ratio of the height to the radius of the cylinder, so that the equilibrium may be

neutral, when the spherical surface rests on a horizontal plane.

Sol.

Let ABC be a hemisphere and ABDE E D

a cylinder. If r is the radius of the hemisphere

and h the height of the cylinder, their weights

will be in the ratio of their volumes i.e. in the G2

ratio (2/3)r3 : r

2h.

For neutral equilibrium, the C.G. of the

body must be at O. A O B

(2/3)r3 OG1 = r

2h OG2

or (2/3)r3 2r/3 = r

2h h/2

or r = h G1

C

Ex.5 A hemisphere rests in equilibrium on a sphere of equal radius; show that the

equilibrium is unstable when the curved and stable when the flat, surface of the

hemisphere rests on the sphere.

Sol. In the figure, when the curved surface of the hemisphere is placed on the sphere,

h = 5r/8 and R = r.

So 1/h = 8/(5r) and 1/r + 1/R = 2/r.

Hence 1/h < 1/r + 1/R, [since 8/(5r) < 2/r] therefore the equilibrium is unstable and if

the plane surface of the hemisphere is placed on the sphere

h = 3r/8 and r =

So 1/h = 8/(3r) and 1/r + 1/R = 1/R = 0.

Hence 1/h > 1/r + 1/R,

Therefore the equilibrium is stable. C

O1 L

G2

G1 O2

A1

A2

90

O

Ex.6 A heavy right cone rests with its base on a fixed rough sphere of given radius, find

the greatest height of the cone if it is in stable equilibrium.

Sol. Let the greatest height of the cone be A

h. If B is the centre of the base of the cone and

C.G. of the cone be at G, then BG = h/4. G

For equilibrium to be stable, by B

particular case 3 of Art. 2.3, h/4 < R or h < 4R.

Ex.7 A heavy uniform cube balances on the highest point of a sphere, whose radius is r.

If the sphere be rough enough to prevent sliding and if the side of the cube be r/2, show

that the cube can rock through a right angle without falling.

Sol.

Let ABC be the given sphere of radius r.

The cube is placed in equilibrium at A. The side

of the cube is r/2. G

In the position of the equilibrium, GA will

be vertical and A

GA = r/4. K

So GA > r.

Hence the equilibrium will be stable. O

C

Now if the cube begins to swing and swings

upto an arc AK, then B

arc AK = r (if AGK = ).

Now if = /4, then AK = r/4 and the value

of AG = r/4.

In this case GK is vertical.

Hence the cone can swing making an angle /4 on both sides, or through one right

angle, without falling.

Ex.8 A uniform beam, of thickness 2b, rests symmetrically on a perfectly rough

horizontal cylinder of radius a; show that the equilibrium of the beam will be stable or

unstable according as b is less or greater then a.

Sol.

By particular case 3, the equilibrium will be stable or unstable if b < a or b > a.

Ex.9 A lamina in the form of an isosceles triangle, whose vertical angle is , is placed on

a sphere, of radius r, so that its plane is vertical and one of its equal sides is in contact

with the sphere; show that, if the triangle be slightly displaced in its own plane, the

equilibrium is stable if sin be less than 3r/a, where a is one of the equal sides of the

triangle.

Sol.

Let ABC be the isosceles triangle in which AB = a and BAC = .

So AX = a cos(/2) [CX = BX]. C

91

If G is the C.G. of triangle ABC and D

the point of contact of the sphere with AB,

then GD will be vertical and AG = (2/3)AX, G X

or AG = (2/3)a cos(/2),

and GD = AG sin(/2) A D B

= (2/3)a cos(/2) sin(/2),

= (a/3) sin

= h T

[By particular case 3 of Art. 2.3] P

Hence for the equilibrium to be stable, h < r.

or (a/3) sin < r

or sin < (3r/a) W

Ex.10 A weight W is supported on a smooth inclined plane by a given weight P,

connected with W by means of a string passing round a fixed pulley whose position is

given. Find the position of equilibrium of W on the plane and show that it is stable.

Sol.

If the inclination of the plane is and is the angle between the string and the

plane and tension of the string is P, then resolving the forces parallel to the plane,

W sin = P cos

or cos = (W sin)/P

or = cos– 1

[(W sin)/P]

Hence when we know the position of the pulley and the value of , the position of

W can be found.

Now, if W is displaced slightly downwards, will decrease; hence P cos will

increase.

So W will move upwards and if W is slightly moved upwards, then increasing,

P cos will decrease and hence W will move downwards. Therefore the equilibrium will

be stable.

Ex.11 A rough uniform circular disc, of radius r and weight p, is movable about a point

distance c from its centre. A string, rough enough to prevent any slipping, hangs over the

circumference and carries unequal weights W and w at its ends. Find the position of

equilibrium and determine whether they are stable or unstable. A

Sol.

Let ABD be the circular disc with centre

O. C is the point round which the disc can rotate. C

If CO (= c) is inclined at angle to vertical, O

taking moments about C, p

W(r – c sin) = w(r + c sin) + pc sin r

or Wr – wr = sin (Wc + wc + pc) B D

So (W – w)r . w W

(W + w + p)c

So (W – w)r .

sin =

= sin– 1

[ ]

92

(W + w + p)c

The moment increases with increase in and decreases with decrease in . Hence

the equilibrium is stable.

Ex.12 A solid sphere rests inside a fixed rough hemispherical bowl of twice its radius.

Show that, however large a weight is attached to the highest point of the sphere, the

equilibrium is stable.

Sol.

Let w be the weight of the sphere of radius r and let W be the weight placed at the

highest point of the sphere. If h is the length of C.G. of the body from the lowest point of

the sphere, then

w.r + W.2r

w + W

Therefore the equilibrium will be stable if

1/h > 1/r – 1/(2r)

or 2r > h

or (w + 2W)r

w + W

or 2w + 2W > w + 2W

which is always true.

Therefore the equilibrium is stable.

Ex.13 A thin hemispherical bowl, of radius b and weight W, rests in equilibrium on the

highest point of a fixed sphere, of radius a, which is rough enough to prevent any sliding.

Inside the bowl is placed a small smooth sphere of weight W. Show that the equilibrium

is not stable unless

w < W.[(a – b)/2b]

Sol.

If the bowl is displaced, the small sphere moves in such a way that its weight still

acts through the centre of the bowl.

Hence to know the stability of equilibrium the weight of the small sphere may be

taken to act at the centre of the bowl.

Height of the C.G. of the body,

w.b + W(b/2)

w + W

The equilibrium will be stable if

1/h > 1/a + 1/b

or w + W .

wb + W(b/2)

or (w + W)a > (w + W/2)(a + b)

or Wa/2 > wb + Wb/2

or w < W.[(a – b)/2b]

Ex.14 A square lamina rests in a vertical plane on two smooth pegs, which are in the

same horizontal line. Show that there is only one position of equilibrium unless the

distance between the pegs is greater than one quarter of the diagonal of the square, but

h =

2r >

h =

> 1/a + 1/b

93

that, if this condition is satisfied, there may be three positions of equilibrium and the

symmetrical position will be stable, but the other two positions of equilibrium will be

unstable.

Sol.

Let ABCD be the square and P and Q

the pegs. Let the diagonal AC = 2d and let it C

be inclined at an angle to the horizontal AX.

The height GN(= z) of the centre of gravity G

above PQ is given by

z = AG sin – AP sin(- 450) D G B

= d sin – c cos(- 450) sin(- 45

0) , N

if PQ = c, Q P

i.e. z = d sin + (c/2) cos2 (1)

So dz/d = d cosc sin2 (2) A X

and d2z/d

2 = - d sin- 2c cos2 (3)

Now since the pegs are smooth, the equation of virtual work reduces to W.z = 0.

Hence, by (2), the position of equilibrium are given by

cos(d – 2c sin) = 0 (4)

The solutions of this equation are = 900 and sin = d/(2c).

This latter equation has real roots only when 2c > d, i.e. when PQ > ¼ AC.

Take the case when 2c > d.

There are then three positions; the first when AC is vertical and the other two

when AC is inclined at either side of the vertical at an angle sin–1

[d/(2c)] to the

horizontal.

When = 900, then by (3), d

2z/d

2 = - d + 2c = positive.

Therefore z is a minimum and the equilibrium is stable.

When

sin = d/(2c), then

d2z /d

2 = - d sin –2c + 4c sin

2

= (d2 – 4c

2)/(2c) = negative.

In this case z is a maximum and the equilibrium is unstable.

Next take the case when 2c < d.

In this case there is only one position of equilibrium given by = 900, and then

d2z/d

2 = - d + 2c = negative

z is now a maximum and the equilibrium is unstable.

Ex.15 A rod SH, of length 2a and whose centre of gravity G is at a distance d from its

centre, has a string, of length 2c sec, tied to its two ends and the string is then slung

over a small smooth peg P; find the position of equilibrium and show that the position

which is not vertical is unstable.

Sol.

Since SP + PH = 2c sec, the peg P must be somewhere on an ellipse of foci S

and H and semi-major axis c sec.

Also its semi-minor axis = (c2 sec

2 – CH

2) = c tan.

Hence the equation to the ellipse is

94

x2 sin

2 + y

2 = c

2 tan

2, or, referred to polar

coordinates through G, P

sin2 (r cos + d)

2 + r

2 sin

2 = c

2 tan

2(1)

If we find the value of for which r T r T A

is a maximum or minimum and take the

corresponding point P of the ellipse for the G H

position of the peg and make PG vertical, we A S C

shall have the slant position of equilibrium.

(1) gives

cos2r

2 cos

2 - 2 cos.dr sin

2= r

2 - c

2 tan

2+ d

2 sin

2

So d sin tan + [r2 – (c

2 – d

2) tan

2] .

r cos

The least value of r is clearly (c2 – d

2) tanand then

So d tan

(c2 – d

2)

Since in this case r is a minimum the centre of gravity is at its minimum depth

below the peg and therefore at the maximum height above the horizontal and the

equilibrium is unstable. The other two positions of equilibrium are when P is at A or A

and the rod is then clearly vertical.

If GP is a minimum it is clear that GP must be a normal at P; so that P may also

be found from the fact that its normal passes through a known point G on the major axis.

2.7 Unit Summary:

1. Forces in three dimensions may be compounded by the law of parallelopiped of forces,

which is an extension of law of parallelogram of forces in two dimensions. 2. When forces in different direction in space are acting upon a rigid body, we say that the body is

under the action of forces in three dimensions.

3. The moment of a forces F about any point is equal to F.p, where p is perpendicular distance of

the point from the line of action of force. It is said to be +ive or –ive according as the force has a

tendency to rotate the body about the point in anticlockwise or clockwise direction.

4. However, in three dimension those forces which try to rotate the body from axis of x to

axis of y, from axis of y to axis of z and from axis of z to axis of x will have + ive

moments about the axis, similarly the moments of couples.

5. We already know that two equal and unlike parallel forces form a couple whose

moment is equal to the product of the force and the arm of the couple.

6. Any given system of forces acting at any given point of a rigid body can be reduced to

a single force acting through an arbitrarily chosen point and a couple whose axis passes

through that point.

7. The system will be in equilibrium when R = 0 and G = 0, which in other words means

that X = 0, Y = 0, Z = 0. Also L = 0, M = 0, N = 0.

cos

cos

95

8. The sum of the resolved parts of the system of forces parallel to any three axes of co-

ordinates must vanish separately, and also the sums of their moments about the three axes

must vanish separately.

9. Any system of forces acting on a rigid body can be reduced to a single force together

with a couple whose axis is along the direction of the force.

10. A single force R together with a couple K whose axis coincides with the direction of

the force when taken together is called a wrench of the system. The single force R is

called the intensity of the wrench.

11. The ratio K/R i.e. moments of the couple about central axis divided by forces is called

the pitch and is of a linear magnitude. Also it is clear that when the pitch is zero the

wrench reduces to a single force. When the pitch is infinite then R = 0 and hence the

wrenches reduce to a couple K only.

12. The straight line, along which the single forces acts, when considered together with

the pitch, is called a screw. Hence screw is a definite straight line associated with a

definite pitch.

13. Corresponding to a given system of forces referred to a base or origin O, all those

lines which radiate from O, such that the moment of the given system of forces about

these lines is zero are called null lines at O.

14. The plane in which all these null lines lie is called null plane of the point O.

15. The point O itself is called null point.

16. If, when a body is slightly displaced from an equilibrium position, the force acting on

the body tend to make it return towards its position of equilibrium, then the equilibrium is

said to be STABLE. If the force acting on the body tends to move the body away from

the position of equilibrium, the equilibrium is said to be UNSTABLE. If the forces acting

on the body in the displaced position are in equilibrium, the equilibrium is said to be

NEUTRAL.

17. If the lower surface be concave instead of convex near A, then R is negative and the

equilibrium is stable or unstable according as

(1/h) > or < 1/r + 1/R.

18. If the lower surface near A be plane, R is infinite and the equilibrium is stable or

unstable according as

h < or > r.

19. If the upper body have a plane face in contact with the lower body, r is infinite and

the equilibrium is stable or unstable according as

h < or > R.

In the critical case, when

1/h = 1/r + 1/R, the determination of stability is difficult.

2.8 Assignments:

1. Two forces P and Q act along the straight lines whose equations are y = x tan, z = c

and y = – x tan, z = – c respectively. Show that their central axis lies on a straight line

x(P – Q) tan z P2 – Q

2

P + Q c P2 + 2PQ cos2 + Q

2

For all values of P and Q prove that this line is a generator of the surface

y = =

96

(x2 + y

2) z sin2 = 2cxy.

2. Two equal forces act along generators of the same system of the hyperboloid

(x2 + y

2)/a

2 – z

2/b

2 = 1 and cut the plane z = 0 at the ends of perpendicular diameters of

the circle x2 + y

2 = a

2. Show that the pitch of the equivalent wrench is (a

2b)/(a

2 + 2b

2).

3. A force parallel to the axis of z acts at the point (a, 0, 0) and an equal force

perpendicular to the axis of z acts at the point (– a, 0, 0). Show that the central axis of the

system lies on the surface z2 (x

2 + y

2) = (x

2 + y

2 – ax)

2.

4. A force F acts along the axis of Z and a force mF along a straight line intersecting the

axis of x at a distance c from the origin and parallel to the plane of yz. Show that as this

straight line turns round the axis of x, the central axis of the force generates the surface

[m2z

2 + (m

2 – l)y

2] (c – x)

2 = x

2z

2.

5. Three forces act along the straight lines x = 0, y – z = a; y = 0, z – x = a; z = 0, x – y =

a. Show that they cannot reduce to a couple. Prove also that if the system reduces to a

single force, its line of action must lie on the surface

x2 + y

2 + z

2 – 2yz – 2zx – 2xy = a

2.

6. Forces act along generators of the same system of a hyperboloid. Shew that two

generators of the same system are null lines of the system of forces.

7. Shew that null planes of a series of points which lie in a straight line AB passes

through a second straight line CD; and that if the series of lines AB be generators of a

hyperboloid, the lines CD will also be generators of a hyperboloid.

8. A system of forces is reduced to two forces one of which acts along as assigned line.

Shew (i) that the four lines of action of two such pairs of forces are generators of the

same system of a hyperboloid of one sheet; (ii) that lines meeting to such forces and the

central axis generate a hyperbolic paraboloid, one set of whose generators is

perpendicular to the central axis.

9. A solid homogeneous hemisphere of radius r has a solid right cone of the same

substance constructed on its base; the hemisphere rests on the convex side of a fixed

sphere of radius R, the axis of the cone being vertical. Show that the greatest height of the

cone consistent with stability for a small rolling displacement is

r[{(3R + r)(R – r)} – 2r]

R + r

10. Three equal particles repelling each other forces proportional to the nth power of the

distance connected together by three equal elastic strings. Find the position of

equilibrium and shew that it is stable if n < p/(p – a), where a is the unstretched and p the

stretched length of any string.

11. A solid ellipsoid, whose axes are of lengths 2a, 2b, 2c, rests with the “c-axis” vertical

on a rough horizontal plane. The centre of gravity is on the vertical axis at a distance h

from the bottom vertex. Show that the equilibrium is stable if h is less than both a2/c and

b2/c.

12. A solid frustum of a paraboloid of revolution, of height h and latus rectum is 4a, rests

with its vertex on the vertex of a paraboloid of revolution, whose latus rectum is 4b; show

that the equilibrium is stable if h < 3ab/(a + b).

13. Show that a sphere partially immersed in a basin of water cannot rest in stable

equilibrium on the summit of any convex part of the base.

97

2.9 References:

1. Loney, S. L.: An Elementary Treatise on Statics, Cambridge University Press, 1956

2. Varma, R. S.: Text-Book on Statics, Pothishala (Private) Limited Lajpat Road,

Allahabad – 2, 1962.

UNIT –III Velocity and Acceleration, Simple

Harmonic Motion and Elastic String

STRUCTURE

1.1 Introduction

1.2 Objectives

1.3 Velocity and Acceleration Along Radial and Transverse Direction

1.4 Along Tangential and Normal Direction

1.5 Simple Harmonic Motion

1.6 Elastic String

1.7 Unit Summary

1.8 Assignments

1.9 References

98

1.1 Introduction:

This unit introduces the basics of Velocity and Acceleration Along Radial and

Transverse Direction, Along Tangential and Normal Direction, Simple Harmonic Motion,

Elastic String. It also help us to understand the basic concepts of above topics.

In this unit we shall study the basic ideas of Velocity and Acceleration Along

Radial and Transverse Direction, Along Tangential and Normal Direction, Simple

Harmonic Motion, Elastic String. It is hopped the unit help students in studying.

1.2 Objectives:

At the end of the unit the students would be able to understand the concept of:

Velocity and Acceleration

Radial and Transverse Velocity and Acceleration

Angular Velocity and Acceleration

Tangential and Normal Direction

Simple Harmonic Motion

Geometrical Representation of the S. H. M.

Hooke‟s Law

Horizontal Elastic String

Vertical Elastic String

99

1.3 Velocity and Acceleration Along Radial and Transverse Direction

Velocity and Acceleration:

Let a particle move along a straight line starting from a given point O on the line

and let it come to the position P in time t where OP = x. Further suppose in a subsequent

interval t where t is small, the particle moves through a distance PQ(= x) and comes

to the position Q. Thus (x/t) is the average velocity of the particle during the interval

t. As t and consequently x becomes smaller and smaller, the point Q approaches the

point P and the quotient (x/t) measure the rate of displacement of the particle. This

quotient gives the velocity of the particle in the limit when t0. Thus if v be the

velocity of the particle at time t, we have

displacement in time t

v = lim ----------------------------

t0 t

. . . .

x dx O P Q X

= lim ----- = -------

t0 t dt

It is usual to denote differential coefficients with regard to time by dots; thus x

means (dx/dt) and x means (d2x/dt

2).

Thus v = (dx/dt) = x

Velocity of a particle has got magnitude as well as direction. The magnitude is

called the speed. If both speed and direction remain the same throughout a certain

interval the velocity is uniform throughout that interval. If either of these changes, the

velocity becomes variable.

Acceleration of a moving particle is defined as the rate of change of velocity. If v

be the velocity of the particle at time t when it is at the point P and v + v be its velocity

at time t + t when it is at Q, then v is the change of velocity in interval t. thus if f be

the acceleration of the particle at time t, we have

change of velocity in time t

f = lim ----------------------------------

t0 t

v dv

= lim ----- = -------

t0 t dt

But v = (dx/dt) so f = (d2x/dt

2) = x

Also f = lim(v/t) = lim (v/x) . lim (x/t)

. ..

.

..

100

= (dv/dx) . (dx/dt) = v (dv/dx)

Thus any one of the three expressions (dv/dt) or (d2x/dt

2) or v (dv/dx) may be

taken as the acceleration of the moving particle.

Acceleration also has magnitude as well as direction. Negative acceleration is also

known as retardation. Retardation implies decreases in the magnitude of velocity.

It should be very carefully noted that (dx/dt) or x is the velocity of the particle in

the sense in which x increases, similarly, v (dv/dx) or x is the acceleration of the particle

in the sense x increasing.

Radial and Transverse Velocity and Acceleration:

A particle is moving in a plane curve, to find components of velocity and

acceleration at time t along and perpendicular to the radius vector drawn from a fixed

point in the plane.

Consider the fixed point O as the

pole and line OX as the initial line. Let P v + v

be the position of the particle at time t, its u + u

coordinates be (r,) and Q be the position v Q

at time t + t, its coordinates (r + r, + ), u

so that the chord PQ is the displacement in r + r

time t. Draw QM perpendicular from Q to P M

OP so that PM and QM are the components

of the displacement PQ along and perpendicular O

to OP. Let u, v be the components of velocity A

along and perpendicular to OP.

Then

Displacement along OP in time t

u = lim -----------------------------------------

t 0 t

PM OM – OP

= lim ------ = lim -----------

t 0 t t 0 t

OQ cos – OP

= lim ------------------

t 0 t

(r + r) 1 – r

= lim ---------------, Small quantities of above the first order

t 0 t being neglected.

r dr

= lim --- = ---- = r

t 0 t dt

displacement perp. to OP in time t

.

.

..

r

101

v = lim ---------------------------------------------

t 0 t

QM OQ sin

= lim ------ = lim ---------

t 0 t t 0 t

(r + r) sin (r + r) sin

= lim ------------- = lim -------------- . ----

t 0 t t 0 t t

sin

= lim (r + r) ---- because lim -------- = 1

t 0 t 0

r

= lim -----, neglecting the other term

t 0 t

d

= r ----- = r

dt

Thus the components of velocity along and perpendicular to the radius vector are

r and r, in the sense in which r and increase. These are called the radial and transverse

or cross-radial components of velocity.

Now let the components of velocity along and perpendicular to OQ be u + u, v +

v; (u, v) being those along and perpendicular to OP.

Thus the change of velocity along OP in time t

= (u + u) cos – ( v + v) sin – u

= (u + u) . 1 – ( v + v) . – u, neglecting higher power of

= u – v, neglecting the other term.

Similarly the change of velocity perpendicular to OP in time t

= (u + u) sin + (v + v) cos – v

= (u + u) + (v + v) . 1 – v, as before

= u + v, neglecting the other term.

Therefore, radial acceleration

Change of velocity along OP in time t

= lim -----------------------------------------------

t 0 t

u – v

= lim ----------

.

. .

102

t 0 t

du d

= ------ v -------

dt dt

d dr d d

= --- --- r ---- . ----

dt dt dt dt

d2r d

= --- r ---- = r – r 2

dt2 dt

Transverse acceleration

Change of velocity perp. to OP in time t

= lim -------------------------------------------------

t 0 t

u + v

= lim ------------

t 0 t

d dv

= u ----- + ---

dt dt

dr d d d

= ----- ---- + --- r ----

dt dt dt dt

dr d dr d d2

= --- . --- + ---- . --- + r ----

dt dt dt dt dt2

d2 dr d

= r --- + 2 --- . ---

dt2 dt dt

= r + 2r

1 d d

also = --- ---- r2 ----

r dt dt

.. .

. . ..

103

1 d

= --- ---- (r2)

r dt

Thus the components of acceleration are r – r along OP in the sense r increasing

1 d

and --- --- (r2) perp. to OP in the sense increasing.

r dt

Cor. If the particle describe a circle of radius a, then r = a = constant, so that u = r = 0, v =

r = a;

Radial acc. = r – r = 0 – a

2 = - a

2, i.e., a

2 towards the center and

Transverse acc. = r + 2r = a, i.e., tangentially.

Ex.1 If the radial and transverse velocities of a particle are always proportional to each

other, show that the path is an equiangular spiral.

