116 CH A P T E R 13 KI N E T I C S O F A PA RT I C L E : FO R C E A N D AC C E L E R AT I O N

13

The 50-kg crate shown in Fig. 136a rests on a horizontal surface forwhich the coefficient of kinetic friction is If the crate issubjected to a 400-N towing force as shown, determine the velocity ofthe crate in 3 s starting from rest.

SOLUTIONUsing the equations of motion, we can relate the crates accelerationto the force causing the motion. The crates velocity can then bedetermined using kinematics.

Free-Body Diagram. The weight of the crate is As shown in Fig. 136b, the frictional

force has a magnitude and acts to the left, since it opposes themotion of the crate. The acceleration a is assumed to act horizontally, inthe positive x direction.There are two unknowns, namely and a.

Equations of Motion. Using the data shown on the free-bodydiagram, we have

(1)

(2)

Solving Eq. 2 for substituting the result into Eq. 1, and solvingfor a yields

Kinematics. Notice that the acceleration is constant, since theapplied force P is constant. Since the initial velocity is zero, thevelocity of the crate in 3 s is

Ans.= 15.6 m>s :v = v0 + act = 0 + 5.1851321:+ 2

a = 5.185 m>s2NC = 290.5 N

NC ,

NC - 490.5 + 400 sin 30 = 0+ cFy = may ;400 cos 30 - 0.3NC = 50a:+ Fx = max ;

NC

F = mkNC50 kg 19.81 m>s22 = 490.5 N. W = mg =

mk = 0.3.

EXAMPLE 13.1

P 400 N

30

(a)

30

400 N

490.5 N

F 0.3 NC

NC

(b)

y

x

a

30

400 N

490.5 N

F 0.3NC

NC (c)

50a

Fig. 136

NOTE: We can also use the alternative procedure of drawing thecrates free-body and kinetic diagrams, Fig. 136c, prior to applyingthe equations of motion.

13.4 EQUATIONS OF MOTION: RECTANGULAR COORDINATES 117

13

EXAMPLE 13.2

A 10-kg projectile is fired vertically upward from the ground, with aninitial velocity of Fig. 137a. Determine the maximum heightto which it will travel if (a) atmospheric resistance is neglected; and(b) atmospheric resistance is measured as where isthe speed of the projectile at any instant, measured in

SOLUTIONIn both cases the known force on the projectile can be related to itsacceleration using the equation of motion. Kinematics can then beused to relate the projectiles acceleration to its position.

Part (a) Free-Body Diagram. As shown in Fig. 137b, the projectilesweight is We will assume theunknown acceleration a acts upward in the positive z direction.

Equation of Motion.

The result indicates that the projectile, like every object having free-flight motion near the earths surface, is subjected to a constantdownward acceleration of

Kinematics. Initially, and and at the maximumheight Since the acceleration is constant, then

Ans.

Part (b) Free-Body Diagram. Since the force tends to retard the upward motion of the projectile, it acts downwardas shown on the free-body diagram, Fig. 137c.

Equation of Motion.

Kinematics. Here the acceleration is not constant since dependson the velocity. Since we can relate a to position using

Separating the variables and integrating, realizing that initially (positive upward), and at we have

Ans.

NOTE: The answer indicates a lower elevation than that obtained inpart (a) due to atmospheric resistance or drag.

h = 114 mL

h

0dz = -

L

0

50

v dv

0.001v2 + 9.81= -500 ln1v2 + 98102 `

50 m>s0

v = 0,z = h,v0 = 50 m>sz0 = 0,

-10.001v2 + 9.812 dz = v dv1+ c2 a dz = v dv;a = f1v2, FD

a = -(0.001v2 + 9.81)-0.01v2 - 98.1 = 10a,+ cFz = maz ;

FD = 10.01v22 Nh = 127 m 0 = 15022 + 21-9.8121h - 02v2 = v02 + 2ac1z - z021+ c2

v = 0.z = h,v0 = 50 m>s,z0 = 0

9.81 m>s2.

a = -9.81 m>s2-98.1 = 10a,+ cFz = maz ;

W = mg = 1019.812 = 98.1 N.

