I. Newton’s first law.

II. Newton’s second law.

III. Particular forces:-Gravitational- Weight- Normal - Friction- Tension

IV. Newton’s third law.

Chapter 5 – Force and Motion I

I. Newton’s first law: If no net force acts on a body, then the body’s velocitycannot change; the body cannot accelerate v = constant in magnitude and direction.

- Principle of superposition: when two or more forces act on a body, the netforce can be obtained by adding the individual forces vectorially.

Newton mechanics laws cannot be applied when:

1) The speed of the interacting bodies are a fraction of the speed oflightEinstein’s special theory of relativity.

2) The interacting bodies are on the scale of the atomic structure Quantum mechanics

- Inertial reference frame: where Newton’s laws hold.

II. Newton’s second law: The net force on a body is equal to the product of the body’s mass and its acceleration.

)1.5(amFnet

)2.5(,, ,,, zznetyynetxxnet maFmaFmaF

- The acceleration component along a given axis is caused only by the sum of the force components along the same axis, and not by force componentsalong any other axis.

- System: collection of bodies.

- External force: any force on the bodies inside the system.

III. Particular forces:

-Gravitational: pull directed towards a second body, normally the Earth

)3.5(gmFg

)4.5(mgW

- Weight: magnitude of the upward force needed to balance the gravitationalforce on the body due to an astronomical body

)5.5(mgN

- Normal force: perpendicular force on a body from a surface against whichthe body presses.

- Frictional force: force on a body when the bodyattempts to slide along a surface. It is parallelto the surface and opposite to the motion.

-Tension: pull on a body directed away from thebody along a massless cord.

IV. Newton’s third law: When two bodies interact, the forces on the bodiesfrom each other are always equal in magnitude and opposite in direction.

)6.5(CBBC FF

QUESTIONS

Q2. Two horizontal forces F1, F2 pull a banana split across a frictionless counter.Without using a calculator, determine which of the vectors in the free body diagrambelow best represent: a) F1, b)F2. What is the net force component along (c) thex-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) thesplit’s acceleration vector point?

jNiNF

jNiNFˆ)2(ˆ)1(

ˆ)4(ˆ)3(

2

1

jNiNFFFnetˆ)6(ˆ)2(21

Same quadrant, 4 F1F2

I. Frictional force

-Kinetic: (fk) appears after a large enough external force is applied and the body loses its intimate contact with the surface, sliding along it.

No motion

Acceleration

Constant velocity

Counter force that appears when an external force tends to slide a body along a surface. It is directed parallel to the surface and opposite to the sliding motion.

F(applied

force)

//Ffs

-Static: (fs) compensates the applied force, the body doesnot move.

max,sk ff )1.6(max, Nf ss

)2.6(Nf kk

Friction coefficientsslidesbodyfFIf s max,//

After the body starts sliding, fk decreases.

Q1. The figure below shows overhead views of four situations in which forces acton a block that lies on a frictionless floor. If the force magnitudes are chosenproperly, in which situation it is possible that the block is (a) stationary and(b) moving with constant velocity?

ay≠0a=0

ay≠0

a=0

Fnet

Fnet

Q5. In which situations does theobject acceleration have (a) anx-component, (b) a y component?(c) give the direction of a.

Q. A body suspended by a rope has a weigh of 75N. Is T equal to, greater than,or less than 75N when the body is moving downward at (a) increasing speed and(b) decreasing speed?

Fg

)( agmTamTFF gnet

Movem

ent

(a) Increasing speed: vf >v0a>0 T< Fg

(b) Decreasing speed: vf < v0 a<0 T> Fg

Q8. The figure below shows a train of four blocks being pulledacross a frictionless floor by force F. What total mass isaccelerated to the right by (a) F, (b) cord 3 (c) cord 1? (d) Rank theblocks according to their accelerations, greatest first. (e) Rank thecords according to their tension, greatest first.

T1 T2 T3

(a) F pulls mtotal= (10+3+5+2)kg = 20kg

(b) Cord 3 T3 m=(10+3+5)kg = 18kg

(c) Cord 1 T1 m= 10kg

(d) F=ma All tie, same acceleration

(e) F-T3= 2aT3-T2= 5aT2-T1=3aT1=10a

F-T3= 2a F=18a+2a=20aT3-13a= 5a T3=18aT2-10a=3a T2=13aT1=10a

Q. A toy box is on top of a heavier dog house, which sits on a wood floor. Theseobjects are represented by dots at the corresponding heights, and six verticalvectors (not to scale) are shown. Which of the vectors best represents (a) thegravitational force on the dog house, (b) on the toy box, (c) the force on the toy boxfrom the dog house, (d) the force on the dog house from the toy box, (e) force on thedog house from the floor, (f) the force on the floor from the dog house? (g) Which ofthe forces are equal in magnitude? Which are (h) greatest and (i) least inmagnitude? (a) Fg on dog house: 4 or 5