Sol.

dr d

Here --- = kr --- where k is some constant

dt dt

dr

or ---- = kd

r

Integrating we get, log r = k + C, where C is some constant.

or r = aek

, where a is also a constant.

This is an equiangular spiral.

Ex.2 The velocities of a particle along and perpendicular to the radius from a fixed origin

are r and ; find the path and show that acceleration, along and perpendicular to the

radius vector, are 2r – (

2

2)/r and (+ /r)

Sol.

dr d

Here ---- = r and r ----- =

dt dt

rd dr d

Dividing we get, ----- = ----- or --- ---- = -----

dr r r2

Integrating we get

1

- ---- . ---- = log + C, where C is a constant.

r

This gives the path.

Radial acceleration = r – r

.

.

.

. . .

..

.. .. ..

. . . .

. .

.. .

104

2

2

= r – r . ------, since r = r so r = r = 2r

r2

= 2r – ----, and = -----

r r

Transverse acceleration

= 2r + r

= 2r . ---- + r

r

By differentiating r =

we get r + r =

i.e. r = ---- - r . ------ = ------ . ( - r )

r r r

Hence transverse acceleration = 2 + ----- ( - r)

r

= [2 + (/r) - ] = [+ (/ r)]

Ex.3 A straight line of constant length moves with its ends on two fixed rectangular axes

OX, OY and P is the foot of the perpendicular from O on the straight line. Show that

velocity of P perpendicular to OP is OP. and along OP is 2CP. where C is the middle

point of the line and is the angle COX.

Sol. Since AOB is a right angled triangle, C is the middle point of the hypotenuse, OC =

CA = CB = a if AB = 2a

COA = = CAO

POX = 90 – Y

and PO = OC cos (90 – 2) = a sin2 B

so polar coordinate of P are (a sin2, 90 – )

Also CP = a cos2

so Velocity of P along OP P

d

= ---- (a sin2) = 2a cos2 C

dt

O X

= 2.CP A

d

Velocity of P perp. to OP = a sin2 ----- (90 – )

dt

= - OP.,

i.e. velocity along PA = OP.

. . .

.

. .

.

. . .

..

..

..

..

..

. .

.

.

.

.

105

Ex.4 A boat which is rowed with constant velocity U starts from a point A on the bank of

a river which flows with a constant velocity V; and it points always towards a point B on

the other bank exactly opposite to A; find the equation of the path of the boat. If V = U,

show that the path is a parabola whose focus is B.

Sol.

Let P be the position of the boat at time t. It has two velocities, U towards B and

V downstream. Take the opposite bank as initial line and B as pole and P the point (r, ),

its components of velocity are r along BP produced and r perpendicular to BP.

Hence resolving along and perp. to BP, we get

r = V cos – U

and r = - V sin B X

rd V sin u r

Dividing we get ----- = ------------- P v

dr V cos – U

or (dr/r) = [- cot+ (U/V)cosec]dA

Integrating we get log r = log cosec + (U/V) log tan/2 + C1 where C1 is

constant.

This gives the path.

If V = U, it becomes log r = log cosec + log tan/2 + C1

or log r = log [1/(2cos2/2)] + C1

or 2r cos2/ 2 = A where A is also a constant

i.e., A/r = 2 cos2/2 = 1 + cos which is a parabola, with B as focus.

Ex.5 A and B are two fixed points on the circumference of a circle and the distances from

A and B of any other point P on the circumference are r and s respectively. If u and v are

the components of P‟s velocity, as it moves round the circumference, along AP and BP

respectively prove that being the angle APB.

u sin2 = r – s cos, u v

v sin2 = s – r cos r s

Sol.

Resolving along AP, s r

r = u – v cos(180 - ) = u + v cosB A

and resolving along BP

s = v – u cos(180 - ) = v + u cos

so r – s cos = u sin2

and s – r cos = v sin2

Angular Velocity and Acceleration:

The angular velocity of a point P about another point O is the rate of change of

the angle, which OP makes with some fixed direction.

A particle is moving in a plane curve.

Take a line OX fixed in the plane of the curve

as initial line and O as pole. Let P be the Q

. .

. .

. .

. .

.

.

. .

. .

106

position of the particle at time t, being given

by the angle XOP = . P

Let Q be the position at time t + t, so

that the angle described in time t is .

Thus the average angular velocity of P O X

about O is (/t).

As t becomes smaller and smaller, Q approaches P and (/t) becomes the rate

of change of . This is angular velocity.

Thus angular velocity of P about O = lim (/t) = (d/dt) =

t0

Similarly angular acceleration is d(d/dt) = (d2/dt

2) = v

dt r

Let v be the velocity of the particle r

at P. It is along the tangent to the path at P. P

Components of velocity of P along and perp.

to OP are r and r. r

Resolving perp. to OP, we get

r = v sin O

X

where is the angle the tangent at P makes p

with OP.

Thus = (vsin/r)

i.e. angular velocity of P about O = [(velocity of P resolved perp. to OP)/OP]

Also = (vsin)/r and sin = (p/r) where p is perp. from O to the tangent at P. So

= (vp)/r2.

Ex.1 A particle describe an equiangular spiral r = ae in such a manner that its

acceleration has no radial component. Prove that its angular velocity is constant and that

the magnitude of the velocity and acceleration is each proportional to r.

Sol.

Here r – r 2 = 0

From r = ae, r = ae

= r

r = r + r = r

+ r

so r = r – r2 = 0 as given above.

Hence = 0

so = constant = K say

Then r = Kr so that v2

= r2 + r

= 2K

2r2

so v varies as r

Since radial acc. is zero, only acc. is transverse

= 2r + r = 2r = 2K2r

so acc. Varies as r.

Ex.2 A rod moves with its ends on rectangular axes OX, OY. If x, y be a point P on the

rod and if the angular velocity of the rod is constant, show that components of

.

..

. .

.

.

.

.

.

.

.

. . .

. . .

.

.

. . .

. . . .

..

.. .. ..

.. .. ..

..

107

acceleration of P along the axes are – x and – y

and the resultant acceleration is

OP.2 towards O.

Sol.

If C be the middle point then OC = CA = CB = a,

so that AOC = = OAC.

It is given that = const. =

Let CP = fixed = c for any particular

point P chosen. Then B

x = a cos + c cos = (a + c) cos

y = (a – c) sin C

x = - (a + c) sin = - (a + c) sin

y = (a – c) cos

x = - (a + c) 2cos = - x

2 P

y = - (a – c) 2sin = - y

A

Resultant acc. makes angle with x-axis as O

tan-1

(y/x) = tan-1

(y/x) =

where is the angle OP makes with x-axis.

Hence acc. is along PO as both x and y are negative, and is equal to

(x2 + y

2) =

2(x

2 + y

2) =

2.OP.

1.4 Along Tangential and Normal Direction

A particle is moving in a plane curve; to find components of its acceleration along

the tangent and the normal to the curve at any instant.

Let A be a fixed point on the curve and P be the position of the particle at time t

where AP = s. Let v be the velocity of the particle at P, it being entirely along the tangent

at P; so that v = s.

Let Q be the position of the particle at time t + t and v + v be the velocity there

i.e. along the tangent at Q.

Let the tangents at P and Q make angles

and + with a fixed line OX so that

is the angle between the tangents.

Thus the change of velocity along the Y

tangent at P in times t v + v

= (v + v)cos – v

= (v + v).1 – v, N Q v

neglecting terms of second order

= v

and the change of velocity along the normal at

P in time t A P

= (v + v)sin – 0 +

= (v + v), neglecting second order terms O X

= v, neglecting the other terms.

Thus tangential acceleration

change of velocity along the tangent in time t

..

..

.. ..

.. ..

.

. .

.

.

.. ..

108

= lim ------------------------------------------------------

t0 t

v dv d2s

= lim ----- = ----- = -----

t0 t dt dt2

or = (dv/ds).(ds/dt) = v. (dv/ds)

Normal acceleration

change of velocity along the normal in time t

= lim -------------------------------------------------------

t0 t

v v s

= lim ------- = lim -------- . ------

t0 t s t

v s

= lim ------- . lim ----

s0 s t0 t

= (v/).v = (v2/)

where is the radius of curvature at P.

Thus for a particle moving in a plane curve, component of acceleration along the

tangent is (dv/dt) or (d2s/dt

2) or v(dv/ds), in the sense in which s increase; and the

component of acceleration along the normal is (v2/) in the inward sense.

Cor. For a particle moving in a circle of radius a, s = a, so that tangential acc. = s

=a, in the sense increasing and normal acc. = (v2/) = (s

2/a) = (a

/a) = a

, towards

the center.

Ex.1 Prove that the acceleration of a point in a curve with uniform speed is 2. A point

describes a cycloid s = 4a sin with uniform speed v. Find its acceleration at any point s.

Sol. If v = const.

Components of acc. are (dv/dt) = 0 and (v2/).

Now = (d/dt) = (d/ds).(ds/dt) = (v/)

So v =

Hence acc. = (

/) =

2.

For s = 4a sin, = (ds/d) = 4a cos

So acc. = (v2/) = (v

2/4acos) = [v

/{4a(1 – sin

2)}]

= [v2/{4a(1 – s

2/16a

2)}] = [v

2/(16a

2 – s

2)]

Ex.2 A point P describes, with a constant angular velocity of OP, an equiangular spiral of

which O is the pole. Find its acceleration and that its direction makes the same angle with

the tangent at P as the radius vector OP makes with the tangent.

Sol.

. . . ..

..

.

.

. .

.

109

Equation to equiangular spiral is r = aecot

Here = constant = c, say

So r = a cot ecot

= cr cot

r = cr cot = c2cot

2r

So r – r = c

2 cot

2 r – c

2r = (cot

2 – 1)c

2r

Also 2r + r = 2c2r cot + 0 = 2c

2r cot, so = 0

These are the components of acceleration in polar coordinates hence the resultant

acc. = (cot2 – 1)

2 c

4r2 + 4c

4r2

cot2] = c

2r cosec

2. If be the angle the direction of the

resultant acc. makes with OP then tan = (2c2r cot)/[(cot

2 – 1)c

2r] = (2cot)/(cot

2 –

1) = tan2, so =2 but the tangent makes an with OP, so direction of the acceleration

makes the same angle with the tangent.

Ex.3 A particle is moving in a parabola with uniform angular velocity about the focus;

prove that its normal acceleration at any point is proportional to the radius of curvature of

its path at that point.

Sol.

Equation to parabola with the focus as pole is p2 = ar.

Here = const. = c, say but = (vp/r2), so c = (vp/r

2); so v = (cr

2/p)

Also = r(dr/dp)

Differentiating p2 = ar with respect to p, we get

2p = a(dr/dp), so (dr/dp) = (2p/a)

So = r.(2p/a) = (2pr/a) = [(2r.ar)/(ap)] = (2r2/p)

Now normal acc. = (v2/) = [(c

2r4/p

2)/(2pr/a)] = [(ac

2r3)/(2p

3)] = [(ac

2r3)/(2ar.p)]

= [(c2r2)/(2p)] = (c

2/4). i.e. varies as.

Ex.4 A particle describes a parabola with uniform speed; show that its angular velocity

about the focus S, at any point P, varies inversely as SP3/2

.

Sol.

Here v = const. = c, say.

Equation to parabola with S as pole is p2

= ar.

Now = (vp/r2) = (cp/r

2) = [c(ar)/r

2] = (ca/r

3/2) (1/r

3/2)

Ex.5 Prove that the angular acceleration of the direction of motion of a point moving in a

plane is (v/).(dv/ds) – (v2/

2).(d/ds).

Sol.

If the tangent at any point of the path makes angle with the fixed line OX, then

gives the direction of motion.

To find

Now = (d/ds).(ds/dt) = v.(d/ds) = (v/)

d v. ds d v. d. v.

dt dt ds ds

= (v/).(dv/ds) – (v2/

2).(d/ds).

.

.

.

. .

. .

..

.. ..

. .

.

.

..

.. = = v

110

Ex.6 Prove that if the tangential and normal accelerations of a particle describing a plane

curve be constant throughout the motion the angle through which the direction of

motion turns in time t is given by = A log(1 + Bt).

Sol.

Here s = const. = a say and (s2/) = (v

2/) = const. = b, say

So s = at + c and [s2/(ds/d) = b or s.[(ds/dt)/(ds/d)] = b or s.(d/dt) = b

So (at + c).(d/dt) = b. Hence (d/dt) = [b/(at + c)]

Therefore = A log (1 + Bt).

Ex.7 A particle moves in a catenary (s = c tan), the direction of its acceleration at any

point make equal angles with the tangent and the normal to the path at the point. If the

speed at the vertex where = 0 be u, show that the velocity and acceleration at any other

point are given by ue

and (2/c).ue

cos2

Sol.

Since the acc. bisects the angle between the tangent and the normal, the tangential

and normal components of acc. are equal.

Hence v.(dv/ds) = (v2/) or (dv/ds) = (v/) or (dv/ds) = v.(d/ds)

So (dv/v) = d. Integrating log v = + const.

Therefore v = Ae

but when = 0, v = u

Hence A = u therefore v = ue

Tang. acc. = v.(dv/ds) = v.(dv/d).(d/ds) = (ue

.ue

)/(c sec2), since s = c tan

= (u2e

2/c).cos

2 so (ds/d) = c sec

2 =

Normal acc. = (v2/) = (u

2e

2/(c sec

2) = (u

2e

2.cos

2)/c

Hence resultant acc. = [{(u2e

2 cos

2)/c}

2 + {(u

2e

2 cos

2)/c}

2]

= (2/c).u2e

2 cos

2

1.5 Simple Harmonic Motion:

A particle is to say to execute Simple Harmonic Motion if it moves in a straight

line such that its acceleration is always directed towards a fixed point in the line and is

proportional to the distance of the particle from the fixed point.

Let O be the fixed point on a line BOA

and P be the position of the particle at time t

where OP = x, so that the acceleration of the

particle in the sense OP is x. . . .

Now the given acceleration is towards B O x P A

O and is proportional to x. Let it be x, where

is constant.

Since x is in the direction of OP produced and x is towards O, the equation of

motion is

x = - x.

Taking v(dv/dx) instead of x; we can write the above equation as

v(dv/dx) = - x. (1)

.. .

. . . .

..

..

..

..

111

Integrating with respect to x, we get

(v2/2) = - (x

2/2) + (C/2) where C is a constant

or v2 = - x

2 + C.

If A be the extrme position of the particle i.e., it is at rest at A i.e., when x = a, v =

0 where OA = a, we get

0 = - a2 + C, so C = a

2

Hence v2 = (a

2 – x

2)

i.e. v = (a2 – x

2) (2)

If the particle moves from A towards O, v is negative

Hence x = v = - (a2

– x2)

or (dx/dt) = - (a2 – x2)

or dt = - [dx/(a2 – x

2)]

Integrating we get .t = cos-1

(x/a) + C1, where C1 is a constant. Initially at A, t =

0, x = a i.e., the particle started from A, then

0 = cos-1

1 + C1, so C1 = 0

Hence .t = cos-1

(x/a)

or x = a cos.t (3)

If the particle moves from O towards A, v is positive so that

x = (a2 – x

2)

or .dt = [dx/(a2 – x

2)]

Integrating, we get, .t = sin-1

(x/a) + C2, where C2 is a constant.

If the particle starts from O, t = 0, x = 0, 0 = sin-1

0 + C2, so C2 = 0

Hence x = a sin.t (4)

Thus the solution of (1) is x = a cos.t or x = a sin.t according as the starting

point is A or O.

From (2), v = 0 when x = a.

Thus if B is a point on the other side of O such that OB = OA = a, the particle

comes to rest also at B. When x = 0, v = .a, i.e., at O, the velocity is a.

Consider the solution x = a cos.t

The motion starts from A under an attraction towards, O. When the particle

reaches O, x = 0. So cos.t = 0. Hence t = (/2) i.e., t = [/(2)] is the time required

in moving from A to O.

As the particle reaches O, the attraction ceases but the particle has a velocity .a

towards the negative side of O hence the particle passes O and moves towards the

negative side. As soon as the particle comes to the left side of O, attraction changes

direction and becomes towards O; hence the velocity will go on decreasing as the particle

moves towards the left, till at B, the velocity becomes zero so that the particle stops. But

the particle is being attracted towards O hence starts moving towards O and reaches O

with a velocity .a, due to which it passes O and moves towards A and again stops at A

where its velocity becomes zero. The motion is then repeated. Thus the motion is from A

to B and back to A and so on. The motion is oscillatory. Time from O to B is equal to

that from A to O hence the period i.e., the time from A to B and back to A is 4.[/(2)

= (2/). The distance a (= OA) i.e., the distance of the center from one of the position

of rest is called the amplitude.

.

.

112

Thus the period which is equal to (2/) is independent of the amplitude i.e.,

whatever be the amplitude the period is the same. Thus the simple harmonic motion is

oscillatory and periodic, the period being independent of amplitude.

The frequency is the number of complete oscillation in the one second, so that if

n be the frequency and T the periodic time, then

n = (1/T) = /(2)

The equation (1), namely x = - x, can be solved as a differential equation. The

most general solution of this equation is

x = A cos.t + B sin.t (5)

A, B are constants to be determined from initial conditions. In the first case when the

motion starts from A, the initial conditions are t = 0, x = a, x = 0.

Now t = 0, x = a given a = A.

Differentiating (5), x = - Asin.t + Bcos.t (6)

The condition t = 0, x = 0 gives 0 = 0 + B, so B = 0.

Hence the solution is x = a cost

In the second case when the motion starts from O, the first condition is t = 0, x =

0. So 0 = A, A = 0

Hence x = B sint.

To determine B, we must know the velocity of projection from O.

Let us take the case of particle, projected from A with velocity V along OA

produced, so that the initial conditions are t = 0, x = a, x = V.

Hence from (5) and (6), we get

a =A

V = B. So B = (V/)

Hence the solution is x = a cos.t + (V/) sin.t

Also the general solution of (1) can be written as

x = a cos (t + )

This is the periodic with period (2/).

The quantity is called the epoch, the angle (t + ) is called the argument.

The particle is at its maximum distance at time t0 where (t0 + ) = 0 i.e., t0 = - (/).

Hence the time that has elapsed since the particle was at its maximum distance is equal to

t – t0 = t + (/) = (t + )/. This is the phase at time t.

Geometrical Representation of the S. H. M.:

Let a particle P move on a circle with

constant angular velocity and let M be the P

foot of the perpendicular from P on any diameter

OA. If a be the radius of the circle, the only

acceleration of P is 2a towards O. If AOP =

and OM = x, the component of this acceleration O M A

along OA

= 2a.cos =

2a.(x/a) =

2x towards O.

Hence the equation of motion of the point M is

x = - 2x.

..

.

.

.

.

..

113

Thus is S. H. M.

Thus if a particle describes a circle with constant angular velocity the foot of the

perpendicular from it on any diameter executes a simple harmonic motion.

Ex.1 A particle is moving with S. H. M. and while making an excursion from one

position of rest to the other, its distances from the middle point of its path at three

consecutive seconds are observed to be x1, x2, x3, prove that the time of a complete

revolution is 2/cos-1

[(x1 + x3)/2x2].

Sol.

From x = a cost

x1 = a cost, x2 = a cos (t + 1), x3 = a cos(t + 2)

So x1 + x3 = 2a cos(t + 1) cos

Hence (x1 + x3)/(2x2) = cos

Therefore = cos-1

[(x1 + x3)/(2x2)]. T = (2/)

Ex.2 A particle starts from rest under an acceleration K2x directed towards a fixed point

and after time t another particle starts from the some position under the same

acceleration. Show that the particles will collide at time (/K) + (t/2) after the start of the

first particle provide t < (2/K).

Sol.

x = - K2x, So period = (2/K)

The condition t < (2/K) indicates that the second starts before the first has made

one complete oscillation. Let them meet after time t of the starts of the second then

a cos K(t + t) = a cos Kt

So K(t + t) = 2 – Kt, hence t = (/K) – (t/2).

Therefore t + t = (/K) + (t/2)

Ex.3 A horizontal shelf is moved up and down with S. H. M. of period ½ sec. What is the

amplitude admissible in order that a weight placed on the shelf may not be jerked off?

Sol.

(2/) = ½. So = 162

Weight will be jerked off when the max. acc. of S. H. M. is greater than g and if it

is not to be jerked off, max. acc. of S. H. M. must be g i.e.,

a = g. So a = (g/162)

Ex.4 In a S. H. M. of amplitude a and period T, prove that

T

v2dt = (

2a/T)

0

Sol.

x = a cost, v = x = - asint,

T = (2/)

T T T

v2dt = a

2sin

2tdt = a

2sin

2(t / T)dt

..

.

114

0 0 0

2

= a2sin

2z (T/2)dz, Put (2t / T) = z

0

= [a2T/(2)]. = (a

2T/2).(4

2/ T

2 ) = (2

2a

2/T)

Ex.5 Show that, if a smooth straight turned be cut between any two points on the earth‟s

surface; it should be traversed by a particle starting from rest under the action of gravity

in about 42 ½ minutes.

Sol.

Let O be the center of the earth, AB the

tunnel and OC perpendicular from O to AB.

Let P be the position of the particle at

time t where CP = x. As the force of attraction B C P A

inside the earth varies as its distance from the x

center, the force on P = .OP along PO. O

Its component along PC

= PO cos OPC

= PO. (x/PO) = - x

Hence the equation of motion of P is x = - x.

It is S. H. M. of periodic time T = (2 /)

So Time from A to B = ½ T = (/).

Now attraction at A is g, so that a = g, where a = earth‟s radius.

So = (g/a)

Hence Required time = (/) = (a/g)

= 3.1416 [(4000 x 1760 x 3)/32.2] secs.

= 42 ½ minutes nearly.

Ex.6 If in a simple harmonic motion u, v, w be the velocities at distances a, b, c from a

fixed point on the straight line, which is not the center of force, show that the period T is

given by the equation

u2 v

2 w

2

(/ T

2).(b – c).(c – a).(a – b) = a b c

1 1 1

Sol.

Let the intensity of the force be and amplitude be so that velocity at distance x

from center of force is given by

v2 = (

2 – x

2)

Let d be the distance of the fixed point from the center of force.

Hence

u2 = [

2 – (a + d)

2]

v2 = [

2 – (b + d)

2] and T =(2/), so = (

/ T

2)

w2 = [

2 – (c + d)

2]

Thus

(u2/) + a

2 + 2ad + d

2 –

2 = 0

..

115

(v2/) + b

2 + 2bd +d

2 –

2 = 0

and (w2/) + c

2 + 2cd + d

2 –

2 = 0

Eliminating d and (d2 –

2), we get

(u2/) + a

2 a 1

(v2/) + b

2 b 1 = 0

(w2/) + c

2 c 1

u2 a 1 a

2 a 1

or (1/) v2 b 1 + b

2 b 1 = 0, which gives the answer.

w2 c 1 c

2 c 1

Ex.7 A body is attached to one end of an inextensible string and the other end moves in a

vertical straight line with n complete oscillations per second. Show that the string will not

remain tight during the motion unless

n2 < g /(

2a)

where a is the amplitude of the motion.

Sol. The maximum acceleration of the upper end executing S. H.M. is a and its period is

T = (2/). Also T = (1/n).

Hence = (42/ T

2) = 4

2n

2

So that the maximum acc. = 42n

2a.

The maximum acc. of the particle is g which is possible when the string is not

tight hence the string will not remain tight if the acc. of the upper end is greater than g

i.e., if 42n

2a > g

i.e., the string will not remain tight unless n2 < g/(4

2a)

Ex.8 A particle P of unit mass free to slide on a straight wire is attracted towards a point

O of the wire with a force .OP. If the wire be mad to revolve about O with constant

angular velocity in a horizontal plane, show that the motion of the particle on the wire

is simple harmonic with a period 2/(- 2) provided

2 < ; and prove that when

2 =

½ , the path of P in space is a circle.