m>s. vFD = 10.01v22 N,

50 m>s,z

(a)

z

98.1 N

a

(b)

z

98.1 N

a

(c)

FD

Fig. 137

118 CH A P T E R 13 KI N E T I C S O F A PA RT I C L E : FO R C E A N D AC C E L E R AT I O N

13

The baggage truck A shown in the photo has a weight of 900 lb andtows a 550-lb cart B and a 325-lb cart C. For a short time the drivingfrictional force developed at the wheels of the truck is where t is in seconds. If the truck starts from rest, determine its speedin 2 seconds. Also, what is the horizontal force acting on the couplingbetween the truck and cart B at this instant? Neglect the size of thetruck and carts.

FA = 140t2 lb,

EXAMPLE 13.3

A B C

NA NBNC

FA

900 lb550 lb 325 lb

(a)

SOLUTIONFree-Body Diagram. As shown in Fig. 138a, it is the frictionaldriving force that gives both the truck and carts an acceleration. Herewe have considered all three vehicles as a single system.

Equation of Motion. Only motion in the horizontal direction hasto be considered.

Kinematics. Since the acceleration is a function of time, the velocityof the truck is obtained using with the initial condition that

at We have

Ans.

Free-Body Diagram. In order to determine the force between thetruck and cart B, we will consider a free-body diagram of the truck sothat we can expose the coupling force T as external to the free-bodydiagram, Fig. 138b.

Equation of Motion. When then

Ans.

NOTE: Try and obtain this same result by considering a free-bodydiagram of carts B and C as a single system.

T = 39.4 lb

40122 - T = a 90032.2

b [0.7256122];+ Fx = max ;t = 2 s,

v = 0.3628t2 `0

2 s

= 1.45 ft>sL

v

0dv =

L

2 s

00.7256t dt;

t = 0.v0 = 0a = dv>dt

a = 0.7256t

40t = a900 + 550 + 32532.2

ba;+ Fx = max ;

NA

T

FA

900 lb

(b)

Fig. 138

120 CH A P T E R 13 KI N E T I C S O F A PA RT I C L E : FO R C E A N D AC C E L E R AT I O N

13

The 100-kg block A shown in Fig. 1310a is released from rest. If themasses of the pulleys and the cord are neglected, determine the speedof the 20-kg block B in 2 s.

SOLUTION

Free-Body Diagrams. Since the mass of the pulleys is neglected,then for pulley C, and we can apply as shown inFig. 1310b. The free-body diagrams for blocks A and B are shownin Fig. 1310c and d, respectively. Notice that for A to remainstationary whereas for B to remain static Hence A will move down while B moves up. Although this is thecase, we will assume both blocks accelerate downward, in thedirection of and The three unknowns are T, and

Equations of Motion. Block A,

(1)

Block B,

(2)

Kinematics. The necessary third equation is obtained by relating to using a dependent motion analysis, discussed in Sect. 12.9. Thecoordinates and in Fig. 1310a measure the positions of A and Bfrom the fixed datum. It is seen that

where l is constant and represents the total vertical length of cord.Differentiating this expression twice with respect to time yields

(3)

Notice that when writing Eqs. 1 to 3, the positive direction was alwaysassumed downward. It is very important to be consistent in thisassumption since we are seeking a simultaneous solution of equations.The results are

Hence when block A accelerates downward, block B acceleratesupward as expected. Since is constant, the velocity of block B in 2 sis thus

Ans.

The negative sign indicates that block B is moving upward.

= -13.1 m>s= 0 + 1-6.542122

v = v0 + aBt1+ T2aB

aB = -6.54 m>s2aA = 3.27 m>s2T = 327.0 N

2aA = -aB

2sA + sB = l

sBsA

aB

aA

196.2 - T = 20aB+ TFy = may ;

981 - 2T = 100aA+ TFy = may ;

aB .aA ,+sB .+sA

T = 196.2 N.T = 490.5 N,

Fy = 0ma = 0

EXAMPLE 13.5

T

196.2 NsB

(d)

aB

Fig. 1310

Datum

A

CsB

sA

(a)

B

TT

2T

(b)

2T

981 NsA

(c)

aA