Sol. When the wire has turned through an angle , P .

let P be the position of the particle where OP = r so r .

that attraction on P is r towards O. Equation of r

motion of P is r – r2 = - r. But = constant = .

So r – r2 = - r. or r = - (-

2)r. O

This is the S. H. M. of periodic time 2/(- 2) if >

2.

If = 22 then the equation becomes r = -

2r

Solution of which is r = a cos(t + )

But = So = t + C. But C = 0 since = 0 = t, = t.

Hence r = a cos( + ) which is a circle.

Hooke’s Law:

If an elastic string is fixed at one end and pulled at the other, it is found to

increase in length. The extension is directly proportional to the product of the tension and

.. . .

.. ..

..

.

116

the natural length and inversely as the area of the cross-section. Thus if x be the

extension, l the natural length, A the area of the cross-section and T the tension, then

x = [(T l)/(A)] or T = (Ax)/ l]

where is a constant depending on the material of the string.

If we take A = unit area, then T = (x/l); i.e. the tension of the elastic string is

proportional to the extension. This Hooke‟s Law and is called the modulus of elasticity.

1.6 Elastic String

Horizontal Elastic String:

One end of an elastic string is fixed to x

a point on a smooth horizontal table and . . . . . .

to the other end a particle is attached; the B A O A T P B

particle is pulled out to a distance and then

let go; to discuss the motion.

Let O be the fixed point and a be the natural length of the string (= OA). Let the

particle be pulled out to a point B where AB = b and then released.

Let P be the position of the particle at any subsequent time t, and let AP = x, so

that x is the extension at time t.

If T be the tension of the string towards A, then by Hooke‟s law

T = (x/a), where is the modulus of elasticity.

If m be the mass of the particle, the equation of motion is

mx = - T

i.e. mx = - (x/a)

or x = - [/(am)].x

This shows that the motion is simple harmonic about A.

The period of oscillation = 2(am/

Writing the above equation in the form v(dv/dx) = - [/(am)].x and integrating

with respect to x, we get (v2/2) = - [/(am)].(x

2/2) + (C/2) where C is a constant

i.e. v2 = - [/(am)]x

2 + C

At the extreme point B where x = b, v = 0

So 0 = [/(am)].b2 + C, therefore C = [/(am)].b

2

Hence v2 = [/(am)](b

2 – x

2).

When the particle reaches A, for which x = 0, v = [/(am)].b.

As soon as the particle reaches A, the string gets back its natural length,

consequently T = 0 and therefore simple harmonic motion ceases but the particle has a

velocity b.[/(am)] at A and hence with this velocity the particle moves backwards till it

comes to a point A on the other side of O where OA = OA = a.

This velocity remains constant through out, as the string remains slack and hence

the time taken by the particle to move from A to A, a distance = 2a, is (2a/b).(am/).

As soon as the particle goes beyond A, the string becomes extended so that the

tension comes into play and the motion is again simple harmonic till the particle reaches

a point B where OB = OB, where the motion stops. The particle then retraces its path

under S. H. M. till it comes to A and then under uniform motion till the point A and then

again S. H. M. till the point B where the motion again stops. This motion is repeated.

..

..

..

117

Thus the motion from B to A, A to B, B to A and A to B is simple harmonic

and the total period for which is 2[(am)/].

The time from A to A and A to A is (4a/b).[(am)/].

Hence the total time for one complete oscillation is

2[(am)/] + (4a/b)[(am)/].

Ex.1 A particle of mass m executes S. H. M. in the line joining the points A and B on the

smooth table and is connected with these points by elastic strings whose tension in

equilibrium are each T; show that the time of an oscillation is 2(mll)/{T(l + l)}]

where l, l are the extension of the strings beyond their natural lengths.

Sol.

Let a, b be the natural lengths of two

strings and O the equilibrium position of the A T O T B

particle. a + l b + l

In equilibrium position

T = 1(l/a) = 2(l/b) (1)

When the particle is slightly displaced A T1 T2 B

from O, let P be its position at time t where O x P

OP = x. Then if T1, T2 be the tension in the

strings PA and PB, we have

T1 = 1[(l + x)/a], T2 = 2[(l – x)/b].

Equation of motion is mx = T2 – T1 = - x [(1/a) + (2/b)] from (1)

= - Tx[(1/l) + (1/l)], also from (1)

i.e., x = - [T(l + l)/(mll)].x

This is a S. H. M. of periodic time = 2[(mll)/{T(l + l)}].

Ex.2 A particle m is attached to a light wire, which is stretched tightly between two fixed

points with a tension T. If a, b are the distances of the particle from the two ends, prove

that the period of a small transverse oscillation of m is

2[(mab)/{T(a + b)}]

Sol.

When the particle is slightly displaced

at right angles to AB and it is at a position P at P

time t where OP = x, the tension T remains T T

practically the same. Sum of components of x x

x

two T‟s along PO

= T.[x/(a2 + x

2)] + T.[x/(b

2 + x

2)] B b O a A

= T.[(x/a) + (x/b)] neglecting other terms.

Hence the equation of motion is

mx = - Tx[(1/a) + (1/b)] = - T.[(a + b)/(ab)].x

i.e. x = - [T(a + b)/(mab)].x

..

..

..

..

118

This is a S. H. M. of periodic time = 2[(mab)/{T(a + b)}].

If the length of the wire be given, i.e. (a + b) is given, then time period is

maximum when ab is maximum, i.e. when a = b. Hence for a wire of given length the

period is longest when the particle is attached to the middle point.

Vertical Elastic String:

A particle is suspended from a fixed O O O

point by an elastic string; to discuss the motion.

O is the fixed point, OA = a the natural a

length of the string. A particle of mass m is

attached to the free end and hangs in equilibrium

at a depth b below A and is at B so that AB = b. A A A

If T0 be the tension in that position then b T0

T0 = mg mg

i.e. (b/a) = mg (1) B B

The particle is then taken to a depth c below T x

B to the position C so that BC = c and then released. mg P

Let P be the position of the particle at any

subsequent time t when BP = x; so that if T be the C

tension in that position

T = [(b + x)/a] = mg + (x/a), from (1).

The equation of motion is mx = mg – T = mg – mg – (x/a) = - (x/a)

i.e. x = - [/(am)]x. (2)

This is a S. H. M. having B, the position of equilibrium, as the centre of

oscillation.

The periodic time = 2[(am)/] = 2 (b/g) from (1).

The amplitude of this S. H. M. is c and if BC<AB, i.e. c<b, the motion is

completely simple harmonic about the point B.

If, however, c<b, the particle in its upward motion goes above A, but at A the

string gets its natural length and so the tension becomes zero hence the S. H. M. ceases,

but the particle has a velocity at A with which it rises against gravity.

Now integrating (2), we get

v2 = C – [/(am)]x

2 = C – (g/b)x

2; where C is a constant.

But when x = c, v = 0, then 0 = C – (g/b)c2. So C = [(gc

2)/b]

Hence v2 = (g/b)(c

2 – x

2).

Therefore at A, where x = - b, v2 = (g/b)(c

2 – b

2).

This is the velocity at A.

Hence the height through which the particle rises above A is

[v2/(2g)] = [(c

2 – b

2)/(2b)]

This will be the case provided the height risen above O does not exceed a, the

natural length of the string, otherwise the motion again S. H. M.

Hence the condition is [(c2 – b

2)/(2b)] < 2a, i.e. c

2 < (b

2 + 4ab).

..

..

119

Ex.1 A heavy particle is attached at one point of a uniform light elastic string. The ends

of the string are attached to two points in a vertical line. Show that the period of vertical

oscillation in which the string remains taut is 2[(mh)/(2)], where is the coefficient

of elasticity of the string and h the harmonic mean of the unstretched lengths of the two

parts of the string.

Sol.

O, B the points where the ends are O

attached and A the point where the particle

is attached and let OA = a, AB = b and let T1

l1, l2 be the natural lengths of these two

portions. A

Therefore the equilibrium x

[(a – l1)/l1] = mg + [(b – l2)/l2] (1) P

Let the particle be slightly displaced. T2

When it is at P where AP = x, the lengths of

the two portions are (a + x), (b – x) and T1 B

and T2 be their tensions

T1 = [(a + x – l1)/l1], T2 = [(b – x – l2)/l2]

Therefore the equation of motion is

mx = mg + T2 – T1 = mg + [(b – x – l2)/l2] – [(a + x – l1)/l1]

= - x [(1/l1) + (1/l2)] = - x [(l1 + l2)/l1l2] = - (2/h).x

So period = 2[(mh)/(2)].

Ex.2 Two bodies M and M are attached to the lower end of an elastic string whose upper

end is fixed and are hung at rest, M falls off; show that the distance of M from the upper

end of the string at time t is

a + b + c cos(g/b).t

where a is unstretched length of the string, b and c the distances by which it would be

extended when supporting M and M respectively.

Sol.

Mg = (b/a), Mg = (c/a)

Mx = Mg – T = Mg – (x/a) = Mg – (Mg/b).x O O

So x = - (g / b).(x – b)

Therefore x – b = A cos(g/b).t + B sin(g/b).t

Initially x = b + c, t = 0, x = 0, So B = 0, A = c A A

Hence x = b + c cos(g/b).t b + c M

Required depth = a + x = etc.

M M

Ex.3 A tight elastic string of natural length l has one extremity fixed at a point A and the

other attached to a stone the weight of which in equilibrium, would extend the string to a

..

..

..

.

120

length l1; show that if the stone be dropped from rest at A, it will come to instantaneous

rest at a depth (l12 – l

2) below the equilibrium position and this depth is attained in time

(2l/g) + [(l1 – l)/g].[ - cos-1(l1 – l)/(l1 + l)}]. Prove also that if the greatest depth

below A be lcot2(/2), the modulus of elasticity is ½ mg tan

2 and the above time is

(2l/g)[1 + ( - ) cot].

Sol.

Let P be the position of the particle at time t, where AP = x. A

The equation of motion is

mx = mg – T = mg – [(x – l)/l].

In equilibrium position, x = 0, x = l1 l

So 0 = mg – [(l1 – l)/l].

Hence mx = mg – [(mg)/(l1 – l)].(x – l) = - [(mg)/(l1 – l)].(x – l1)

i.e. x = - [g/(l1 – l)].(x – l1). B

Integrating this equation we get

x2 = - [g/(l1 – l)].(x – l1)

2 + C T

where C is constant.

Now the simple harmonic motion beings P

when the particle has fallen freely through a

distance l so that when x = l, x2 = 2gl.

So 2gl = - [g/(l1 – l)].(l – l1)2 + C

or C = 2gl – g(l – l1) = g(l + l1)

Hence x2 = g(l + l1) – [g/(l1 – l)].(x – l1)

2 (1)

The particle will come to rest when x = 0, so

(x – l1)2 = (l1

2 – l

2)

i.e. x – l1 = (l12 – l

2), which is the answer.

Time for free fall is given by l = ½gt2

i.e. t = (2l/g)

From (1) x2

= [g/(l1 – l)].[l12 – l

2 – (x – l1)

2]

Time from B to the lowest point is given by

[g/(l1 – l)].t

dx

l [l12 – l

2 – (x – l1)

2]

= [sin-1{(x –l1)/(l12 – l

2)] l

= sin-1

1 + sin-1[(l1 – l)/(l1 + l)]

= (/2) + (/2) – cos-1[(l1 – l)/(l1 + l)] = cos

-1[(l1 – l)/ (l1 + l)]

Therefore total time of fall

= (2l/g) + [(l1 – l)/g][ - cos-1(l1 – l)/(l1 + l)}]

Further if the greatest depth below the centre be l cot2(/2) we have

l cot2(/2) = l1 + (l1

2 – l

2)

or l12 – l

2 = l

2 cot

4(/2) + l1

2 – 2ll1cot

2(/2)

So l1 = (l/2).[1 + cot4(/2)]/cot

2(/2)

= l1 + (l1

2 – l

2)

l1 + (l12 – l

2)

..

..

..

..

.

.

.

.

.

121

Hence = [(mgl)/(l1 – l)] = [2mgcot2(/2)]/[1 – cot

2(/2)]

2 = ½ mg tan

2

Also time of fall = (2l/g).(1 + {(l1 – l)/2l}.[- cos-1(l1 – l)/(l1 + l)}])

Now [(l1 – l)/2l] = [cot2(/2) – 1]/[2cot(/2)] = cot

And [(l1 – l)/(l1 + l)] = [{cot2(/2) – 1}/{cot

2(/2) + 1}]

2 = cos

Hence time = (2l/g)[1 + (- )cot]

Ex.4 A particle is attached to the mid-point of a light elastic string of natural length a.

The ends of the string are attached to fixed points A and B, A being at a height 2a

vertically above B, and in equilibrium the particle rests at a depth (5a/4) below A. The

particle is projected vertically downwards from this position with velocity (ga). Prove

that the lower string slackens after a time (/g)(a/12) and that the particle comes to rest

after a further time (a/2g) where is the acute angle defined by the equation tan =

(3/2).

Sol.

In eqn./m position C0, A A A

T0 = mg +

[(5a/4) – (a/2)]/(a/2) T0

= mg + [(3a/4) – (a/2)]/(a/2) T1

giving = mg. C0 C0 T

When the particle has gone x

through x below C0, T0 C y

mx = mg + – T1 1

= mg + [(3a/4) – x – (a/2)]/(a/2)

– [(5a/4) +x – (a/2)]/(a/2)

= - 4 (mg/a).x using the value of. B B B

so x = - (4g/a).x (1)

Solution is x = A cos2(g/a).t + B sin 2(g/a).t

Initially t = 0, x = 0, x = (ga)

So x = (a/2) sin2(g/a).t,

Lower slackens when = 0, so x = (a/4)

Therefore (a/4) = (a/2) sin2(g/a).t, hence t = (/12)(a/g).

Integrating (1) x2 = (g/a).(a

2 – 4x

2) since

when x = 0, x = (ga).

Therefore when x = (a/4), x2 = (3/4).ga.

At the instant the string slackens, its depth = (5a/4) + (a/4) = (3/2).a.

After this instant, let the particle go down through y, then

my = mg – T = mg – [(3/2).a + y – (a/2)].(a/2)

= mg – 2mg.[(a + y)/a] = - mg – 2(mg/a).y

y = - 2(g/a).[y + (a/2)]

Solution is y + (a/2) = A cos (2g/a).t + B sin(2g/a).t

Initially t = 0, y = 0, y = [(3/4).ga]

So A = (a/2), B = [(3/8).a]

Hence y = 0 when tan(2g/a).t = (3/2)

i.e. tan = (3/2) where = (2g/a).t i.e. t = [a/(2g)]

..

..

.

.

.

.

..

..

. .

.

122

Ex.5 A heavy particle is attached to one end of a fine elastic string, the other end of

which is fixed. The unstretched length of the string is a and its modulus of elasticity is n

times the weight of the particle. The particle is pulled vertically downwards till the length

of the string is a and is then let go from rest. Show that the time it returns so this position

is 2[a/(ng)]1/ 2

(- + + tan – tan)where and are positive acute angles given by

sec = (na/a) – n – 1, sec2 = sec

2 – 4n.

Sol.

When the depth of the particle below Q is x, we have

mx = mg – T = mg – nmg.[(x – a)/a]

i.e. x = - (ng/a).[x – {(n + 1)/n}.a] (1)

So x – [(n + 1)/n].a = A cos (ng/a).t + B sin(ng/a).t

When t = 0, x = a and x = 0, giving

x = [(n + 1)/n].a + [a – {(n + 1)/n}.a]cos (ng/a).t

When x = a, we have

a = [(n + 1)/n].a + [a – {(n + 1)/n}.a]cos(ng/a).t

i.e. - 1 = [(na/a) - (n + 1)]cos(ng/a).t (2)

= sec cos(ng/a).t

So cos(ng/a).t = - cos, therefore (ng/a).t = -

t = (a/ng).( - )

Integrating (1) with the condition x = 0 when x = a, we have

x2 = (ng/a).[{a – a(n + 1)/n}

2 – {x – a(n + 1)/n}

2]

So when x = a, x = u say

u2 = (ng/a)[{a – a(n + 1)/n}

2 – (a

2/n

2)] = (ag/n)tan

2

The particle with the velocity u goes against gravity till a height a above O. Then

velocity v is

v2 = u

2 – 4ag = (ag/n) tan

2 – 4ag = (ag/n) tan

2

Time to this point is

t = [(u – v)/g] = (a/ng).(tan – tan) (3)

The motion thereafter is given by

mx = - mg – T = - mg – nmg [(x – a)/a]

So x = - (ng/a).[x – a +(a/n)]

Hence x – a + (a/n) = A cos(ng/a).t + Bsin(ng/a).t

with conditions t = 0, x = a, x = v = (ag/n).tan

x – a + (a/n) = (a/n) cos(ng/a).t + (a/n)tan sin(ng/a).t

Hence x = 0 gives

tan(ng/a).t = tan, so t = (a/ng). (4)

Hence from (2), (3) and (4), we get the answer being twice the sum of three.

Ex.6 A body of mass 5lb. is hung on a light spring and is found to stretched it 6 ins. The

mass is then pulled down a further 2 ins. and released. Find the period of the oscillations

and the kinetic energy of the mass as it passes through its equilibrium position.

Sol.

Since a stretch of 0.5 ft. is due to a force of 5 lb. weight therefore a stretch of x ft.

would correspond to a tension of 10x lb. weight.

..

..

.

.

..

..

.

.

123

In the equilibrium position the weight is balanced by the tension and when the

body is at a distance x ft. below the equilibrium position there is therefore an extra

tension of 10xg poundals acting upwards on the body; and it is this extra unbalanced

tension which causes acceleration, so that

5x = - 10xg,

or x = - 64x

Hence the period = 2/64 = ¼ = 0.785 sec.

Again the velocity is greatest as the body passes through the equilibrium position

and is then amplitude, where 2 = 64 and amplitude = 2 ins. = ½ ft.

Therefore the velocity v = 4/3 f.s., and the kinetic energy

½ mv2 = ½ 5 16/9 absolute units = 40/9 1/32 = 5/36 foot-pound.

Ex.7 A cage of mass M lb. is being pulled up with uniform velocity u by a long steel

cable when the upper end of the cable is suddenly fixed. Having given that a weight of m

lb. would extend the cable 1 ft., shew that the amplitude of the oscillation of the cage is

u(M/mg).

Sol.

If the cage were hanging from the cable at rest, the cable, though extended by the

weight of the cage, would have a definite length, which we may call the equilibrium

length, and such a position of the cage may be called an equilibrium position. When the

upward motion is uniform, since there is no acceleration, the length of the cable will

remain the equilibrium length and the position of the cage relative to the upper end of the

cage will still be the equilibrium position. We assume that any vertical displacement of

the cage which extends or compresses the cable results in an extra tension or thrust

proportional to the extension or compression. Hence, after the fixing of the upper end of

the cable, the motion of the cage becomes a simple harmonic motion about its

equilibrium position, starting from this position with velocity u.

Since a weight of m lb. would extend the cable 1 ft., therefore a displacement of

the cage through x ft. from the equilibrium position would be opposed by an unbalanced

force of mx pounds weight, so that

Mx = - mgx,

or x = - 2x,

where 2 = mg/M

But if a is the amplitude of the oscillation the velocity at the centre of the

harmonic motion is a.

Therefore a = u; but = (mg/M),

so that a = u(M/mg).

Ex.8 A heavy particle is supported in equilibrium by two equal elastic strings with their

other ends attached to two points in a horizontal plane and each inclined at an angle of

600 to the vertical. The modulus of elasticity is such that when the particle is suspended

from any portion of the string its extension is equal to its natural length. The particle is

displaced vertically a small distance and then released. Prove that the period of its small

oscillation is 2(2l/5g), where l is the stretched length of either string in equilibrium.

Sol. Let m be the mass of the particle and

the modulus of elasticity. Then by supposing ½3l

..

..

..

..

124

the particle to be suspended from any portion B A

of the string, since the extended length is double ½ l l

the natural length we find that = mg.

If l0 be the natural length of either string, O

we have, in the equilibrium position, x y

mg = 2[(l – l0)/l0].cos600 = [(l – l0)/l0];

but = mg, therefore l0 = ½ l. P

Let x denote the vertical displacement and y the length of either string at time t.

To find the period of small oscillation we want to obtain an equation of the form

x = - x,

where is a constant. It will therefore be sufficient for our purpose to write down the

equation of motion at time t and neglect all powers of x higher than the first.

We have mx = mg – 2[(y – l0)/l0] cosOPA,

where P is the particle at time t, O is its equilibrium position and PA = PB = y are the

strings.

Now y2 = (x + ½ l)

2 + ¾ l

2 = l

2 + lx + x

2;

Therefore y = l[1 + (x/l)]1/ 2

= l + ½ x,

correct to the first power of x, and

cos OPA = [ ½ l + x]/y = (½ l + x)/(l + ½ x)

= ½ [1 + (2x/l)][1 – x/(2l)]

= ½ [1 + 3x/(2l)],

to the first power of x.

Hence mx = mg – [2(l + ½ x – ½ l)/( ½ l)].[ ½ {1 + 3x/(2l)}],

therefore x = g – g (1 + x/l).[1 + 3x/(2l)],

or x = - 5/2(gx/l);

which represents a simple harmonic motion of period 2(2l/5g).

Ex.9 A warship is firing at a target 3000 yards away dead on the beam, and is rolling

(simple harmonic motion) through an angle of 30 on either side of the vertical in a

complete period of 16 secs. A gun is fired during roll 2 secs. after the ship passes the

vertical. The gun was correctly aimed at the moment of firing, but the shell does not

leaves the barrel till 0.03 sec. later. Shew that the shell will miss the centre of the target

by about 4 feet.

Sol.

Let denote the angle turned through by the ship in t seconds after passing the

vertical. Then the change in is simple harmonic, so that it is connected with t by an

equation

= - n2 (1)

where 2/n = the complete period = 16 secs.

The complete solution of (1) is

= A sin nt + B cos nt,

but vanishes for t = 0, therefore B = 0,and

= A sin nt.

Also A is the amplitude of the oscillation, i.e. an angle of 30 or /60 radians.

The angular velocity is therefore given by

..

..

..

..

..

..

.

125

= nA cos nt;

and 2 secs. after passing the vertical the value of this is

(/8).(/60).cos (/4);

or taking 2 as equal to 10, the angular velocity of the ship at the instant of firing the gun

is (1/482) radians per sec.

It follows that during the 0.03 sec. before the shell leaves the barrel the gun

receives an additional angular elevation = 0.03/482 radians.

Now if V be the velocity and the angular elevation of projection,

(V2 sin2)/g = 9000 feet;

and, if denote the additional range when the elevation is + ,

V2 sin 2( + )/g = 9000 + ;

or V2(sin2 + 2 cos2)/g = 9000 + ,

so that = 18000 cot2.

Also the shell will pass over the centre of the target at a height tan ( + )

approximately; and if we neglect the square of the angular elevation, this is equal to

9000, and substituting the value found for this gives 4 feet as the approximate result.

1.7 Unit Summary:

1. Velocity of a particle has got magnitude as well as direction.

2. The magnitude is called the speed.

3. If both speed and direction remain the same throughout a certain interval the velocity is

uniform throughout that interval. If either of these changes, the velocity becomes

variable.

4. Acceleration also has magnitude as well as direction.

5. Negative acceleration is also known as retardation.

6. Retardation implies decreases in the magnitude of velocity.

7. A particle is to say to execute Simple Harmonic Motion if it moves in a straight line

such that its acceleration is always directed towards a fixed point in the line and is

proportional to the distance of the particle from the fixed point.

8. Thus if a particle describes a circle with constant angular velocity the foot of the

perpendicular from it on any diameter executes a simple harmonic motion.

9. Simple harmonic motion is oscillatory and periodic, the period being independent of

amplitude.

10. The frequency is the number of complete oscillation in the one second, so that if n be

the frequency and T the periodic time, then n = (1/T) = /(2)

1.8 Assignments:

1. If the radial and transverse velocities of a point are always proportional to each other

and this holds for acceleration also, prove that its velocity will vary as some power of the

radial vector.

126

2. The velocities of a particle along and perpendicular to a radius vector from a fixed

origin and r2 and

2. Show that the equation to the path is

(/) = ( / 2r2) + C,

and the components of acceleration are

2r3 –

2 (

4/r) and r

2 +

(

3/r)

3. Prove that the path of a point, which possesses two constant velocities, one along a

fixed direction and the other perpendicular to the radius vector drawn from a fixed point,

is a conic section.

4. A particle moves along a circle r = 2acos in such a way that its acceleration towards

the origin is always zero. Prove that

(d2/dt

2) = - 2 cot.

2

5. A particle is performing a S. H. M. of period T about a centre O and it passes through a

point P (OP = b) with velocity v in the direction OP; prove that the time which elapses

before its return to P is (T/).tan-1

[(vT)/(2b)].

6. A point in a straight line with S. H. M. has velocities v1 and v2 when its distances from

the centre are x1 and x2. Show that the period of motion is 2[(x12 – x2

2)/(v2

2 – v1

2)].

7. A particle P moves in a straight line OCP being attracted by a force m.PC, always

towards C whilst C moves along OC with a constant acceleration f. If initially C was at

rest at the origin O and P was at a distance c from O moving with velocity V, prove that

the distance of P from O at any time t is

[(f /) + c]cos.t + (V/).sin t – (f /) + ½ f t2.

8. A point executes S. H. M. such that in two of its positions the velocities are u, v and

the corresponding accelerations are ; show that the distance between the positions is

(v2 –u

2)/(+

and the amplitude of the motion is [(v2 – u

2)(

2v

2 –

2u

2)]

1/ 2 /(

2 –

2).

9. A body moving in a straight line OAB with S. H. M. has zero velocity when at points

A and B whose distance from O are a and b respectively and has a velocity v when half-

way between them. Show that the complete period is (b – a)/v.

10. The particle of masses m1 and m2 are tied to the ends of an elastic string of natural

length a and modulus . They are placed on a smooth table so that the string is just taut

and m2 is projected with any velocity directly away from m1. Prove that the string will

become slack after the lapse of time [(am1m2)/{(m1 + m2)}].

11. A light elastic string of modulus is stretched to double its length and is tied to two

fixed points distant 2a apart. A particle of mass, m, tied to its middle point, is displaced in

the line of the string through a distance equal to half its distance from the fixed points and

released. Prove that the time of a complete oscillation is [(am)/] and the maximum

velocity is [(a)/m] where is the modulus of elasticity.

12. A particle of mass m resting on a smooth horizontal plane is attached to two fixed

points A and B on the plane by elastic string of unstretched lengths a and b respectively

(a > b), the points A and B being (a + b) apart. The particle is held at B and then released.

Prove that the particle will oscillate to and fro through a distance b[(a + b)/a] in a

periodic time (a + b)(m/), where is the modulus of each string.

13. One end of an elastic string, whose modulus of elasticity is and whose unstretched

length is a, is fixed to a point on a smooth horizontal table and the other end is tied to a

particle of mass m which is lying on the table. The particle is pulled to a distance where

.

127

the extension of the string is b and then let go; shew that the time of a complete

oscillation is 2[+ (2a/b)(am/).

14. An endless cord consists of two portions of lengths 2l and 2l respectively, knotted

together, their masses per unit of length being m and m. It is placed in stable equilibrium

over a small smooth peg and then slightly displayed. Shew that the time of a complete

oscillation is 2[(ml + ml)/{(m – m)g}].

15. A smooth light pulley is suspended from a fixed point by a string of natural length l

and modulus of elasticity ng. If masses m1 and m2 hang at the ends of a light inextensible

string passing round the pulley. Show that the pulley executes S. H. M. about a centre

whose depth below the point of suspension is

l.[1 + (2M/n)]

where M is the harmonic mean between m1 and m2.

16. One end of an elastic string is fixed and to the other end is fastened a particle heavy

enough to stretch the string to double its natural length a. The string is drawn vertically

down till it is four times its natural length and then let go. Show that the particle returns

to this point in time

(a/g).[ + (4/3)].

1.9 References

1. Loney, S.L.: An Elementary Treatise on the Dynamics of a Particle and of Rigid

Bodies, S.Chand & Company (Pvt.) Ltd, Ram Nagar, New Delhi,

1952.

2. Lamb, Horace: Dynamics: Cambridge University Press, 1961

3. Ramsey, A.S.: Dynamics Part I & II, : Cambridge University Press,1954

UNIT –IV Smooth and Rough plane Curves, Resting

Medium and Particle of Varying Mass

STRUCTURE

2.1 Introduction

2.2 Objectives

2.3 Motion on Smooth and Rough plane Curves

2.4 Motion in a Resting Medium

128

2.5 Motion of Particle of Varying Mass

2.6 Unit Summary

2.7 Assignments

2.8 References

2.1 Introduction:

This unit introduces the basics of Motion on Smooth and Rough plane Curves,

Motion in a Resting Medium, Motion of Particle of Varying Mass. It also help us to

understand the basic concepts of above topics.

In this unit we shall study the basic ideas of Motion on Smooth and Rough plane

Curves, Motion in a Resting Medium, Motion of Particle of Varying Mass. It is hopped

the unit help students in studying.

2.2 Objectives:

At the end of the unit the students would be able to understand the concept of:

Motion on Smooth Plane Curve

Motion on Rough Plane Curve

129

Motion in Resisting Medium

Particle falls under Gravity

Particle Projected Upwards

Particle Projected under Gravity

Particle Moving under Gravity

Bead Moves on a Smooth Wire

2.3 Motion on Smooth and Rough plane Curves

Motion on Smooth Plane Curve:

Theorem:

A heavy particle is made to move on a smooth curve in a vertical plane; Discuss

the motion.

Proof:

The external forces are the weight mg Y

of the particle downward and the normal

reaction R of the curve. If P be the particle at

time t, the equations of motion (tangential and R

normal) are

mv(dv/ds) = - mg sin A P

m(v2/) = R – mg cos

if s be measured from a fixed point A down the mg

curve below P, the radius of curvature of the O X

130

curve at P. Since sin = (dy/ds) we have from the first equation

mv (dv/ds) = - mg (dy/ds)

Integrating ½ mv2 = - mgy + C, where C is a constant. If the particle be projected

from a point where y = y1 with a velocity u then

½ mu2

= - mgy1 + C

Hence ½ mv2 – ½ mu

2 = - mg(y – y1) = -mgh

where h is the height of P above the point of projection. This is in fact the equation of

energy and follows at one from the principle of energy namely the change in kinetic

energy is equal to the work done.

Thus for upwards projection

v2 = u

2 – 2gh

and for downward projection v2 = u

2 + 2g where h is the vertical distance between the

two points. These equations hold good whatever be the paths followed. The second

equation gives R, the reaction on the particle. The particle will leave the curve when R=0.

Ex.1 A particle slides down the smooth curve y = a sinh(x/a), the axis of x being

horizontal and the axis of y downwards, starting from rest at the point where the tangent

is inclined at a to the horizon; shew that it will leave the curve when it has fallen through

a vertical distance a sec.

Sol. Curve is y = a sinh(x/a)

The equations of motion are

m.(d2s/dt

2) = mv(dv/ds) = mg sin

and m(v2/) = mg cos – R,

i.e. v(dv/ds) = g sin, (1)

and (v2/) = - (R/m) + g cos. (2)

Now y = a sinh(x/a), so (dy/dx) = cosh(x/a) = tan,

i.e. tan = cosh(x/a) = [1 + sinh2(x/a)]

= [1 + (y2/a

2)] = [(a

2 + y

2)/a

2].

Also (d2y/dx

2) = (1/a).sinh(x/a) = y/a

2, so that

[1 + (dy/dx)2]3/2

[1 + cosh2(x/a)]

3/2 (2a

2 + y

2)3/2

(d2y/dx

2) (y/a

2) ay

At = , y = b, tan = [1 + (b2/a

2)].

From (1) v(dv/ds) = g sin = g.(dy/ds), so 2vdv = 2gdy

Integrating, we get v2 = 2gy + c

Initially when v = 0, y = b, then c = - 2gb,

so that v2 = 2g.(y – b). (3)

The particle leaves the curve when R = 0.

Therefore from (2) v2

= g cos. (4)

From (3) and (4)

2g.(y – b) = g cos = g/(1 + tan2)

g[1 + cosh2(x/a)]

3/2(a

2/y) g[2 + sinh

2(x/a)]

3/2(a

2/y)

(2 + y2/a

2) (2 + y

2/a

2)

= g.( 2 + y2/a

2).(a

2/y)

or 2(y – b) = (2a2/y) + y

or y2 – 2yb + b

2 = 2a

2 + b

2, i.e. (y – b)

2 = b

2 + 2a

2

= = =

= 2g(y – b)=

131

or (y – b)2 = a

2 + a

2tan

2 = a

2sec

2

as tan2 = 1 + (b

2/a

2), so b

2 = a

2 tan

2 – a

2

so y – b = a sec.

Ex.2 A particle descends a smooth curve under the action of gravity, describing equal

vertical distances in equal times, and starting in a vertical direction. Shew that the curve

is a semi-cubical parabola, the tangent at the cusp of which is vertical.

Sol.

Let the axis of x be horizontal and axis of y vertically downwards. „Describing

equal vertical distances in equal times‟ means velocity, i.e. (dy/dt) is constant, say k.

i.e. (dy/dt) = k (1)

Initially the particle starts in a vertical direction with velocity k.

Therefore v2 = k

2 + 2gy; [x = 0; u = y = k],

i.e. (dx/dt)2 + (dy/dt)

2 = k

2 + 2gy; so (dx/dt) = (2gy) from (1)

Now (dx/dy) = [(dx/dt)/(dy/dt)] = (2gy)/k

or dx = [(2gy)/k]dy, integrating x = [(2g)/k].(2/3).(y)3/2

+ C

Initially when x = 0, y = 0; so C = 0.

Hence x = (2/3).[(2g)/k].y3/2

which represents a semi cubical parabola whose tangent at the cusp is vertical.

Ex.3 A particle is projected with velocity V from the cusp of a smooth inverted cycloid

down the arc; shew that the time of reaching the vertex is 2(a/g)tan-1

[(4ag)/V].

Sol.

Equation of motion is

(d2s/dt

2) = g.(s/4a) or [D

2 + g/(4a)].s = 0, i.e. D = i(g/4a).

Its solution is s = A cos(g/4a).t + B sin(g/4a).t.

Initially when t = 0, s = 4a, (ds/dt) = - V; so A = 4a

and as (ds/dt) = - A(g/4a)sin(g/4a).t + B(g/4a) cos(g/4a).t

so - V = B(g/4a); i.e. B = - V(4a/g).

Hence s = 4a cos(g/4a).t – V(4a/g) sin(g/4a).t

When it comes to the vertex, s = 0.

So tan(g/4a).t = [4a/V(4a/g)] = (4ag)/V

Therefore t = 2(a/g) tan-1[(4ag)/V].

Ex.4 A particle slides down the arc of a smooth cycloid whose axis is vertical and lowest;

prove that the time occupied in falling down the first half of the vertical height is equal to

the time of falling down the second half.

Sol.

Let the equation of cycloid be s = 4asin, where s is measured from the vertex.

Also s = (8ay).

At t, let P be the position of the particle. Let P(x, y); at cusp A, s = 4a.

So vertical height of A = 2a.

Equation of motion is

m(d2s/dt

2) = - mg sin Y

or (d2s/dt

2) = - (g/4a).s A

132

[as s = 4a sin. A

Integrating (ds/dt)2 = - (g/4a).s

2 + C.

R

Initially when s = 4a.(ds/dt) = 0; P

so C = (2/4a).16a2 = 4ag. mg

Hence (ds/dt)2 = (g/4a).(16a

2 – s

2); O X

so (ds/dt) = - (g/4a).(16a2 – s

2)

- ve sign being taken as the particle is moving in direction of s decreasing.

Therefore t = - 2(a/g) [1/(16a2 – s

2)].ds + D = 2(a/g) cos

-1(s/4a) + D

Initially when s = 4a, t = 0; so D = 0.

So that t = 2(a/g)cos-1

[(8ay)/4a].

If t1, be the time taken in falling first half of the vertical distance and t2 that in

falling second half of vertical distance. a

So t1 = 2(a/g).[cos-1

{(8ay)/4a}] = 2(a/g)[/4 – 0] = 2(a/g).(/4)

2a

and t2 = 2(a/g).[cos-1

{(8ay)/4a}] = 2(a/g)[/2 – /4] = 2(a/g).(/4) = t1.

Ex.5 A particle is placed very close to the vertex of a smooth cycloid whose axis is

vertical and vertex upwards and is allowed to run down the curve. Shew that it leaves the

curve when it is moving in a direction making with the horizontal an angle of 450.

Sol.

Let at any time t the, the particle be at P. So the equations of motion are

m(d2s/dt

2) = mg sin (1)

and m(v2/) = mg cos - R (2)

(1) may be written as

v(dv/ds) = g(s/4a) as s = 4a sin.

Integrating, v2 = g(s

2/4a) + A.

Initially when s = 0, v = 0, A = 0, so that v2 = g(s

2/4a) (3)

From (2) and (3),

R = mg cos – [(mgs2)/(4a.4a cos)] as = (ds/d) = 4a cos

or R = (mg/cos).(cos2 – sin

2) = (mg/cos).cos2.

The particle will leave the curve when R = 0, i.e. when

cos2 = 0 or 2 = /2; so = /4.

O T X

R

P

mg ms

Motion on Rough Plane Curve:

a

0

133

Let s be the arc measured from a fixed point A on the curve, in the direction of

motion, to the position of the particle P whose mass is m. Let R be the pressure measured

inwards, the coefficient of friction and T, N the components of the external force on the

particle along the tangent and the outward normal. Then the equations of motion are

mv(dv/ds) = T – R (1)

and mv2/ = R – N (2)

In the case in which a heavy particle is sliding upwards on a curve in a vertical

plane, the equations are

mv.(dv/ds) = - R - mg sin (3)

and mv2/ = R - mg cos (4)

where is the inclination of the tangent to the horizontal.

Eliminating R, we get

v(dv/ds) + (v2/ = - g(sincos),

or dv2/d + 2v

2 = - 2g (sincos), (5)

From the intrinsic equation of the curve, say s = f(), we get = f and the

equation (5) has an integrating factor e2

.

y mv2/ y T

mv(dv/ds) R

P P

A s A R N

O x

Note:

In applying the method of this article it should be noted that the friction must

always oppose the motion and that in the case of a two-sided constraint (i.e. bead on a

wire), if R changes sign during the motion, the motion will cease to be represented by

equation (5), because it will become necessary to change the sign of R in (4) but in (3).

Ex.1 The base of a rough cycloidal arc is horizontal and its vertex downwards; a bead

slides along it starting from rest at the cusp and coming to rest at the vertex. Show that

2e

= 1.

Sol.

The intrinsic equation of cycloid is

s = 4a sin;

So = ds/d = 4a cos

The equations of motion are

mv.(dv/ds) = R - mg sin

i.e. ½ m.(dv2/ds) = R - mg sin (1)

and (mv2)/ = R - mg cos (2)

Eliminating R between (1) and (2), we get

dv2/d - 2v

2 = 2g(cos - sin)

= 8ga cos(cos - sin), since = 4a cos

134

The equation being linear in v2, the I.F. = e

– 2 d = e

– 2

and therefore the solution is

v2e

– 28ga e

– 2cos(cos - sin) d + A

ga e– 2

cos2) – sin2] d + A

ga - ½ e– 2

e– 2

sin2) + 2(1 – 2)cos2 + A

4 + 42

At the cusp when = /2, v = 0.

So A = - 4ga [- ½ e–

+ e–

{- 2 (1 – 2 )}/(4 + 4

2)].

Also at the vertex when = 0, v = 0.

So A = - 4ag [ - ½ + {2 (1 – 2 )}/(

4 + 4

2)].

Elimination of A between the last two equation gives

e–

2 (1 – 2 ) 2 (1 –

2 )

4 + 42 4 + 4

2

or e–

2

.

1 + 2

1 + 2

or 2e

= 1.

Ex.2 A particle slides in a vertical plane down a rough cycloidal arc whose axis is

vertical and vertex downwards, starting from a point where the tangent makes an angle

with the horizon and coming to rest at the vertex. Shew that e

= sin – cos.

Sol.

The intrinsic equation of cycloid is

s = 4a sin;

So = ds/d = 4a cos

The equations of motion are

mv.(dv/ds) = R - mg sin

i.e. ½ m.(dv2/ds) = R - mg sin (1)

and (mv2)/ = R - mg cos (2)

Eliminating R between (1) and (2), we get

dv2/d - 2v

2 = 2g(cos - sin)

= 8ga cos(cos - sin), since = 4a cos

The equation being linear in v2, the I.F. = e

– 2 d = e

– 2

and therefore the solution is

v2e

– 28ga e

– 2cos(cos - sin) d + A

ga e– 2

cos2) – sin2] d + A

ga [- ½ e– 2

e– 2

sin2) + 2(1 – 2)cos2] + A

4 + 42

At the cusp when = , v = 0.

So A = - 4ga [- ½ e– 2

+ e– 2

sin2) + 2(1 – 2)cos24 + 4

2)]

Also at the vertex when = 0, v = 0.

So A = - 4ag [ - ½ + {2 (1 – 2 )}/(

4 + 4

2)].

Elimination of A between the last two equations gives

e– 2

sin2) + 2(1 – 2)cos2 2 (1 –

2 )

4 + 42

4 + 42

- ½ e–

- = - ½ +

=

- ½ e– 2

- = - ½ +

135

so that e

= sin – cos.

Ex.3 A bead moves along a rough curved wire, which is such that it changes its direction

of motion with constant angular velocity. Shew that a possible form of the wire is an

equiangular spiral.

Sol.

The equations of motion are

mv(dv/ds) = - R (1)

and mv2/ = R (2)

Now we are given that

d/dt = (constant).

So v = ds/dt = (ds/d).(d/dt) = i.e. = v/

Thus from (2), v = R/m and therefore from (1), we have

v(dv/ds) = - v or dv = - ds.

Integrating v = A - s.

so that v = = (ds/d) = A - s.

or ds

A - s

Integrating further

log (A - s) =

i.e. (A - s) = e

or s = (A e

)/() = a + b e

(say)

which represents an equiangular spiral.

Ex.4 A rough cycloid has its plane vertical and the line joining its cusps horizontal. A

heavy particle slides down the curve from rest at a cusp and comes to rest again at the

point on the other side of the vertex where the tangent is inclined at 450

to the vertical.

Shew that the coefficient of friction satisfies the equation 3 + 4 log (1 + ) = 2 log2

Sol.

The intrinsic equation of cycloid is

s = 4a sin;

So = ds/d = 4a cos

The equations of motion are

mv.(dv/ds) = R - mg sin

i.e. ½ m.(dv2/ds) = R - mg sin (1)

and (mv2)/ = R - mg cos (2)

Eliminating R between (1) and (2), we get

dv2/d - 2v

2 = 2g(cos - sin)

= 8ga cos(cos - sin), since = 4a cos

The equation being linear in v2, the I.F. = e

– 2 d = e

– 2

and therefore the solution is

v2e

– 28ga e

– 2cos(cos - sin) d + A

ga e– 2

cos2) – sin2] d + A

ga [- ½ e– 2

e– 2

sin2) + 2(1 – 2)cos2] + A

= d

136

4 + 42

At the cusp when = /2, v = 0.

So A = - 4ga [- ½ e–

+ e–

{- 2 (1 – 2 )}/( 4 + 42)].

Also when = - /4, v = 0.

So A = - 4ag e/ 2

[ - ½ - (4)/( 4 + 4

2)].

Elimination of A between the last two equations gives

e–

2 (1 – 2 ) 4 e

/ 2

4 + 42 4 + 4

2

or 2 = (1 + 2 + 2) e

3/ 2 = (1 + )

2e

3/ 2

Taking logarithm both sides, we get

log 2 = 2 log (1 + ) + (3)/2

or 3 + 4 log (1 + ) = 2 log2

Ex.5 A particle is projected along the inner surface of a rough sphere and is acted on by

no forces; show that it will return to the point of projection at the end of time

a(e2

– 1)/(V)

where a is the radius of the sphere, V is the velocity of projection and is the coefficient

of friction.

Sol.

Since there is no forces, the particle will continue to move in the plane section,

i.e. a circle in which it starts.

The equations of motion are

mv(dv/ds) = - R (1)

and mv2/a = R (2)

Eliminating of R between (1) and (2) gives

1 dv2 v

2

2 ds a

or 1 dv2 d v

2

2 d ds a

or 1 dv2 v

2

a d a [since s = a, so that d/ds = 1/a]

or dv2

d (3)

The equation being linear in v2, its integrating factor = e

2 d = e

2

Thus the solution of the differential equation (3) is

v2 e

2 = A, where A is the some constant.

Initially when = 0, v = V; so A = V2.

Hence v2 e

2 = V

2 or v e

= V

or a d

dt [since v = ds/dt = a (d/dt) from s = a]

So a 2

V 0

a 2

- ½ e–

- = - ½ e/ 2

-

+ = 0

+ = 0

+ 2 = 0

+ 2v2 = 0.

e

= V,

e

d = t

[e

] =

137

V 0

a

V

Ex.6 A bead of mass m slides on a rough circular wire of radius a in a vertical plane, the

coefficient of friction being 1/2. The bead is projected from the lowest point; show that

in order that it may just reach the highest point the velocity of projection must be

[(2/3)6gae2(/ 2 + )

],

where is the acute angle tan –12.

Sol.

If v is the velocity when is the angular distance of the bead from its lowest

position, the equations of motion, in the early stages of the motion, are

mv.(dv/ds) = - mg sinR (1)

and mv2/a = R - mg cos (2)

leading to

dv2/d + 2v

2 = - 2ga(sin cos), (3)

which gives, on integration,

3 sin + (22 – 1) cos

42 + 1 (4)

For = 1/2, this becomes

v2e2

= v02 – 2gae

2 sin (5)

where v0 is the value of v when

According to the data v decreases steadily and vanishes when so there will

be an angle between 0 and for which cosv2/(ag) and from (2) for this value of

the pressure R vanishes and changes sign. From this point onwards (3) ceases to

represent the motion, because (1) still holds good, but instead of (2) we have

mv2/a = R – mg cos (6)

which with (1) leads to

dv2/d – 2v

2 = - 2ga(sin – cos), (7)

giving, on integration,

– 3 sin + (22 – 1) cos

42 + 1 (8)

For = 1/2, this becomes

v2e

–2 = C + 2gae

–2 sin

But v is to vanish when , therefore C = 0 and in the later stages of the

motion v2 = 2ga sin (9)

But the motion being continuous the values of v2 given by (5) and (9) are the

same at the point at which R vanishes, i.e. when v2 is also – ga cos. Substituting this

value in (9) gives

cos = 2 sin,

or cos sin 1.

- 2 1 3

so that = /2 + , where is the acute angle tan–12. Then from (5) and (9) we get

v02 = 22gae

2 sin

[e2

- 1] =

v2e

2 = C – 2gae

2

v2e

–2 = C – 2gae

–2

= =

138

= (2/3) 6gae2(/ 2 +

2.4 Motion in Resisting Medium:

Particle falls under Gravity:

Theorem: A particle falls under gravity (supposed constant) in a resisting medium whose

resistance varies as the square of the velocity. Find the motion if the particle starts from

rest.

Proof:

Let v be the velocity when the particle has fallen a distance x in time t from rest.

The equation of motion is

d2x/dt

2 = g – v

2.

Let = (g/k2), so that d

2x/dt

2 = g (1 – v

2/k

2) (1)

From (1) it follows that if v equaled k, the acceleration be zero; the motion would

then be unresisted and the velocity of the particle would continue to be k. For this reason

k is called the „terminal velocity‟.

From (1) v (dv/dx) = g (1 – v2/k

2),

so that (2g/k2).x = [2v/(k

2 – v

2)].dv = - log(k

2 – v

2) + A.

Since v and x are both zero initially, so A = log k2.

Therefore k2 – v

2 = k

2exp(-2gx/k

2).

Hence v2 = k

2[1 – exp(– 2gx/k

2)] (2)

It follows that x = when v = k. Hence the particle would not actually acquire the

„terminal velocity‟ until it had fallen an infinite distance.

Again (1) can be written

dv/dt = g(1 – v2/k

2).

So gt/k2 = [1/(k

2 – v

2)] dv = (1/2k) log [(k + v)/(k –v)] + B.

Since v and t were zero initially, so B = 0.

Hence [(k + v)/(k – v)] = e2gt/ k

So v = k.[(e2gt/ k

– 1) /(e2gt/ k

+ 1)] = k tanh(gt/k) (3)

From (2) and (3), we have

exp(- 2gx/k2) = 1 – v

2/k

2 = 1 – tanh

2(gt/k) = 1/cosh

2(gt/k)

Therefore exp(gx/k2) = cosh(gt/k) and x = (k

2/g).log cosh(gt/k) (4)

Particle Projected Upwards:

Theorem: If the particle were projected upwards instead of downwards. Find the motion.

Proof:

Let V be the velocity of projection.

The equation of motion now is

d2x/dt

2 = - g – v

2 = - g(1 + v

2/k

2) (1)

where x is measured upwards.

139

Hence v(dv/dx) = - g(1 + v2/k

2).

So 2g/k2 = - [2v/(v

2 + k

2)]dv = - log(v

2 + k

2) + A,

where 0 = - log(V2 + k

2) + A.

Therefore 2gx/k2 = log[(V

2 + k

2)/(v

2 + k

2)]. (2)

Again (1) gives

dv/dt = - g(1 + v2/k

2).

So - gt/k2 = [1/(k

2 + v

2)].dv = (1/k).tan

-1(v/k) + B,

where 0 = (1/k).tan-1

(V/k) + B.

Hence gt/k = tan-1

(V/k) – tan-1

(v/k) (3)

Equation (2) gives the velocity when the particle has described any distance and

(3) gives the velocity at the end of any time.

Ex.1 A heavy particle is projected vertically upwards with velocity u in a medium, the

resistance of which is gu2

tan2a times the square of the velocity, a being a constant.

Shew that the particle will return to the point of projection with velocity u cosa, after a

time

ug1

cota[a + log{cosa/(1 – sina)}].

Sol.

Resistance = gu2

tan2a.v

2.

Consider the motion of the rising particle, the equation of motion is

v(dv/dx) = - g – gu-2

tan2a.v

2

or x = - (u2/g) [v/(u

2 + v

2tan

2a)].dv

Integrating the greatest height attained, will be given by

x = - [u2/(2gtan

2a)].[log(u

2 + v

2tan

2a)] = [(u

2cot

2a)/(2g)] log sec

2a (1)

When the particle falls, the equation of motion is

v(dv/dx) = g – v2gu

2tan

2a = (g/u

2).[u

2 – v

2tan

2a].

The velocity attained by the particle in falling through the distance given by (1),

will be obtained after integration as follows

[v/(u2 – v

2tan

2a)]dv = (g/u

2) dx

i.e. ½ cot2a.[log(u

2 – v

2tan

2a)] = (g/u

2).[x]

or ½ cot2a.log[(u

2 – v

2tan

2a)/u

2] = (g/u

2).[(u

2cot

2a)/2g].log sec

2a

or [(u2 – v

2tan

2a)/u

2] = cos

2a

or v2tan

2a = u

2(1 – cos

2a) = u

2sin

2a.

or v2 = u

2 cos

2a.

which gives the velocity attained v = u cosa i,e, the particle reaches the point of

projection with velocity u cosa.

Now to find the time, we have

dv. d2x g

dt dt2 u

2 for upward motion.

0

u

0

u

0

v (u2/2g) cot

2a log sec

2a

0

0

v

0

(u2/2g) cot

2a log sec

2a

= = – (u2 + v

2 tan

2a)

140

gt 0 dv

u2 u

u2 + v

2 tan

2a

1 v tana 0 – a

u tana u u u tana

i.e. the time taken in reaching the greatest height say t1 is given by t1 = (au cota)/g.

Again to find the time of falling to the point of projection we have

dv/dt = (g/u2)[u

2 – v

2 tan

2a].

So the time from the greatest height to the point of projection, say t2 is given by

u2

u cosa dv

g 0

u2 v

2 tan

2a

u cota cosa

g

1 sina

Hence the total time taken by the particle in returning to the point of projection is

u cota cosa au cota

g

1 sina g

= ug1

cota[a + log{cosa/(1 – sina)}].

Ex.2 A particle of mass m is projected vertically under gravity, the resistance of the air

being mk times the velocity; shew that the greatest height attained by the particle is

(V2/g).[ - log(1 + )], where V is the terminal velocity of the particle and V is its initial

vertical velocity.

Sol.

Resistance = mkv

The equation of motion for falling particle is

m(d2x/dt

2) = mg – mkv

In case the particle falls under gravity,

(d2x/dt

2) = 0.

So 0 = mg – mkv or v = g/k = V (given as terminal velocity)

Thus V = g/k (1)

When the particle is projected upwards, the equation of motion is

m(d2x/dt

2) = mg – mkv

or d2x/dt

2 = – g – (gv/V) from (1)

or v(dv/dx) = – (g/V).(V + v)

So – (g/V).dx = [v/(V + v)].dv = [1 – V/(V + v)].dv

Integrating, (– g /V).x = v – V log(V + v) + A

Initially when x = 0, v = V; A = V log(V + V) – V.

Thus we have (- g/V).x = v – V log(V + v) + V log(V + V) – V.

The particle will be at the greatest height when v = 0, therefore putting v = 0 in

the above equation, we have

(- g/V).x = V + Vlog(1 + )

i.e. x = (V2/g).[ log(1 + )]

Ex.3 A person falls by means of a parachute from a height of 800 yards in 2½ minutes.

Assuming the resistance to vary as the square of the velocity, shew that in a second and a

half his velocity differs by less than one percent from its value when he reaches the

ground and find an approximate value for the limiting velocity.

=

= tan– 1

=

t2 =

log =

logt1 + t2 = +

141

Sol.

When the parachute has fallen a space x in time t, we have, if

= g/k2,

v2 = k

2[1 – exp(– 2gx/k

2)] (1)

v = k tanh(gt/k) (2)

and x = (k2/g)log cosh(gt/k) (3)

Here 2400(g/k2) = log cosh(150g/k)

So exp(2400g/k2) = [e

150g / k + e

– 150g / k]/2 (4)

The second term on the right hand is very small, since k is positive.

Hence (4) is approximate equivalent to exp(2400g/k2) = ½ e

150g / k.

So 2400g/k2 = 150g/k – log2 = 150g/k, nearly.

Hence k = 16 is a first approximation.

Putting k = 16(1 + y), (4) gives, for a second approximation,

300(1 – 2y) = ½ [e300(1 – y)

+ e– 300 (1 – y)

] = ½ e300 (1 – y)

, very approx.

So e-300 y

= ½

Therefore y = (1/300)log2 = 0.693/300 = 0.0023.

Hence a second approx. is k = 16(1 + 0.0023), giving the terminal velocity.

Also the velocity v1, when the particle reaches the ground, is, by (1), given by

v12 = k

2[1 – exp(- 2.32 2400/16

2)] = k

2[1 – e

6 0 0] = k

2, for all practical

purpose.

When v is 99% of the terminal velocity, (2) gives

tanh (gt/k) = 99/100 = 0.99.

So e2gt/ k

= 199 = e5 . 2

, from the tables.

Therefore t = k/(2g) 5.3 = (16/64) 5.3 = 1.235 approx., i.e. t is less than 1½ secs.

Ex.4 A particle of mass m is falling under the influence of gravity through a medium

whose resistance equals times the velocity. If the particle be released from rest, show

that the distance faller through in time t is g(m2/

2).[e

(t/ m) – 1 +(t/m)].

Sol.

Resistance = v for the mass m.

The equation of motion is

m(d2x/dt

2) = mg – v

or d2x/dt

2 = g – (/m).(dx/dt), so v = dx/dt

i.e. d2x/dt

2 + (/m).(dx/dt) = g

This is a linear differential equation in dx/dt and hence its

I.F. = e(/ m)dt

= et/ m

.

Therefore the solution is

(dx/dt).et/m

= g et/ m

dt + A = (mg/).et/ m

+ A

Initially when t = 0, dx/dt = 0. So A = - mg/

so that et/ m

.(dx/dt) = (mg/).[et/m

– 1]

or dx = (mg/).[1 – e– t/ m

]dt.

Integrating, x = (mg/).[t + (m/).e– t/ m

] + B.

Initially when t = 0, x = 0; so B = - m2g/

2.

Hence x = g.(m2/

2).[e

– t/ m – 1 + t/m].

142

Ex.5 If the resistance vary as the fourth power of the velocity, the energy of m lbs. At a

depth x below the highest point when moving in a vertical line under gravity will be

E tan(mgx/E) when rising and E tanh(mgx/E) when falling, where E is the terminal

energy in the medium.

Sol.

Resistance = v4.

When the particle is falling, equation of motion is

d2x/dt

2 = g – v

4.

The acceleration is zero if g – v4 = 0

i.e. v = (g/)1/4

= K the terminal velocity (say). So = g /K4.

If K is the terminal velocity then terminal energy E = ½ mK2.

Now the equation of motion when the particle is rising, is

d2y/dt

2 = - g – (g/K

4).v

4, i.e. vdv/dy = - (g/K

4).[K

4 + v

4]

or [2v/(K4 + v

4)].dv = - (2g/K

4)dy.

Integrating, (1/K2) tan

– 1(v

2/K

2) = - (2gy/K

4) + A.

Initially when y = 0, let v = u; so A = (1/K2).tan

– 1(u

2/K

2).

Thus 2gy = K2 tan

– 1(u

2/K

2) – K

2 tan

– 1(v

2/K

2). (1)

This gives the velocity of the particle at a height y from the ground.

If h be the greatest height to which the particle rises then putting v = 0 in (1), we

get

2gh = K2

tan– 1

(u2/K

2). (2)

The height which is x feet below the heightest point is

= h – x = (K2/2g).tan

– 1(u

2/K

2) – x.

The velocity of the particle at this height is obtained by substituting

(K2/2g). tan

– 1(u

2/K

2) – x for y in (1), whence we get

2g[(K2/2g).tan

– 1(u

2/K

2) – x] = K

2 tan

– 1(u

2/K

2) – K

2 tan

– 1(v

2/K

2).

or 2gx = K2 tan

– 1(v

2/K

2), i.e. v

2 = K

2 tan (2gx/K

2).

Putting K2 = 2E/m, the energy at this point is given by

½ mv2 = E tan (mgx/E).

Further when the particle is falling the equation of motion is

v(dv/dx) = (g/K4).(K

4 – v

4), i.e., [v/(K

4 – v

4)].dv = (g/K

4).dx

or (1/2K2).[{v/(K

2 + v

2)} + {v/(K

2 – v

2)}] dv = (g/K

4)dx.

Integrating, (1/4K2).log[(K

2 + v

2)/(K

2 – v

2)] = (g/K

2).x + B.

Initially when x = 0, v = 0, so B = 0.

Thus log[(K2 + v

2)/(K

2 – v

2)] = 4gx/K

2,

i.e. (K2 + v

2)/(K

2 – v

2) = exp(4gx/K

2).

Applying componendo and dividendo,

v2/K

2 = [exp(4gx/K

2) – 1]/[exp(4gx/K

2) + 1]

= [exp(2gx/K2) – exp(– 2gx/K

2)]/[exp(2gx/K

2) + exp(– 2gx/K

2)]

= tanh (2gx/K2), so that v

2 = K

2 tanh (2gx/K

2).

Putting K2 = 2E/m, the energy at the point is

½ mv2 = E tanh (mgx/E).

Ex.6 A heavy particle is projected vertically upwards in a medium the resistance of

which varies as the square of the velocity. It has a kinetic energy K in its upwards path at

a given point; when it passes the same point on the way down, show that its loss of

143

energy is [K2/(K + K)], where K is the limit to which its energy approaches in its

downwards course.

Sol.

Resistance = v2.

For the falling particle, equation of motion is

d2x/dt

2 = g – v

2.

Acceleration is zero if g – v2 = 0

i.e. v = (g/) = k the terminal velocity; so = g/k2.

If k be terminal velocity, then terminal energy = ½ mk2 = K (given).

When the particle is going upwards, equation of motion is

v(dv/dx) = - (g/k2).(k

2 + v

2), since d

2x/dt

2 = - g – v

2

or [2v/(k2 + v

2)]dv = - 2g/k

2.

Integrating log(k2 + v

2) = - (2g/k

2).x + A.

Initially when x = 0, let v = V; so A = log(k2 + V

2).

Thus 2gx/k2 = log[(k

2 + V

2)/(k

2 + v

2)], (1)

This gives velocity of the particle at a certain height x. If h be the greatest height

attained by the particle, then by putting v = 0 in (1), we have

2gh/k2 = log[(k

2 + V

2)/k

2] (2)

When the particle falls, the equation of motion is

v(dv/dy) = (g/k2).(k

2 – v

2) or – [2v/(k

2 – v

2)].dv = - (2g/k

2)dy.

Integrating log(k2 – v

2) = - (2gy/k

2) + B.

Initially when y = 0, v = 0; so B = logk2.

Thus 2gy/k2 = log[k

2/(k

2 – v

2)] (3)

This gives velocity of particle after falling a distance y. If V2 be the velocity of

the falling particle when distant x1, from the ground i.e. (h – x1) from the highest point,

therefore putting y = h – x1 in (3), we get

(2g/k2).(h – x1) = log[k

2/(k

2 – V2

2)] (4)

If V1 be the velocity of the rising particle at the same height x1 then from (1), we

get

(2g/k2).x1 = log[(k

2 + V

2)/(k

2 + V1

2)] (5)

Adding (4) and (5), we get

2gh/k2 = log[{k

2(k

2 + V

2)}/{(k

2 + V1

2).(k

2 – V2

2)}] (6)

Comparing (2) and (6), we have

log[(k2 + V

2)/k

2] = log[{k

2(k

2 + V

2)}/{(k

2 + V1

2).(k

2 – V2

2)}]

i.e. (k2 + V

2)/k

2 = k

2(k

2 + V

2)/[(k

2 + V1

2).(k

2 – V2

2)]

or - k2V2

2 – V1

2V2

2 + k

2V1

2 = 0, i.e. V2

2 = k

2V1

2/(k

2 + V1

2)

Kinetic energy at P when passing down

= ½ mV22 = ½ (mk

2.V1

2)/(k

2 + V1

2) = (½ mk

2. ½ mV1

2)/( ½ mk

2 + ½ mV1

2)

= (K.K)/(K + K)

and kinetic energy at P when going up

= ½ mV12 = K.

Loss of energy = V – VV/(V + V) = V2/(V + V)

Ex.7 A particle is projected in a resisting medium whose resistance varies as (velocity)n

and it comes to rest after describing a distance s in time t. Find the values of s and t and

144

show that s is finite if n < 2, but infinite if n = or > 2, whilst t is finite if n < 1, but infinite

if n = or > 1.

Sol.

Resistance = vn

The equation of motion is

d2x/dt

2 = v(dv/dx) = - v

n or dx = - dv/v

n – 1

so that x = - (1/vn – 1

)dv = (1/n – 2).(1/vn – 2

) + A

Initially when x = 0, let v = V; so A = – (1/n – 2).(1/Vn – 2

)

Thus x = (1/n – 2).[(1/vn – 2

) – (1/Vn – 2

)]

If n > 2 and v = 0, x = .

If n = 2, then x = - (1/v) dv = log(V/v)

Constant being logV as v = V, when x = 0, so that x = when v = 0.

If n < 2 let n = 2 – , then

x = - (1/v1 –

)dv = (1/).(V – v

) as v = V when x = 0.

Hence s = [x] (at v = 0) = (1/)V = a finite quantity.

Again dv/dt = - vn, so that t = - dv/v

n

= (1/n – 1).[(1/vn – 1

) – (1/Vn – 1

)] as v = V when x = 0.

Let n < 1 and = 1 – p, then

t = (1/p).(Vp – v

p) as v = V when t = 0 so that when v = 0,

t = (1/p)Vp = (1/1 – n)V

1 – n = a finite quantity.

Let n > 1 and = 1 + p, then t = (1/p).[(1/vp) – (1/V

p)] (as v = 0 when t = 0).

Hence when v = 0, t = .

Let n = 1, then t = - dv/v = - logv + logV (as v = V when x = 0)

= log(V/v).

Hence when v = 0, t = .

Conclusively

s is finite if n < 2 but infinite if n = 0or > 2

and t is finite if n < 1 but infinite if n = 0 or > 1

Ex.8 In the previous question if the resistance be k (velocity) and the initial velocity be

V, show that v = Vek t

and s = (V/k).(1 – ek t

).

Sol.

Resistance = kv.

The equation of motion is d2x/dt

2 = dv/dt = kv.

So kt = dv/v = logv + A.

Initially when t = 0, v = V; so A = logV,

So that kt = log(V/v); so v = Ve kt

.

Further, v = ds/dt = Ve kt

Therefore s = Vekt

dt = (V/k).ekt

+ B.

Initially when s = 0; so B = V/k.

Hence s = (V/k).[1 – ekt

].

Ex.9 A particle falls from rest at a distance a from the centre of the Earth towards the

Earth, the motion meeting with a small resistance proportional to the square of the

velocity v and the retardation being for unit velocity; show that the kinetic energy at

145

distance x from the centre is mgr2[(1/x) – (1/a) + 2(1 – x/a) – log(a/x)], the square of

being neglected and r being radius of the Earth.

Sol.

When the particle moves towards the centre of the earth and is at a distance x

from the centre of the earth; attraction of the earth = /x2 towards the centre and

resistance = v2 away from the centre (opposite to the motion).

Therefore the equation of motion of the particle is

v(dv/dx) = v2 – (/x

2) = v

2 – (gr

2/x

2).

So on earth‟s surface g = (/r2) or = gr

2

and acceleration at a distance x = (/x2) = gr

2/x

2

i.e. dv2/dx – 2v

2 = - 2gr

2/x

2.

Being linear differential equation in v2, its I.F. = e

– 2dx = e

– 2x and therefore its

solution is

v2.e

– 2x = - 2gr

2 (1/x2).e

– 2x dx + A

or v2(1 – 2x) = - 2gr

2 [(1 – 2x)/x

2]dx + A (expanding exponential terms

and neglecting squares of )

= - 2gr2 [ - (1/x) – 2log x] + A

= 2gr2[(1/x) + logx

2] + A.

Initially when x = a, v = 0; so A = - 2gr2.[(1/a) + log a

2],

so that v2(1 – 2x) = [(1/x) – (1/a) + log (x

2/a

2)] 2gr

2

i.e. v2 = 2gr

2(1 – 2x)

-1[(1/x) – (1/a) + log (x

2/a

2)]

= 2gr2 (1 + 2x)[(1/x) – (1/a) + 2log (x/a)] (expending binomially and

neglecting 2 etc.)

= 2gr2 [(1/x) – (1/a) + 2(1 – x/a) + 2log(x/a)] (neglecting

2 etc.)

So K.E. = ½ mv2 = mgr

2 [(1/x) – (1/a) + 2(1 – x/a) – 2log(a/x)]

Ex.10 An attracting force, varying as the distance, acts on a particle initially at rest at a

distance a. Show that, if V be the velocity when the particle is at a distance x and V the

velocity of the same particle when the resistance of the air is taken into account, then

V = V[1 – (1/3).k.{(2a + x).(a – x)/(a + x)}]

nearly, the resistance of the air being given to be k times the square of the velocity per

unit of mass, when k is very small.

Sol.

The equation of motion is

v(dv/dx) = x + kv2 or dv

2/dx – 2kv

2 = 2x.

Being linear differential equation in v2, its I.F. = e

2kdx = e

2 k x and therefore its

solution is

(v2/).e

2k x = 2xe

2k xdx

= 2[{(xe2kx

)/( 2k)} + (1/2k) e2k x

dx]

= (xe2kx

)/k – (ae2a k

)/k + (1/2k2)[e

2k x – e

-- 2ak]

= [(1 + 2kx)/2k2].e

2k x – [(1 + 2ak)/2k

2].e

2ak

or v2/ = [(1 + 2kx)/2k

2] – [(1 + 2ak)/2k

2].e

2k(x – a)

a

x

a

x

a

x

146

Since at a distance x, v = V (taking resistance of air into account)

So V2 1 + 2kx 1 + 2ka 4k

2(x – a)

2 8k

3(x – a)

3

2k2 2k

2 2! 3!

1 + 2kx 1 + 2ak (1 + 2ka).(x – a)

2k2 2k

2 k

(1 + 2ka)(x – a)2 – (2k/3)(1 + 2ka)(x – a)

3 + ……

= (1/2k2) + x/k – (2/2k

2) – a/k – (x – a)/k – 2a(x – k) – (x – a)

2

2ak (x – a)2 – (2k/3)(x – a)

3, neglecting k

2 etc.

= 2ax + 2a2 – x

2 + 2ax – a

2 – (2k/3)(x – a)

2(3a + x – a)

= a2 – x

2 – k(x – a)

2. (2/3).(x + 2a). (1)

But as the velocity is V when there is no resistance of air, i.e. k = 0.

Therefore V2/ = a

2 – x

2 (2)

[or thus v(dv/dx) = x; so v2 = 2 xdx giving K

2/ = a

2 – x

2].

Dividing (1) by (2), we get

K2/K

2 = 1 – (2k/3).[(a – x)(2a + x)/(a + x)]

or K = K[1 – (2k/3).{(a – x)(2a + x)/(a + x)}]1/2

= K[1 – ½ (2k/3).{(a – x)(2a + x)/(a + x)} + …… ]

= K[1 – (k/3).{(a – x)(2a + x)/(a + x)}], neglecting k2 etc.

Particle Projected under Gravity:

Theorem: A particle is projected under gravity and a resistance equal to mk (velocity)

with a velocity u at an angle to the horizon. Find the motion.

Proof:

Let the axes of x and y be respectively horizontal and vertical and the origin at the

point of projection. Then the equations of motion are

d2x/dt

2 = k(ds/dt).(dx/ds) = k(dx/dt),

and d2y/dt

2 = k(ds/dt).(dy/ds) – g = k(dy/dt) – g.

Integrating, we have

log (dx/dt) = kt + const. = kt + log (u cos),

and log {k(dy/dt) + g} = kt + const. = kt + log(ku sin + g);

So dx/dt = u cos ekt

(1)

and k (dy/dt) + g = (ku sin + g) ekt

(2)

Therefore x = (u cos/k)ekt

+ const. = (u cos/k)(1 – ekt

) (3)

and ky = + gt = [(ku sin + g)/k].ekt

+ const.

= [(ku sin + g)/k](1 – ekt

) (4)

Eliminating t, we have

y = (g/k2).log[1 – kx/(u cos)] + [x/(u cos)](u sin + g/k) (5)

which is the equation to the path.

The greatest height is attained when dy/dt = 0, i.e. when

ekt

= g/[ku sin + g], i.e. at time (1/k) log(1 + ku sin/g),

and then y = u sin/k – (g/k2)log(1 + ku sin/g).

[1 + 2k(x –a) + + .]

a x

+

147

It is clear from equation (3) and (4) that when t = , x = (u cos/k and y = .

Hence the path has a vertical asymptote at a horizontal distance u cos/k from the point

of projection. Also, then, dx/dt = 0 and dy/dt = g/k i.e. the particle will then have just

attained the limiting velocity.

Cor: If the right-hand side of (5) be expanded in powers of k, it becomes

kx k2x

2 k

3x

3

u cos 2u2 cos

2 3u

3 cos

3

+ (x/ucos)(u sin + g/k),

gx2 gkx

3 gk

2x

4

2u2 cos

2 3u

3 cos

3 4u

4 cos

4

On putting k = 0, we have the ordinary equation to the trajectory for unresisted

motion.

Particle Moving under Gravity:

Theorem: A particle is moving under gravity in a medium whose

resistance = m(velocity)2. Find the motion.

Proof:

When the particle has described a distance s, let its tangent make an angle with

the upward drawn vertical and let v be its velocity.

The equation of motions are then

v(dv/ds) = g cos – v2 (1)

and v2/ = g sin (2)

(1) gives dv2 d

d ds

i.e. from (2), 1 d(sin)

d

So (1/)(d/d).sin + 3 cos = 2sin,

Therefore d 1 1 3 cos 1 2

d sin3 sin

4 sin

3

Hence 1/(sin3) = 21/sin

3d

= (cos/sin2) – log[(1 + cos)/sin] + A (3)

(2) then gives

v2[A – (cos/sin

2) – log {(1 + cos)/sin}] = g/sin

2

Equation (3) gives the intrinsic equation of the path, but cannot integrated further.

Bead Moves on a Smooth Wire:

Theorem: A bead moves on a smooth wire in a vertical plane under a

resistance = k (velocity)2. Find the motion.

Proof:

When the bead has described an arcual distance s, let the velocity be v at an angle

to the horizon and let the reaction of the wire be R.

The equations of motion are

y = (g/k2) [ – - … ]

y = x tan ….

= - 2g cos – v2

= - 2 cos – 2sin

( )

148

v(dv/ds) = g sin – kv2 (1)

and v2/ = g cos – R (2)

Let the curve be s = f().

Then (1) gives

d v2

d2

i.e. d(v2)

d

a linear equation to give v2.

Cor: Let the curve be a circle so that s = a, if s and be measured from the highest

point.

(1) then gives

d(v2)

d

So v2e

2ak = 2ag sin.e

2akd = [2ag/(1 + 4a

2k

2)].e

2ak(2ak sin – cos) + C.

Therefore v2 = [2ag/(1 + 4a

2k

2)].(2ak sin – cos) + Ce

-2ak.

Ex.1 A particle, of mass m, is projected in a medium whose resistance is mk (velocity)

and is acted on by a force to a fixed point (= m..distance). Find the equation to the path,

in the case when 2k2 = 9, show that it is a parabola and that the particle would

ultimately come to rest at the origin, but that the time taken would be infinite.

Sol.

The equations of motion are

d2x dx

dt2 dt (1)

and d2y dy

dt2 dt (2)

(1) may be written as (D2 + kD + )x = 0, where D = d/dt, which gives

k (k2 – 4)

2

Hence putting

k + (k2 – 4)

2

and

- k - (k2 – 4)

2

the solution of (1) is x = Aept

+ Beqt

Similarly the solution of (2) is

y = Cept

+ Deqt

,

The initial conditions determine the constants A, B, C and D and we thus have the

path.

Now if 2k2 = 9, then p = (½) and q = 2(½); hence by putting (½) =

p1, we have

x = Aep

1t + Be

2 p1t and y = Ce

p1

t + De

2p1

t

) = f ()[g sin – kv2] (

+ 2kf ().v2 = 2g sin.f ()

+ 2akv2 = 2ag sin

= - x - k

= - y - k

D =

p =

q =

149

Consider

(Dx – By)2 = (ADe

p1

t + BDe

2p1t – BCe

p1t – BDe

2p1t)2

= (AD – BC)2e 2p

1t

But also as

Ay – Cx = ACep

1t + ADe

p1

t – ACe

p1

t – BCe

2p1

t

= (AD – BC)e2p

1t

Hence (Dx – By)2 = (AD – BC).(AD – BC)e

2p1t

= (AD – BC)(Ay – Cx),

which represents a parabola.

Again the particle will be at the rest at the origin at time t given by

0 = ep

1t[A + Be

p1

t] and 0 = e

p1t[A + 2Be

p1

t]

(on putting x = 0 and dx/dt = 0)

0 = ep

1t[C + De

p1

t] and 0 = e

p1t[C + 2De

p1

t]

(on putting y = 0 and dy/dt = 0)

All the four equations are satisfied by

ep

1t = 0 or e

p1t =, i.e. t =

Hence the time taken would be infinite.

Ex.2 A particle of unit mass is projected with velocity u at an inclination above the

horizon in a medium whose resistance is k times the velocity. Show that its direction will

again make an angle with the horizon after a time (1/k) log [1 + (2ku/g).sin].

Sol.

Resistance = kv = k(ds/dt)

The equation of motions are

d2x ds dx

dt2 dt dt (1)

and d2y ds dy

dt2 dt dt (2)

Integrating (1), log(dx/dt) = kt + A.

Initially when t = 0, dx/dt = u cos; so A = log(u cos), so that

log(dx/dt) = kt + log(u cos), i.e. (dx/dt) = u cos.ek t

(3)

Integrating (2),

log[k(dy/dt) + g] = kt + log(g + ku sin),

so when t = 0, dy/dt = u sin

or k(dy/dt) + g = (g + ku sin)ekt

or dy/dt = (g/k) + (1/k)(g + ku sin)ekt

(4)

Dividing (4) by (3)

dy g/k + (1/k)(g + kusin)ekt

gekt

g + ku sin

dx u cos.e-kt

ku cos ku cos

If the particle again makes an angle with the horizon after time t, then

dy gekt

g + ku sin

dx ku cos ku cos

or gekt

sin g + ku sin

ku cos cos ku cos

or ekt

= (2ku sin)/g + 1

= - k cos = - k

= - g - k sin = - g - k

= = +

= tan( ) = tan = +

+ =

150

So t = (1/k) log[1 + (2ku/g) sin]

Ex3. If a high throw is made with a diabolo spool the vertical resistance may be

neglected, but the spin and the vertical motion together account for a horizontal drifting

force which may be taken as proportional to the vertical velocity. Show that if the spool

is thrown so as to rise to the height h and return to the point of projection, the spool is at

its greatest distance c from the vertical through that point when it is at a height 2h/3 and

show that the equation to the trajectory is of the form 4h3x

2 = 27c

2y

2(h – y).

Sol.

Neglecting the vertical resistance, equation of motion is

d2y/dt

2 = g.

Integrating w.r.t. t, we get

dy/dt = gt + A

Initially when t = 0, dy/dt = v (say); so A = v.

Thus dy/dt = v – gt or dy = (v – gt)dt

Integrating again y = vt – ½ gt2 + B

Initially when y = 0, t = 0; so B = 0

so that y = vt – ½ gt2 (1)

Also when the particle reaches the highest height h,

0 = v2 – 2gh, i.e. v

2 = 2gh (2)

Now the horizontal drifting is given by

d2x/dt

2 = (dy/dt)

Integrating, dx/dt = y + C

Initially when y = 0, dx/dt = u (say); so C = u,

so that dx/dt = – vt + ½ gt2 + u, from (1)

Integrating it again,

x = ut – v(t2/2) + (g/6)t

3 + D

Initially when t = 0, x = 0; so D = 0

Thus x = ut – ½ vt2 + (1/6)gt

3 (3)

Now the spool goes to height h and again comes to the position y = 0 in time

given by

0 = vt – ½ gt2 [on putting y = 0 in (1)]

whence t = (2v/g) and then x is given to be zero, so that from (3)

0 = u(2v/g) – (v/2)(2v/g)2 + (g/6)(2v/g)

3

i.e. u = (v2/g) – (v

2/3g) = (1/3)(v

2/g)

Substituting this value of u in (3), we get

x = (g/6)[t3 – (3vt

2/g) + (2v

2/g

2).t] = (gt/6)[t

2 – 3vt/g + (2v

2/g

2)]

= (gt/6)(v/g – t)(2v/g – t) = (/3)(v/g – t)(vt – ½ gt2)

= (/3)[(2/g)(h – y)]y, {as h – y = (v2/2g) – vt + ½ gt

2 = (g/2)(v/g – t)

2

So v/g – t = [(2/g)(h – y)]}

Squaring both sides,

9gx2 = 2

2y

2(h – y)

Now x = c when y = 2h/3

Therefore 9gc2 = 2

2(4h

2/9)(h/3) = 8

2h

3/27 i.e.

2 = (279gc

2)/8h

3

Hence 9gx2 = 2 (279gc

2)y

2(h – y)/(8h

3)

151

or 4h3x

2 = 27c

2y

2(h – y)

Ex.4 If the resistance vary as the velocity and the range on the horizontal plane through

the point of projection is a maximum, show that the angle which the direction of

projection makes with the vertical is given by (1 + cos)/(cos + ) = log[1 + sec],

where is the ratio of the velocity of projection to the terminal velocity.

Sol.

Resistance varies the velocity. Let be the angle which the direction of projection

makes with the horizontal, so that = /2 .

The equations of motion are

d2x ds dx dx

dt2 dt ds dt (1)

and d2y ds dy dy

dt2 dt ds dt (2)

(1) may be written as x/x = k

Integrating, log(dx/dt) = kt + A

Initially when t = 0, dx/dt = u cos; A = log u cos,

so that log(dx/dt) = kt + log u cos, i.e. x = u cos.ekt

(3)

and (2) may be written as ky/(g + ky) = - k

Integrating, log(ky + g) = kt + B

Initially when t = 0, y = u sin; so B = log(g + ku sin),

so that log(ky + g) = kt + log(g + ku sin),

i.e. k(dy/dt) + g = (g + ku sin)ekt

(4)

Again integrating (3), we have

x = (u cos.ekt

)/k + C

Initially when t = 0, x = 0; so C = (1/k) u cos

Therefore x = (u cos/k)(1 – e-kt

) (5)

Similarly integrating (4), we get

y = [(g + ku sin)/k2](1 – e

kt) – gt/k (6)

When the particle strikes the horizontal plane, y = 0 and therefore the time of

flight is obtained by putting y = 0 in (6),

i.e. kgt = (g + ku sin)(1 – ekt

) (7)

Differentiating (7) w.r.t., we get

kg(dt/d) = ku cos(1 – ekt

) + k(g + ku sin)ekt

(dt/d)

or dt/d = u cos(1 – ekt

)/[g(1 – ekt

) – ku sin ekt

] (8)

From (5) the range x = (u cos/k)(1 – ekt

), where t is the time of flight given by

(7).

So dx/d = (u sin/k)(1 – ekt

) + u cos ekt

(dt/d)

For maximum x,

dx/d = 0, i.e. dt/d = [u sin(1 – ekt

)]/[ku cos ekt

] (9)

Equating the two values of dt/d from (8) and (9), we have

cos sin

g – gekt

– ku sin ekt

ku cos ekt

or ku cos2 = g sin e

kt – g sin – ku sin

2

= - kv cos = -k = - k

= - kv sin - g = -k - g = - k - g

.. .

.. .

.

.

=

.

.

152

or ekt

= [ku + g sin]/(g sin), i.e. kt = log[(ku + g sin)/(g sin)] (10)

Eliminating t between (7) and (10), we get

g log[(ku + g sin)/(g sin)] = (g + ku sin)[1 – (g sin)/(ku + g sin)]

= (g + ku sin)[ku/(ku + g sin)] (11)

In order that acceleration is zero at any time, dx/dt = 0, y = 0, i.e. dx/dt = 0 and

d2y/dt

2 = 0/k, i.e., the terminal velocity = g/k.

Now the given ratio = u/(terminal velocity) = , i.e., ku = g.

Putting this value of ku in (11), we have

log[(+ sin)/sin] = (1 + sin)/(+ sin)

Putting = /2 - , log[(+ cos)/cos] = (1 + cos)/(+ cos)

or log(1 + sec) = (1 + cos)/(+ cos)

Ex.5 If a particle be moving in a medium whose resistance varies as the velocity of the

particle, show that the equation of the trajectory can, by a proper choice of axes, be put

into the form y + ax = b logx.

Sol.

Resistance = kv = k(ds/dt)

Assuming that the tangent to the curve at any point of the trajectory makes an

angle with the horizon, the equations of motion of the particle in horizontal and

vertical direction are

d2x ds dx dx

dt2 dt ds dt (1)

and d2y ds dy dy

dt2 dt ds dt (2)

Integrating and applying the initial conditions that when t = 0, dx/dt = u cos,

dy/dt = u sin, we get

dx/dt = u cos ekt

, (3)

and k(dy/dt) + g = (ku sin + g)ekt

(4)

Integrating again and applying the initial conditions that when t = 0, x = 0, we get

x = [(u cos)/k](1 – ekt

) (5)

and ky + gt = [(ku sin + g)/k](1 – ekt

) (6)

where u is the velocity and the angle of projection.

Eliminating t between (5) and (6), we have

y = (g/k2).log[1 – (kx)/(u cos)] + [x/(u cos)].[u sin + g/k]

g k u cos x

k2 u cos k ku cos

g k g u cos x

k2

u cos k2 k ku cos

g k g u cos u cos ku sin + g

k2

u cos k2 k k ku cos

u cos ku sin + g

k ku cos

or g ku sin + g k

k2 g u cos

u cos ku sin + g g u cos

= - k(ds/dt) cos = -k = - k

= - k(ds/dt) sin - g = -k - g = - k - g

log { ( - x)} + (ku sin + g) =

log ( - x) + (ku sin + g) + log =

log ( - x) – + = log

- x) + (

[ y - ( + log )]

- x) + ( log = ( - x)

153

k ku cos k2 k (7)

Putting

g ku sin + g k

k2 g u cos

and

u cos

k

the equation (7) becomes

Y + [(ku sin + g)/(ku cos)].X = (g/k2).logX

which is of the form Y + aX = b logX,

where

a = [ku sin + g]/(ku cos) and b = g/k2

Generalising, we get y + ax = b logx

Ex.6 A particle moves in a resisting medium with a given central acceleration P; the path

of the particle being given, show that the resistance is

1 d dr

2p2 ds dp

Sol.

Let Q be the resistance.

Also let be the angle which the radius vector makes with the tangent, so that p =

r sin

Acceleration along the tangent is given by

v(dv/ds) = Q – P cos = P(dr/ds) – Q (1)

and normal acceleration is v2/ = P sin = P.(p/r)

or v2 = P.(p/r). = P.(p/r).r(dr/dp) = P.p(dr/dp) (2)

From (1), we have

Q = P(dr/ds) – v(dv/ds) = P(dr/ds) – (1/2).(dv2/ds),

or 2Q = 2P(dr/ds) + dv2/ds

i.e. 2Qp2 = 2Pp

2(dr/ds) + p

2(dv

2/ds)

= 2Pp2(dr/ds) + p

2 d Pp dr [from (2)]

ds dp

= 2p(dp/ds)[Pp(dr/dp)] + p2 d Pp dr since dr/ds = (dr/dp).(dp/ds)

ds dp

d p2 Pp dr

ds dp

so 1 d dr

2p2 ds dp

Ex.7 If a point travel on an equiangular spiral towards the pole with uniform angular

velocity about the pole, show that the projection of the point on a straight line represents

a resisted simple vibration.

Sol.

[ y - ( + log )]

- x) ( = X

= Y

- ( p3 P )

) (

) (

) ( =

- ( p3 P ) Q =

154

Taking the given straight line as initial line, let the equation of the equiangular

spiral be r = ae

and the angular velocity d/dt = - (- ve sign being taken as the point

moves in the direction of decreasing).

Let be the projection of r, so that

= r cos = ae

.cos

So = ae

.cos.(d/dt) – ae

.sin.(d/dt)

= ae

[cos – sin], since d/dt =

and = ae

(d/dt)[cos – sin] + ae

[- sin – cos].(d/dt)

= a2e

[( 2 - 1)cos – 2 sin] since d/dt =

= – (1 – 2)

2 + 2.ae

.[cos – sin] – 2

2a

2e

cos

= – (1 – 2)

2 + 2 –

= – (1 + 2)

2 + 2

which is of the form

d2/dt

2 + (d/dt) + n

2 = 0

and therefore it represents a resisted simple harmonic motion.

Ex.8 A particle acted on by gravity is projected in a medium of which the resistance

varies as the velocity. Show that tits acceleration retains a fixed direction and diminishes

without limit to zero.

Sol.

Resistance = kv.

The equations of motion are

x = - k(dx/dt) (1)

and y = - k(dy/dt) – g (2)

Integrating (1) and (2) and applying the initial conditions that when t = 0, x = u

cos and y = u sin, we get

dx/dt = u cos.ekt

(3)

and k(dy/dt) + g = (ku sin + g)ekt

(4)

Direction of acceleration

y k(dy/dt) + g

x k(dx/dt) [from (1) and (2)]

(ku sin + g)ekt

ku sin + g

ku cos ekt

ku cos [from (3) and (4)]

which is independent of t; hence the acceleration relations a fixed condition.

Now substituting from (3) and (4) to (1) and (2), we have

Total acceleration = [(d2x/dt

2)2 + (d

2y/dt

2)2]

1/2

= [{ k (dx/dt)}2 + {k (dy/dt) + g}

2]

1/2 [from (1) and (2)]

= [(ku cos ekt

)2 + {(ku sin + g)e

kt}

2] [from (3) and (4)]

= ekt(k

2u

2 + 2kug sin + g

2),

which decreases as t increases and is ultimately zero.

Ex.9 A heavy bead, of mass m, slides on a smooth wire in the shape of a cycloid, whose

axis is vertical and vertex upwards, in a medium whose resistance is m(v2/2c) and the

distance of the starting point from the vertex is c; show that the time of descent to the

cusp is [8a(4a – c)/gc], where 2a is the length of the axis of the cycloid.

= =

= =

.

..

..

..

..

..

.

.

155

Sol.

Resistance = (mv2)/(2c)

The intrinsic equation of the cycloid is s = 4a sin (1)

Now the equation of motion is

v(dv/ds) = g sin – (v2)/(2c) = g.(s/4a) – (v

2)/(2c) [from (1)]

or 2v(dv/ds) = (gs)/(2a) – v2/c

or dv2 g.s v

2

ds 2a c

or dv2 v

2 gs

ds c 2a

Being linear differential equation in v2, its I.F. = e

(1/c)ds = e

s/c and therefore its

solution is

v2. e

s/c = [g/(2a)] s. e

s/c ds + A

= [g/(2a)][s.c es/c

– c2 e

s/c ] + A

Initially when s = c, v = 0; so A = 0

Hence v2. e

s/c = [g/(2a)].c(s – c) e

s/c, i.e. v

2 = (ds/dt)

2 = [(gc)/(2a)](s – c)

or ds/dt = [(gc)/(2a)](s – c)

So t = [(2a)/(gc)] [1/(s – c)]ds = [(2a)/(gc)][2(s – c)]

= [(2a)/(gc)].2(4a – c) = [8a(4a – c)/(gc)]

Ex.10 A particle, moving in a resisting medium, is acted on by a central force (/rn); if

the path be equiangular spiral of angle , whose pole is at the centre of force, show that

the resistance is [(n – 3)/2].(cos/rn).

Sol.

Let P be the resistance per unit mass; then the equations of motion are

v(dv/ds) = – (/rn)cos – P (1)

and v2/ = (/r

n)sin (2)

We know that in an equiangular spiral r = aecot

, we always have

= (constant). So p = r sin = r sin

and therefore = r(dr/dp) = r cosecHence from (2),

v2 = (/r

n)sin = /r

n – 1 since = r cosec

Differentiating it w.r.t. s, we get

2v(dv/ds) = – (n – 1) (/rn)(dr/ds)

v(dv/ds) = – [(n – 1)/2].(/rn)cos since dr/ds = cos = cos

Substituting this value of v(dv/ds) in (1), we have

P = – (/rn) cos + ½ [(n – 1)/r

n].cos= [(n – 3)/2].[cos/r

n]

2.5 Motion of Particles of Varying Mass:

The equation P = mf is only true when the mass m is constant. Newton‟s second

law in its more fundamental form is

d(mv)

dt (1)

= -

+

4a c

4a c

P =

156

Suppose that a particle gains in time t an increment m of mass and that this

increment m was moving with a velocity u.

Then in time t the increment in the momentum of the particle

= m.v + m(v + v – u)

and the impulse of the force in this time is Pt.

Equating these we have, on proceeding to the limit.

m(dv/dt) + v(dm/dt) – u(dm/dt) = P

i.e. d dm

dt dt (2)

When u = 0,we have the result (1)

Ex.1 A falling raindrop has its uniformly increased by access of moisture. If it has given

to it a horizontal velocity, show that it will then describe a hyperbola, one of whose

asymptotes is vertical.

Sol.

Initially taking the radius of the raindrop as unity, its mass

= (4/3).l2 = 4/3 = m (say). (1)

Let the increase of volume at an instant t, when its radius is r, be times its

surface at that instant, so that (dM/dt) = 4r2., where M is the mass at time t, i.e. M =

(4/3)r3.

So dM d{(4/3)r3.)

dt dt

= 4r2. where dr/dt = .

Integrating, r = t + A.

Initially when t = 0, r = 1; so A = 1 and thus r = 1 + t. (2)

As such M = (4/3)r3 = m(1 + t)

3. (3)

Since there is no horizontal force,

d dx

dt dt

Integrating, (1 + t)3(dx/dt) = B.

Initially when t = 0, the horizontal velocity

dx/dt = 2a (say); so B = 2a,

so that (1 + t)3.(dx/dt) = 2a or dx = [2a/(1 + t)

3]dt

Integrating again, x = - a/(1 + t)2 + C

Initially when t = 0, x = 0; so C = a.

Hence x = a[1 – 1/(1 + t)2] (4)

Again considering the vertical motion, we have

d dy

dt dt

Integrating,

(1 + t)3(dy/dt) = (g/4).(1 + t)

4 + D

Initially when t = 0, dy/dt = 0; so D = - g/4

Thus (1 + t)3(dy/dt) = (g/4)[(1 + t)

4 – 1]

(mv) = P + u

=

{m (1 + t)3 }= 0

{m (1 + t)3 }= m (1 + t)

3.g,

157

or dy/dt = (g/4)[(1 + t) – 1/(1 + t)3]

Integrating again

g (1 + t)2 1

4 2 2 (1 + t)2

Initially when t = 0, y = 0; so E = - g/(42)

Hence y = [g/(82)].[(1 + t)

2 + 1/(1 + t)

2 – 2]

= [g/(82)].[1 + t – 1/(1 + t)]

2 (5)

In order to find the equation of the trajectory we have to eliminate t between (4)

and (5) and for this let us put (1 + t)2 = a/(a – x) from (4) to (5), whence we get

g a a – x g x2

82 a – x a 8

2 a (a – x)

or 82ay 8

2a

2y 8

2axy

g g g

or x2 + (8

2a/g).xy = (8

2a

2/g).y,

which represents a hyperbola as

(42a/g) >

Note:

Condition for ax2 + 2hxy + by

2 + 2gx + 2fy + c = 0 to represent a hyperbola is

h2 > ab.

Its asymptote parallel to y-axis is

(82a/g).x – (8

2a

2/g) = 0, i.e. x = a,

which is obviously vertical.

Ex.2 A chain, of great length a, is suspended from the top of a tower so that its lower end

touches the earth; if it be then let fall, show that the square of its velocity, when its upper

end has fallen a distance x, is 2grlog[(a + r)/(a + r –x)], where r is the radius of the earth.

Sol.

Let m be the mass per unit length of the chain. Also let at an instant of time t, the

chain have fallen a distance x, so that the mass of the chain deposited per unit of time

d(mx) dx

dt dt

Now d dx

dt dt

= change in the momentum per unit of time

= the difference of the momentum of the deposited mass and the mass of

the vertical portion of the chain.

dx dx a – x E.m.dy

dt dt 0 (r + a – x – y)2

where is constant of gravitation, E the mass of the earth and y an element of the

vertical portion of the chain at a height y from the surface.

or m (a – x)(d2x/dt

2) – m (dx/dt)

2 = – m(dx/dt)

2

a – x dy .

0 (r + a – x –y)2 [since g = (E.1)/r

2]

or d2x 1 a – x

dt2 r + a – x – y 0

y = [ + ] + E

y = [ + - 2 ] =

x2 = (a – x) = -

= = m.

[m (a – x) ]

= - (m. ) +

+ gr2m

(a – x) = gr2 [ ]

158

= gr2[1/r – 1/(r + a – x)] = [gr/(r + a – x)].[r + a – x –r]

= gr(a – x)/(r + a – x)

or = d2x/dt

2 = gr/(r + a – x).

Multiplying both sides by 2 (dx/dt)dt and integrating.

(dx/dt)2 = – 2gr log(r + a – x) + A

Initially when x = 0, dx/dt = 0; so A = 2gr log(r + a)

Hence (dx/dt)2 = 2gr log[(a + r)/(a + r – x)]

Ex.3 A spherical raindrop of radius a cms. Falls from rest through a vertical height h,

receiving throughout the motion an accumulation of condensed vapour at the rate of k

grammes per square cm. per second, no vertical force but gravity acting; show that when

it reaches the ground its radius will be

k(2h/g)[1 + {1 + (ga2)/(2hk

2)}]

Sol.

Let the radius and the mass of the raindrop be respectively r and M when it has

fallen through a distance x in time t. Then using the relation P = d(mv) , we have

d dx dt

dt dt (1)

Now M = (4/3).r3., where is the density/c.c.

and dM/dt = 4r2.k as the vapour accumulates at the rate of k gm./sp.cm./sec.

Hence from (1),

d dx

dt dt

or d dx

dt dt (2)

Again dM d

dt dt

or r2(dr/dt) = r

2k or dr/dt = k (3)

Integration of (3) gives r = kt + A whilst initially when t = 0, r = a; so A = a, so

that r = kt + a (4)

With the help of (3), (2) may be written as

d dx g dr

dt dt k dt [since dr/dt = k from (3)].

Integrating,

r3(dx/dt) = (gr

4)/(4k) + B,

while initially when r = a, dx/dt = 0; so B = – (ga4)/(4k),

so that r3(dx/dt) = g(r

4 – a

4)/(4k)

or dx g a4 dr

dt 4k2 r

3 dt from (3)

Integrating again,

x = [g/(4k2)].[r

2/2 + a

4/(2r

2)] + C

Initially when r = a, x = 0; so C = - (ga2)/(4k

2)

Thus,

g a4 g r

2 – a

2

8k2 r

2 8k

2 r

( M ) = Mg

[(4/3)r3. ] = .(4/3)r

3g

[r3 ] = r

3g

= .(4/3)r3] = .4r

2k

(r3 ) = r

3

= (r – )

x = (r2 + - 2a

2) = ( )

2

159

When x = h, we have [(r2 – a

2)/r]

2 = 8k

2h/g

or (r2 – a

2)/r = 2k(2h/g)

or r2 – 2k(2h/g).r – a

2 = 0, which gives

2k(2h/g) [(8k2h)/g + 4a

2]

2

= k(2h/g)[1 + {1 + (ga2)/(2hk

2)}] taking + ve sign.

Ex.4 A particle of mass M is at rest and begins to move under the action of a constant

force F in a fixed direction. It encounters the resistance of a stream of fine dust moving in

the opposite direction with velocity V, which deposits matter on it at a constant rate .

Show that its mass will be m when it has traveled a distance

(k/2)[m – M{1 + log(m/M}]

where k = F – V.

Sol.

When the mass of the particle is m, let us suppose that it gains in time t an

increment m of mass and this increment m was moving with a velocity V in the

direction opposite to that of the mass. Hence

mv + m[v + v – (– V)] = increases in the momentum in time t, i.e.

mv + m(v + v + V) = Ft

Dividing throughout by t and proceeding to the limit t 0, we get

m(dv/dt) + (dm/dt).(v + V) = F,

So

Lim m

t 0 t

where m = M + t, since the deposition of matter in time t is t. So

dv d

dt dt

dv d d

dt dt dt

or d d

dt dt

= F – V[0 + ], since dM/dt = 0, M being a constant quantity

= F – V = k as given, F – V = k.

Integrating with regard to t,

(M + t)v = k dt = kt + A.

Initially when t = 0, v = 0; so A = 0, so that

(M + t)v = kt

or v = ds/dt = (kt)/(M + t), i.e. ds = (kt dt)/(M + t)

Integration again,

kt dt k (t + M – M) dt

M + t M + t

k M .

(M + t)

= (k/)[t – (M/).log(M + t)] + B

Initially when t = 0, s = 0.

r =

v. = 0

(M + .t) + (v + V) (M + .t) = F

(M + .t) + v (M + .t) = F (M + .t) + V

(M + .t) {(M + .t)v} = F – V

s = =

= ] dt

160

So B = (Mk/2)logM

Hence s = (k/)[t – (M/).log(M + t)] + (Mk/2).logM

= (k/2)[t – M.logm] + (Mk/

2).logM, since M + t = m

= (k/2)[m – M – M log(m/M)], since t = m – M

= (k/2)[m – M{1 + log(m/M)}] where k = F – V.

Ex.5 Show slides off a roof clearing away a part of uniform breadth; show that, if it all

slide at once, the time in which the roof will be cleared is

6a (2/3) 1/ 2

g sin (1/6)

but that, if the top move first and gradually set the rest in motion, the acceleration is

(1/3)gsin and the time will be [6a/(gsin)], where is the inclination of the roof and a

the length originally covered with snow.

Sol.

Let b be the breadth of the roof and the surface density of snow. Then if y be the

length on the roof at time t, the mass of snow = y.b.

Using the relation d(mv) = P, we have

dt

d dy

dt dt

Since the roof is inclined at an angle to the horizon

or d dy

dt dt

or dy d dy

dt dy dt

or d dy

dy dt

Integrating with regard to y,

dy

dt

Initially when y = a, dy/dt = 0; so A = (2/3)g sin a3.

Thus

dy

dt

or dy/dt = (1/y).[(2gsin)/3].(a3 – y

3)

So

2g sin T a y dy

3 0 0 (a3 – y

3) [put y = az

1/3, so dy = (1/3)az

2/3dz]

or 2g sin 1 az1/3

a dz

3 0 a3/2(1 – z) 3z

2/3

a 1

3 0

a 1

3 0

[yb (- )] = yb.g sin

[y (- )] = yg sin

[y ] = - 2y2g sin 2y

[y ]2 = - 2y

2.g sin

[y ]2 = – (2/3)y

3.g sin + A

y2 ( )

2 = (2/3).g sin (a

3 – y

3)

( ). dt =

( ).T =

= z –1/3

(1 – z) –1/2

dz

= z 2/3 – 1

(1 – z)1/2

– 1

dz

161

a

3

a (2/3) (1/2)

3 (2/3 + 1/2)

Hence

3a (2/3)

2g sin (1/6)(1/6)

6a (2/3)

g sin (1/6)

For the second part, supposing a length x to be in motion, we have

d dx

dt dt

or

d dx

dt dt (1)

or dx d dx

dt dx dt

or d dx

dx dt

Integrating,

dx

dt

Initially when x = 0, dx/dt = 0; so C = 0, so that

dx

dt

or dx/dt = [(2gsin)/3].(x) (2)

so 2g sin T a dx

3 0 0 (x)

or 2g sin x a

3 (1/2) 0

3 .

2g sin

6a .

g sin

Further to find the acceleration, we have

d dx

dt dt from (1)

or (dx/dt)2 + x(d

2x/dt

2) = gx sin

or x(d2x/dt

2) = gx sin – (2/3)gx sin, [from (2), (dx/dt)

2 = (2/3)gx sin

= (1/3)xg sin

So acceleration d2x/dt

2 = (1/3)g sin

= (2/3, 1/2)

=

T = (1/3) ( )

= ( )

[bx ] = bx.g sin

[x ] = x.g sin

[x ] = 2x2.g sin 2x

[x ]2 = 2x

2.g sin

[x ]2 = (2/3)x

3.g sin + C

[x ]2 = (2/3)x

3.g sin

( ). dt =

( ).T = [ ] = 2a)

T = 2a) ( )

= ( )

[x ] = x.g sin

162

Ex.6 A spherical raindrop, falling freely, receive in each instant an increase of volume

equal to time its surface at that instant; find the velocity at the end of time t and the

distance fallen through in that time.

Sol.

When the raindrop has fallen through a distance x in time t, let its radius be r and

its mass M. Then

d dx

dt dt (1)

Now M = (4/3)r3, so that 4 r

2(dr/dt) = dM/dt = 4 r

2, by the question.

So dr/dt = , and r = a + t,

where a is the initial radius.

Hence (1) gives

d dx

dt dt

so dx (a + t)4 a

4

dt 4

since the velocity was zero to start with.

so dx g a4

dt 4 (a + t)3

and g (a + t)2 a

4 g .

42 2 2(a + t)

2 4

2

since x and t vanish together, so

g a4

82 2(a + t)

2

g a2

82 (a + t)

and gt2 2a + t

8 a + t

Ex.7 If a rocket, originally of mass M, throw off every unit of time a mass eM with

relative velocity V and if M be the mass of the case etc., show that it cannot rise at once

unless eV > g, nor at all unless eMV/ M > g. If it just rises vertically at once, show that

its greatest velocity is

V log(M/M) – (g/e).(1 – M/M),

and that the greatest height it reaches is

(V2/2g).[log(M/M)]

2 + (V/e).[1 – (M/M) – log(M/M)].

Sol.

Let u, u be the velocities communicated to the rocket and the mass thrown off in

unit time by the explosion; then the relative velocity of the mass V = u – (– u) = u + u.

When a mass eM is thrown off the remaining mass of the rocket = M – eM = (1 – e)M.

At the start of the explosion:

Momentum of rocket = momentum of mass thrown off, i.e.

(1 – e)Mu = eMu = eM(V – u), since V = u + u.

So u = eV (1)

[M ] = Mg

[(a + t)3 ] = (a + t)

3g

(a + t)3 = g –

g

[a + t – = ]

+ x =

–

a3

+ x =

[(a + t)2 – 2a

2 ]

x =

[(a + t) – ]2

x =

[ ]

2

163

But we know that the gravity in unit time gives a velocity g. Hence the rocket will

not rise at once unless

g < u, i.e. g < eV.

Now when all the powder is burnt (except the last element), the law of

conservation of momentum gives

Mu = eMu, where M is the mass of the case

or Mu = eM(V – u), i.e. u = (eMV)/(M + eM), as u = V – u (2)

Again the velocity imparted by the gravity in unit time being g, it will not start at

all unless

g < (eMV)/(M + eM), i.e. eMV > (M + eM)g > Mg

or unless (eMV)/M > g

Further when the rocket has risen a distance x in time t, then the mass of the

rocket

= initial mass – mass thrown off in time t.

= M – eM.t = (1 – e.t)M

Thus

d dx

dt dt

= change in the momentum

= – eM.(dx/dt – V) – (1 – e.t)M.g

or d2x dx

dt2 dt

= – e(dx/dt) + eV – g + etg

or d2x

dt2

= eV – g(1 – et);

or d2x eV .

dt2 1 - et

Integrating with respect to t,

dx/dt = - V log(1 – et) – gt + A

Initially when t = 0, dx/dt = 0; so A = 0.

Thus dx/dt = - V log(1 – et) – gt (3)

Integrating again,

x = - V log(1 – et).1 dt – ½ gt2 + B

- e.t gt2

1 – et 2

1 gt2

1 – et 2

= – V[t log(1 – et) – t – (1/e) log(1 – et)] – gt2/2 + B

= (V/e).[(1 – et).{log(1 – et)} + et] – gt2/2 + B

Initially when t = 0, x = 0; so B = 0.

Hence x = (V/e)[(1 – et) log(1 – et) + et] – gt2/2

= (V/e)[(1 – et) log(1 – et) – (1 – et) + 1] – gt2/2

= (V/e)[(1 – et){log(1 – et) – 1}] – gt2/2 + V/e (4)

[(1 – et)M ]

(1 – et) - e

(1 – et)

– g =

= - V[log(1 – et)t - dt] – + B

= - V[t log(1 – et) - )dt] – + B

164

Now the greatest velocity is attained when the powder is all burnt, i.e. when mass

of the rocket = mass of the case, i.e. when (1 – et)M = M or 1 – et = (M/M) and then

from (3) and (4), we have

V1 = dx/dt = - V log(M/M) – (g/e).(1 – M/M), since t = (1/e)(1 – M/M), (5)

and x1 = x = (V/e).[(M/M){log(M/M) – 1}] – (g/2).(1/e2).(1 – M/M)

2 + V/e

= (V/e).(1 – M/M) – [(MV)/(eM)].log(M/M) – [g/(2e2)].(1 – M/M)

2

[since log(M/M) = - log(M/M)]

So total height attained

= x1 + V12/(2g)

= (V/e)(1 – M/M) – [(MV)/(eM)].log(M/M) – [g/(2e2)].(1 – M/M)

2

+ [1/(2g)].[V log(M/M) + (g/e).(1 – M/M)]2

= (V/e)(1 – M/M) – [(MV)/(eM)].log(M/M) – [g/(2e2)].(1 – M/M)

2

+ [V2/(2g

2)].[log(M/M)]

2 – (V/e).(1 – M/M).log(M/M)

+ [g/(2e2)].(1 – M/M)

2

{since log(M/M) = - log(M/M) and [log(M/M)]2 = [log(M/M)]

2}

= [V2/(2g)].[log(M/M)]

2

+ (V/e).[1 – M/M – (M/M).log(M/M) – log(M/M) + (M/M).log(M/M)]

= [V2/(2a)].[log(M/M)]

2 + (V/e).[1 – M/M – log(M/M)]

Ex.8 A uniform string, whose length is l and whose weight is W, rests over a small

smooth pulley with its end and just reaching to a horizontal plane; if the string be slightly

displaced, show that when a length x has been deposited on the plane the pressure on it is

W[2log{l/(l – x)} – (x/l)],

and that the resultant pressure on the pulley is W(l – 2x)/(l – x).

Sol.

Initially for equilibrium the length ½ l of the string hangs on either side of the

pulley. But when a length x has been deposited on the plane in an interval of time t (say),

then the length (½ l – x) of the string hangs on one side and ½ l hangs vertically on the

other side. Let T be the tension of the string and d2x/dt

2 the acceleration, then equations

of motion are

m.(l/2).g – T = m.(l/2).(d2x/dt

2) (1)

where m is the mass per unit length

and T – m(l/2 – x)g = m (l/2 – x)(d2x/dt

2) (2)

Adding,

[l/2 – (l/2 – x)]g = [l/2 + (l/2 – x)](d2x/dt

2)

giving d2x/dt

2 = xg/(l – x) (3)

or d2x/dt

2 = – g[(l – x – l)/(l – x)] = g [l/(l – x) – 1]

Multiplying both side by 2.(dx/dt).dt and integrating, we get

(dx/dt)2 = 2g.[ – l log(l – x) – x] + A.

Initially when x = 0, dx/dt = 0; so A = 2gl logl.

Thus (dx/dt)2 = 2g[l log{l/(l – x)} – x] (4)

Now if M be the mass of the whole string, then its mass per unit length = M/l = m

and its weight W = Mg.

165

Again if R be the part of reaction due to the impact, then R.t = m.x.(dx/dt).

Dividing by t and proceeding to the limit, we get

R = m.(dx/dt)2

Total pressure = R + mg.x = m[(dx/dt)2 + gx]

Mg l l .

l g l – x from (4) and also m = M/l

= W[2log{l/(l – x)} – x/l]

Further to find the pressure on the pulley, which is = 2T, let us subtract (2) from

(1) after multiplying (1) by (l/2 – x) and (2) by l/2, whence get

– T[l/2 – x + l/2] + mg[2.(l/2).(l/2 – x)] = 0

or mgl (l – 2x)

2 (l – x)

hence the pressure on the pulley = 2T

mgl (l – 2x)

(l – x)

Mgl (l – 2x)

l (l – x)

(l – 2x)

(l – x)

Ex.9 A weightless string passes over a smooth pulley. One end is attached to a coil of

chain lying on a horizontal table and the other to a length l of the same chain hanging

vertically with its lower end just touching the table. Show that after motion ensues the

system will first be at rest when a length x of chain has been lifted from the table, such

that (l – x)e(2x/ l)

= l. Why cannot the Principle of energy be directly applied to find the

motion of such a system?

Sol.

Let m be the mass per unit of length, so that a mass ml is always in motion and

the mass deposited per unit of time

Lim mx dx

t 0 t dt

So d dx

dt dt

= change in the momentum per unit of time

= – m(dx/dt).(dx/dt) + m(l – x).g – m.xg,

[since the gravity in unit time gives a velocity g]

or l(d2x/dt

2) = – (dx/dt)

2 + (l – 2x)g

or l(d2x/dt

2) + (dx/dt)

2 = (l – 2x)g (1)

or d dx dx 2 dx 2

dx dt dt l dt l

or d dx 2 dx 2

dx dt l dt l (2)

which is linear differential equation in (dx/dt)2.

So its I.F. = e(2/ l)dx

= e2x/ l

Solution is

(dx/dt)2e

2x/ l = (2g/l) (l – 2x)e

2x/ ldx + A

= [ {2g(l log - x)} + x]

T =

=

=

= W

= m =

[ml ]

2 ( ) + ( )2 = (l – 2x)g

( )2 + ( )

2 = (l – 2x)g

166

= (2g/l)[(l – 2x).(l/2).e2x/ l

+ 2.(l/2) e2x/ l

dx ] + A

= (2g/l)[(l/2).(l – 2x)e2x/ l

+ (l2/2)e

2x/ l] + A

= g[l – 2x + l]e2x/ l

+ A = 2g(l – x)e2x/ l

+ A

Initially when x = 0, dx/dt = 0; so A = -2gl.

Hence

(dx/dt)2e

2x/ l = 2g(l – x)e

2x/ l – 2gl.

Again the string comes to rest when dx/dt = 0,

i.e. 2gl(l – x)e2x/ l

– 2gl = 0

or (l – x)e2x/ l

= l.

Further since energy is lost by the deposit of the portions of the chain on the table,

the principle of energy cannot be directly applied.

Ex.10 A mass M is fastened to a chain of mass m per unit length coiled up on a rough

horizontal plane (coefficient of friction = ). The mass is projected from the coil with

velocity V; show that it will be brought to rest in a distance

M 3mV2

m 2Mg

Sol.

Let a length x of the coiled up chain lifts in time t, then equation of motion is

d dx

dt dt

= – (M + mx)g, since P = – R = – (M + x)g

or dx d dx

dt dt dt

= – 2(M + mx)2.g

or d dx

dt dt

= – 2(M + mx)2.g

Integrating with regard to x, we get

(M + mx)2.(dx/dt)

2 = – [2/(3m)].g(M + mx)

3 + A

Initially when x = 0, dx/dt = V; so A = M2V

2 + (2M

3g)/(3m)

Hence,

(M + mx)2.(dx/dt)

2 = – [2/(3m)].g(M + mx)

3 + M

2V

2 + (2M

3g)/(3m)

The mass will be brought to rest when dx/dt = 0

i.e. when

0 = – (2g)/(3m).(M + mx)3 + M

2V

2 + (2M

3g)/(3m)

or (M + mx)3 = M

3 + (3M

2mV

2)/(2g)

= M3[1 + (3mV

2)/(2Mg)]

Extracting cube root of either side,

M + mx = M[1 + (3mV2)/(2Mg)]

1/3

so that

M 3mV2

m 2Mg

Ex.11 A mass in the form of a solid cylinder, the area of whose cross-section is A, moves

parallel to its axis, being acted on by a constant force F, through a uniform cloud of fine

[(1 + )1/3

– 1]

[(M +mx) ]

[(M + mx) ] 2(M + mx)

[(M + mx) ]2

[(1 + )1/3

– 1] x =

167

dust of volume density which is moving in a direction opposite to that of the cylinder

with constant velocity V. If all the dust meets the cylinder clings to it, find the velocity

and distance described in any time t, the cylinder being originally at rest and its initial

mass m.

Sol.

Let M be the mass at time t and v the velocity. Then

M.v + M(v + v + V) = increase in the momentum in time t = Ft.

So M(dv/dt) + v(dM/dt) + V(dM/dt) = F (1)

in the limit.

Also dM/dt = A(v + V) (2)

(1) gives Mv + MV = Ft + constant = Ft + mV.

Therefore (2) gives

M(dM/dt) = A(Ft + mV)

So M2 = A(Ft

2 + 2mVt) + m

2

Therefore (2) gives

Ft + mV

M

Ft + mV .

(m2 + 2mAVt + AFt

2) (3)

Also if the hinder end of the cylinder has described a distance x from rest, so that

v = dx/dt, then x = - Vt + [1/(A)].(m2 + 2mAVt + AFt

2) – [m/(A)]

From (3) we have that the acceleration

dv m2(F – AV

2) .

dt (m2 + 2mAVt + AFt

2)3/2

so that the motion is always in the direction of the force, or opposite, according as F > or

< AV2.

Ex.12 A machine gun, of mass M, stands on a horizontal plane and contains shot, of mass

M. The shot is fired at the rate of mass m per unit of time with velocity u relative to the

ground. If the coefficient of sliding friction between the gun and the plane is , show that

the velocity of the gun backward by the time the mass M is fired is

M (M + M)2 – M

2

M 2mM

Sol.

Let the gun move backward a distance x in time t, then its velocity is dx/dt.

The mass of the shot fired out in time t = mt.

Then the mass of the machine gun with shot = M + M – mt.

Now the change in momentum per unit of time

= impulsive in one second = impulsive force 1 sec.

i.e. d dx

dt dt

= – (M + M – mt)g

or d dx

dt dt

= – (M + M – mt)g + mu

v = - V +

= - V +

=

u – g

[(M + M – mt) ] – mu

[(M + M – mt) ]

168

Integrating with regard to t, we get

dx

dt

= [(g)/(2m)].(M + M – mt)2 + mut + A

Initially when t = 0, dx/dt = 0; so A = – [(g)/(2m)].(M + M)2

Thus

dx

dt

= [(g)/(2m)].(M + M – mt)2 + mut – [(g)/(2m)].(M + M)

2

= [(g)/(2m)].[(M + M – mt)2 – (M + M)

2] + mut

= [(g)/(2m)].[(M + M)2 – 2m(M + M)t + m

2t2 – (M + M)

2] + mut

= [(g)/(2m)].[– 2(M + M) + mt]mt + mut (1)

Now when whole of the mass M (of shot) is fired, then

M = mt.

Hence from (1), we have

dx

dt

= [(g)/(2m)].[– 2(M + M) + M]M + Mu

= [(g)/(2m)].[– M2 – 2MM– M + M

2] + Mu

= [(g)/(2m)].[(M + M)2 – M

2] + Mu

dx M (M + M)2 – M

2

dt M 2mM

Ex.13 A uniform cord, of length l, hangs over a smooth pulley and a monkey, whose

weight is that of the length k of the cord, clings to one end and that system remains in

equilibrium. If he start suddenly and continue to climb with uniform relative velocity

along the cord, show that he will cease to ascend in space at the end of time

[(l + k)/(2g)]1/2

cosh1

(1 + l/k)

Sol.

Let m be the mass per unit length of the cord; then the mass of monkey = mk.

If V is the given relative velocity and V1 the starting velocity of the cord, then

Mk(V – V1) = pull of the monkey on the cord

= mlV1

so that V1 = (kV)/(k + l) (1)

Initially the lengths of the cord on the two sides of the pulley are

(l – k)/2 and (l + k)/2

Let at time t, a length x of the cord have gone over the pulley and the monkey

have ascended a distance y along the cord so that dy/dt = V and d2y/dt

2 = 0, V being

constant.

Then if P be the weight of the pulley, we have

d l – k l + k dx

dt 2 2 dt

= m[(l – k)/2 + x]g – m[(l + k)/2 – x]g + P,

i.e. ml(d2x/dt

2) = (2x – k)mg + P (2)

d dx dy

[(M + M – mt) ]

[(M + M – mt) ]

[(M + M – M) ]

u – g =

[{( + x) + ( – x)}m. ]

[mk( – )]

169

dt dt dt

= mkg – P,

i.e. d2x d

2y

dt2 dt

2

= mkg – P,

or mk(d2x/dt

2) = mkg – P, since d

2y/dt

2 = 0 (3)

Adding (2) and (3), we get

m(l + k).(d2x/dt

2) = 2mgx, i.e. d

2x/dt

2 = (2gx)/(k + l)

Multiplying both side by 2(dx/dt).dt and integrating, we get

(dx/dt)2 = [2g/(k + l)].x

2 + A

Initially when x = 0, dx/dt = V1 = kV/(k + l) from (1)

So A = k2V

2/(k + l)

2

so that dx 2g k2V

2

dt (k + l) (k + l)2

or dx 2g k2V

2

dt (k + l) 2g(k + l) (4)

or dx 2g k2V

2

dt (k + l) 2g(k + l)

or 2g

(k + l)

dx .

k2V

2

2g(k + l)

x{2g(k + l)}

kV

Initially when t = 0, x = 0, so B = 0.

Hence

or 2g

(k + l)

x{2g(k + l)}

kV (5)

Now the monkey stops ascending in space when dx/dt = V, i.e. when from (4),

2g k2V

2

k + l 2g(k + l)

or 2gx2 = (k + l)V

2 – (k

2V

2)/(k + l) = [V

2/(k + l)].(k

2 + 2kl + l

2 – k

2)

=V2.[(l

2 + 2kl)/(k + l)],

which gives

V(l2 + 2kl)

{2g(k + l)}

Substituting this value of x in (5), we get

2g

(k + l)

{l2 + 2kl}

k

so 2g .

(k + l)

[mk( – )]

( )2 = x

2 +

( )2 = [x

2 + ]

= [ ] [x2 + ]

[ ].t

] [x2 +

=

= sinh-1

[ ] + B

[ ].t

= sinh1

[ ]

V2 = [x

2 + ]

x =

[ ].t

= sinh1

[ ]

sinh [t ( )]

170

(l2 + 2kl)

1/ 2

k2

Hence

2g .

(k + l)

so 2g

(k + l)

= 1 + (2l/k) + l2 /k

2 = (1 + l/k)

2,

which gives

t = [(l + k)/(2k)]1/2

cosh–1

(1 + l/k)

Ex.14 A heavy chain, of length l, is held by its upper end so that its lower end is at a

height l above a horizontal plane; if the upper end is let go, show that at the instant when

half the chain is coiled up on the plane is to the weight of the chain in the ratio of 7:2.

Sol.

Let at time t, a length x of the chain be coiled up on the plane. Then this length x

has to move a distance (l + h) as the lower end is already at a height l above the plane.

Since the moving part of the chain falls with acceleration g, using the relation v2 = u

2 +

2gh, where the initial velocity u = 0, the moving part is given by

v2 = 2g(l + x) (1)

The part of the reaction say R due to the impact is given by

Impulse = the change in momentum

i.e. R.t = m x.v (supposing that x length of chain coils up in time t)

Dividing by t and proceeding to the limit.

R = m.(dx/dt).v = mv2 = 2mg(l + x), from (1).

Hence the total pressure P

= R + weight of the length x of the chain

= R + mgx where m is the mass per unit length when half the chain is

coiled up i.e. x = l/2, we have

Pat x = l/ 2 = 2mg(l + l/2) + mg.(l/2), since R = 2mg(l + x)

= (7/2).mlg

which shows that the pressure on the plane when half the chain is coiled up, is to the

weight of the chain in the ratio 7:2.

Ex.15 A uniform chain, of length l and mass ml, is coiled on the floor and a mass mc is

attached to one end and projected vertically upwards with velocity (2gh). Show that,

according as the chain does or does not completely leave the floor, the velocity of the

mass on finally reaching the floor again is the velocity due to a fall through a height

(1/3)[2l – c + a3/(l + c)

2] or a – c,

where a3 = c

2(c + 3h).

Sol.

The equation of motion is

d dx

dt dt

Supposing that a length x of the coil lifts in time t

or dx d dx

=

cosh2 [t ( )]

= 1 + sinh2 [t ( )]

{(mc + mx) }= - (mc + mx)g

{(c + x) }= - 2(c + x)2g 2(c + x)

171

dt dx dt

or dx

dt

Integrating with respect to x, we get

or dx

dt

Initially when x = 0, dx/dt = (2gh); since A = c2.2gh

Hence

(x + c)2.(dx/dt)

2 = – (2g/3)[ x

3 + 3cx

2 + 3c

2x] + c

22gh (1)

Case I. Let us assume that the chain leaves the floor. Then putting x = l in (1) and

supposing that the velocity then is V, we have

(l + c)2V

2 = – (2g/3).[l

3 + 3cl

2 + 3c

2l] + c

2.2gh

= (2g/3).[3c2h + c

3 – {l

3 + 3l

2c + 3lc

2 + c

3}]

= (2g/3).[a3 – (l + c)

3], since a

3 = 3c

2h + c

3 = c

2(c + 3h) (given)

or V2 = (2g/3).[a

3/(l + c)

2 – (l + c)]

Using the relation v2 = u

2 – 2H, where u = V, the mass will rise till v = 0, i.e. the

height attained H = V2/(2g)

Hence the total height = l + H = l + V2/(2g)

= l + (1/3).[a3/(l + c)

2 – l – c]

= (1/3).[2l – c + a2/(l + c)

2]

Case II. If we assume that the chain does not leave the floor, then the mass will come to

rest when dx/dt = 0, i.e. when from (1),

0 = – (2g/3).(x3 + 3cx

2 + 3c

2x) + c

2.2gh

or 3c2h + c

2 – (x

3 + 3cx

2 + 3c

2x + c

3) = 0

or a3 = (x + c)

3, i.e. when x + c = a or x = a – c.

Ex.16 A uniform chain is coiled up on a horizontal plane and one end passes over a small

light pulley at a height a above the plane; initially a length b, > a, hangs freely on the

other side; find the motion.

Sol.

When the length b has increased to x, let v be the velocity; then in the time t next

ensuing the momentum of the part (x + a) has increased by m(x + a).v, where m is the

mass per unit length. Also a length m.x has been jerked into motion and given a velocity

v + u. Hence

m(x + a)v + mx(v + v) = change in the momentum

= impulse of the acting force = mg(x – a).t

Hence, dividing by t and proceeding to the limit, we have

(x + a).(dv/dt) + v2 = (x – a)g

So v(dv/dx).(x + a) + v2 = (x – a)g

x

So v2(x + a)

2 = 2(x

2 – a

2)g dx = 2[(x

3 – b

3)/3 – a

2(x – b)].g

b

So that 2g (x – b)(x2 + bx + b

2 – 3a

2)

3 (x + a)2

This equation cannot be integrated further.

{(c + x) }2 = – 2g(x

2 + 2cx + c

2)

{(c + x) }2 = – 2g(x

3/3 + cx

2 + c

2x) + A

v2 =

172

In the particular case when b = 2a, this gives v2 = (2g/3).(x – b), so that the end

descends with constant acceleration g/3.

This tension T of the coil is clearly given by Tt = mx.v, so that T = mv2.

Ex.17 A ball of mass m, is moving under gravity in a medium which deposits matter on

the ball at a uniform rate . Show that the equation to the trajectory, referred to horizontal

and vertical axes through a point on itself, may be written in the form

k2uy = kx(g + kv) + gu(1 – e

kx/ u),

where u, v are the horizontal and vertical velocities at the origin and mk = 2.

Sol.

The mass of the ball at time t = m + t. Since there is no horizontal force, using

the relation,

d(mv)

dt

we have,

d dx

dt dt

Integrating,

dx

dt

Initially when t = 0, dx/dt = u, so A = mu, so that

dx

dt (1)

or dx = [mu/(m + t)].dt = (mu/).[/(m + t)].dt

Integrating, x = (mu/) log(m + t) + B

Initially when t = 0, x = 0; so B = – (mu/) logm.

Then x = (mu/).log[(m + t)/m], i.e. m + t = me x/m u

(2)

Substituting the value of m + t from (2) in (1), we get

dx/dt = ue– (x/m u)

(3)

Now considering the vertical motion, we have

d dx

dt dt

or d dy dx dx

dx dx dt dt

or d dy

dx dx

[from (1) and (2)]

or d dy

dx dt

or u2(d

2y/dx

2) = – ge

2x/m u = – g

k x/ u as mk = 2

Integrating,

u2(dy/dx) = – g.(u/k)e

k x/u + C

Initially when x = 0, dy/dx = (dy/dt)/(dx/dt) = v/u, so that

C = u2v/u + (g/k)u

Thus u(dy/dx) = – (g/k)ek x/u

+ (v + g/k)

Integrating again,

= P

[(m + t) ] = 0

[(m + t) ] = A

[(m + t) ] = mu

[(m + t) ] = - (m + t)g

[(m + t) = - (m + t)g ]

[mex/m u

ue–(x/m u)

] ue– (x/m u)

= – mex/m u

g

[mu ] ue– (x/m u)

= – mex/m u

g

173

uy = – (g/k2).ue

k x/u + (v + g/k)x + D

Initially when x = 0, y = 0; so D = gu2/k

2

Hence uy = – (gu/k2)e

k x/u + (v + g/k)x + (g/k

2).u

2

or k2uy = kx(g + kv) + gu(1 – e

k x/u)

Ex.18 A mass of in the form of a solid cylinder, of radius c, acted upon by no forces,

moves to its axis through a uniform cloud of fine dust which of volume density , which

is at rest. If the particles of dust which meet the mass adhere to it, and if M and u be the

mass and velocity at the beginning of the motion, prove that the distance x traversed in

time t is given by the equation (M + c2x)

2 = M

2 + 2uc

2Mt.

Sol.

When the mass M has fallen through a distance x in time t, its mass

= M + mass adhered to it = M + c2.x.

Since there is no force i.e. P = 0, using the relation

d(mv)

dt

we have,

d dx

dt dt

Integrating,

dx

dt

Initially when t = 0, dx/dt = u, so A = mu, so that

dx

dt

or 2c2(M + c

2x)(dx/dt) = 2Mu.c

2

Integrating again, we get

(M + c2x)

2 = 2Mc

2ut + B

Initially when x = 0, t = 0; so B = M2

Hence (M + c2x)

2 = 2Mc

2ut + M

2

Ex.19 A chain, of length l, is coiled at the edge of a table. One end is fastened to a

particle, whose mass is equal to that of the whole chain and the other end is put over the

edge. Show that, immediately after leaving the table, the particle is moving with velocity

½ (5gl/6).

Sol.

Let M be the total mass of the chain, so that the mass per unit length is M/l. When

a length x of the chain has moved over the edge in time t, we have

d dx

dt dt

[since mass of the chain of the length x = (M/l).x]

or d dx dx

dt dt dt

or d dx

dx dt

Integrating,

= P

[(M + c2x.) ] = 0

[(M + c2x) ] = A

[(M + c2x) ] = Mu

[(M/l).x ] = (M/l).xg

= 2x2g ] [x 2x

= 2x2g ]

2 [x

174

or dx

dt

Initially when x = 0, dx/dt = 0; so A = 0, so that

dx

dt

or dx

dt

giving

dx

dt (1)

Hence, just before the mass is jerked, i.e. the chain has run over a length l, the

velocity of the chain say V = [(2/3).gl], by putting x = l in (1).

When the mass is just to run off the plane, let the horizontal velocity

communicated to the mass be V1 and as the mass is just off the plane, let the jerk give it a

vertical velocity V2. Then the law of conservation of momentum gives

(M + M)V1 = MV (2)

and (M + M)V2 = MV1 (3)

(1) gives V1 = V/2 and then (2) gives V2 = V/2

Therefore, total velocity of the particle, then = (V12 + V2

2)

= (V2/4 + V

2/16) = (V/4) 5 = [(2/3).gl] ( 5/4) = ½ (5gl/6)

Ex.20 A uniform chain, of mass M and length l, is coiled up at the top of a rough plane

inclined at an angle to the horizon and has a mass M fastened to one end. This is

projected down the plane with velocity V. If the system comes to rest when the whole of

the chain is just straight, show that V2 = (14gl/3) sec sin(– ), where is the angle of

friction.

Sol.

The mass per unit length of the chain = M/l. Let a length x be uncoiled up the

chain at time t. considering the equilibrium of mass M and a mass of length x of the

chain, we have

The reaction R = [Mg + (M/l).xg] cos

and so R = [Mg + (M/l).xg].cos, where = tan

Resolved part of the weight [Mg + (M/l).xg] along the plane is

[M + (M/l)x]g sin

Therefore the equation of motion is

d dx

dt dt

= [M + (M/l)x]g (sin – cos)

or d dx

dt dt

or dx d dx

dt dx dt

or d dx

dx dt

Integrating,

or dx

= (2/3)x3g + A ]

2 [x

= (2/3)x3g ]

2 [x

= (2/3)xg ]2 [

= (2/3)xg ] [

[{M + (M/l).x} ] = [M + (M/l).x]g sin - R

[(x + l) ] = (l + x)g (sin - cos)

[(x + l) ] = 2(l + x)2g (sin - cos) 2(x + l)

[(x + l) ]2 = 2g(sin - tancos).(x

2 + 2xl + l

2)

[(x + l) ]2

175

dt

= (2g/cos[cos sin – sin cos].[x3/3 + lx

2 + l

2x] + A

Initially when x = 0, dx/dt = V; so A = l2V

2

Hence

dx

dt

= (2g secsin( – ).[x3/3 + lx

2 + l

2x] + l

2V

2

The system again comes to rest when the whole of the chain is just straight, when

x = l, dx/dt = 0. Thus we have

0 = 2g sec sin(- ).[l3/3 + l

3 + l

3] + l

2V

2

or V2 = (14gl/3) sec sin( - )

2.6 Unit Summary:

1. For upwards projection

v2 = u

2 – 2gh

and for downward projection

v2 = u

2 + 2g

where h is the vertical distance between the two points.

2. In the case in which a heavy particle is sliding upwards on a curve in a vertical plane,

the equations are

mv.(dv/ds) = - R - mg sin

and mv2/ = R - mg cos

where is the inclination of the tangent to the horizontal.

3. Suppose that a particle gains in time t an increment m of mass and that this increment

m was moving with a velocity u.

Then in time t the increment in the momentum of the particle

= m.v + m(v + v – u)

and the impulse of the force in this time is Pt.

Equating these we have, on proceeding to the limit.

m(dv/dt) + v(dm/dt) – u(dm/dt) = P

i.e. d dm

dt dt

2.7 Assignments:

1. A particle moves in a smooth tube in the form of a catenary, being attracted to the

directrix by a force proportional to the distance from it. Shew that the motion is simple

harmonic.

2. A smooth parabolic tube is placed, vertex downwards, in a vertical plane; a particle

slides down the tube from rest under the influence of gravity; prove that in any position

the reaction of the tube is 2w.[(h + a)/], where w is the weight of the particle, the

radius of curvature, 4a the latus rectum and h the original vertical height of the particle

above the vertex.

[(x + l) ]2

(mv) = P + u

176

3. A curve in a vertical plane is such that the time of describing any arc, measured from a

fixed point O, is equal to the time of sliding down the chord of the arc; shew that the

curve is a lemniscate of Bernouilli, whose node is at O and whose axis is inclined at 450

to the vertical.

4. A particle, of mass m, moves in a smooth circular tube, of radius a, under the action of

a force, equal to m distance, to a point inside the tube at a distance c from its centre; if

the particle be placed very nearly at its greatest distance from the centre of force, show

that it will describe the quadrant ending at it least distance in time

(a/c) log(2 + 1)

5. A bead is constrained to move on a smooth wire in the form of an equiangular spiral. It

is attracted to the pole of the spiral by a force, =m(distance) 2

and starts from rest at a

distance b from the pole. Shew that, if the equation to the spiral be r = aecot

, the time of

arriving at the pole is (/2).(b3/2).sec. Find also the reaction of the curve at any

instant.

6. A particle is projected horizontally from the lowest point of a rough sphere of radius a.

After describing an arc less than a quadrant it returns and comes to rest at the lowest

point. Shew that the initial velocity must be sin[2ga(1 + 2)/(1 – 2

2)], where is the

coefficient of friction and a is the arc through which the particle moves.

7. A bead slides down a rough circular wire, which is in a vertical plane, starting from

rest at the end of a horizontal diameter. When it has described an angle about the

centre, show that the square of its angular velocity is

2g

a(1 + 42)

where is the coefficient of friction and a the radius of the rod.

8. A ring which can slide on a rough circular wire in a vertical plane is projected from the

lowest point with such velocity as will just take it to the horizontal diameter; if the ring

returns to the lowest point, show that its velocity on arrival is to its velocity on departure

as (1 – 22 – 3e

–)1/ 2

: (1 – 22 + 3e

)1/ 2

,

where is the coefficient of friction.

9. A heavy particle slides, from cusp, down the arc of a rough cycloid, the axis of which

is vertical. Prove that its velocity at the vertex is to its velocity at the same point when the

cycloid is smooth as (e–

– 2) : (1 +

2), where is the coefficient of friction.

10. A particle is projected with velocity V along a smooth horizontal plane in a medium

whose resistance per unit mass in times the cube of the velocity. Shew that the distance

it has described in time t is

(1/V)[(1 + 2V2t) – 1]

and that its velocity then is

V/(1 + 2V2t).

11. A heavy particle is projected vertically upwards with a velocity u in a medium the

resistance of which varies as the cube of the particles velocity. Determine the height to

which the particle will ascend.

12. If the resistance to the motion of a railway train vary as its mass and the square of the

velocity and the engine work at constant H.P., show that full speed will never be attained

and that the distance traveled from rest when half-speed is attained is (1/3) log(8/7),

[(1 – 22) sin + 3 (cos – e

– 2)],

177

where is the resistance per unit mass per unit velocity. Find also the time of describing

this distance.

13. A ship, with engines stopped, is gradually brought to rest by the resistance of the

water. At one instant the velocity is 10 ft. per sec. and one minute later the speed has

fallen to 6 ft. per sec. For speeds below 2 ft. per sec. the resistance may be taken to vary

as the speed and for higher speeds to vary as the square of the speed. Show that, before

coming to rest, the ship will move through 900[1 + log5] feet, from the point when the

first velocity was observed.

14. A particle moves from rest at a distance a from a fixed point O under the action of a

force to O equal to times the distance per unit of mass; if the resistance of the medium

in which it moves be k times the square of the velocity per unit of mass, show that the

square of the velocity when it is at a distance x from O is

(x/k) – (a/k).e2k(x – a)

+ (/2k2)[1 – e

2k(x – a)].

Show also that when it first comes to rest it will be at a distance b given by

(1 – 2kb)e2kb

= (1 + 2ka)e-2ak

.

15. Show that in the motion of a heavy particle in a medium, the resistance of which

varies as the velocity, the greatest height above the level of the point of projection is

reached in less than half the total time of the flight above that level.

16. If the resistance of the air to a particle‟s motion be n times its weight and the particle

be projected horizontally with velocity V, show that the velocity of the particle, when it is

moving at an inclination to the horizontal, is V(1 – sin)(n – 2)/2

(1 + sin) - (n + 1)/2

.

17. A heavy bead slides down a smooth wire in the form of a cycloid, whose axis is

vertical and vertex downwards, from rest at a cusp and is acted on besides its weight by a

tangential resistance proportional to the square of the velocity. Determine the velocity

after a fall through the height x.

18. If a body move under a central force in a medium which exerts a resistance equal to k

times the velocity per unit of mass, prove that

d2u P

d2 h

2u

2

where h is twice the initial momentum of momentum about the centre of force.

19. A particle moves with a central acceleration P in a medium of which the resistance is

k (velocity)2; show that the equation to its path is

d2u P

d2 h

2u

2

where s is the length of the arc described and h is twice the initial moment of momentum

about the centre of force.

20. A uniform chain is partly coiled on a table, one end of it being just carried over a

smooth pulley at a height h immediately above the coil and attached there to a weight

equal to that of a length 2h of the chain. Show that until the weight strikes the table, the

chain uncoils with uniform acceleration g/3 and that, after it strikes the table, the velocity

at any moment is (2gh/3)e(x – h)/2h

, where x is the length of the chain uncoiled.

21. A chain, of mass m and length 2l, hangs in equilibrium over a smooth pulley when an

insect of mass M alights gently at one end and begins crawling up with uniform velocity

V relative to the chain; Show that the velocity with which the chain leaves the pulley will

be

+ u = e2kt

+ u = e2ks

178

M2 2M + m

(M + m)2 M + m

22. One end of a heavy uniform chain, of length 5a and mass 5ma, is fixed at a point O

and the other passes over a small smooth peg at a distance a above O; the whole hangs in

equilibrium with the free end at a depth 2a below the peg. The free end is slightly

displaced downwards; prove that its velocity V, when the length of the free portion is x,

is given by

2(x – 2a)2(x + 10a)g

(x + 6a)2

and find the impulsive tension at O at the instant when the part of the chain between O

and the peg becomes tight.

23. A spherical raindrop, whose radius is 0.04 inches, begins to fall from a height of 6400

feet and during the fall its radius grows, by precipitation of moisture, at the rate of 10-4

inches per second. If its motion be unresisted, show that its radius when it reaches the

ground is 0.0420 inches and that it will have taken about 20 seconds to fall.

24. A string of length l, hangs over a smooth peg so as to be at rest. One end is ignited

and burns away at a uniform rate v. Show that the other end will at time t be at a depth x

below the peg, where x is given by the equation (l – vt).(d2x/dt

2 + g) – v(dx/dt) – 2gx = 0.

2.8 References:

1. Loney, S.L.: An Elementary Treatise on the Dynamics of a Particle and of Rigid

Bodies, S.Chand & Company (Pvt.) Ltd, Ram Nagar, New Delhi,

1952.

2. Lamb, Horace: Dynamics: Cambridge University Press, 1961

3. Ramsey, A.S.: Dynamics Part I & II, : Cambridge University Press,1954

V2 + [ gl]

1/2

V2 =