(c) Ftoy from dog house: 1

(b) Fg on toy box: 2

(d) Fdog-house from toy box: 4 or 5

(e) Fdog-house from floor: 3

(f) Ffloor from dog house: 6

(g) Equal: 1=2, 1=5, 3=6

(h) Greatest: 6,3(i) Smallest: 1,2,5

5. There are two forces on the 2 kg box in the overhead view of the figure belowbut only one is shown. The figure also shows the acceleration of the box. Find thesecond force (a) in unit-vector notation and as (b) magnitude and (c) direction.

213331803332

78.20tan27.382132

)ˆ78.20ˆ32(

78.20322012

ˆ20

)ˆ78.20ˆ12(/)ˆ39.10ˆ6(2

/)ˆ39.10ˆ6(/)ˆ240sin12ˆ240cos12(

222

2

2

22

221

2

22

orNF

NjiF

FNFNFNFNF

FiFFF

NjismjikgamF

smjismjia

yTy

xxTx

T

T

F2

Rules to solve Dynamic problems

- Select a reference system.

- Make a drawing of the particle system.

- Isolate the particles within the system.

- Draw the forces that act on each of the isolated bodies.

- Find the components of the forces present.

- Apply Newton’s second law (F=ma) to each isolated particle.

9. (a) A 11kg salami is supported by a cord that runs to a spring scale, which issupported by another cord from the ceiling. What is the reading on the scale, which ismarked in weigh units? (b) Here the salami is supported by a cord that runs around apulley and to a scale. The opposite end of the scale is attached by a cord to a wall.What is the reading on the scale? (c) The wall has been replaced by a second salamion the left, and the assembly is stationary. What is the reading on the scale now?

T

Fg

T

T

TNFTaa

NsmkgmgFW

g

g

8.1070)(

8.107)/8.9)(11( 2

T

T

Fg

T

NFTab g 8.1070)(

T

Fg

T

Fg

T T

NFTac g 8.1070)(

In all three cases the scale is not accelerating, which means that the twocords exert forces of equal magnitude on it. The scale reads themagnitude of either of these forces. In each case the tension force ofthe cord attached to the salami must be the same in magnitude as theweigh of the salami because the salami is not accelerating.

23. An electron with a speed of 1.2x107m/s moves horizontally into a region where aconstant vertical force of 4.5x10-16N acts on it. The mass of the electron ism=9.11x10-31kg. Determine the vertical distance the electron is deflected during thetime it has moved 30 mm horizontally.

nsttsmmtvd xx 4.2)/102.1(03.0 7

dx=0.03m

dyF

v0Fg

mssmtatvd yoyy 0015.0)105.2()/1094.4(5.05.0 292142

21431

23116

/1094.4)1011.9(

)/8.9)(1011.9(105.4

smaakgF

smkgNFFmaF

yynet

gynet

13. In the figure below, mblock=8.5kg and θ=30º. Find (a) Tension in the cord.(b) Normal force acting on the block. (c) If the cord is cut, find the magnitudeof the block’s acceleration.

TN

Fg

2/9.45.865.410)( smaaNmaFTc gx

NsmkgmgFT

FTaaa

gx

gxx

65.415.0)/8.9)(5.8(30sin

000)(2

NmgFNFNab gygyy 14.7230cos00)(

55. The figure below gives as a function of time t, the force component Fx that acts on a 3kg ice block,which can move only along the x axis. At t=0, the block is moving in the positive direction of the axis, witha speed of 3m/s. What are (a) its speed and (b) direction of travel at t=11s?

sx

fxxx

x

f

dtmFvvdt

dtdv

dtdv

mFa

vstsmvt

11

00

0

?11/30

smsmkg

skgmv

kgsmvmvvdtFNsareagraphTotal

f

f

s

fx

/8/33

/15

3)/3()(1511

00

Midterm1_extra_Spring04. Two bodies, m1= 1kg and m2=2kg are connected over a massless pulley. Thecoefficient of kinetic friction between m2 and the incline is 0.1. The angle θ of the incline is 20º.Calculate:

(a) Acceleration of the blocks. (b) Tension of the cord.

NTsmaaAdding

aTamFfTBlock

aTamTgmBlock

NgmNf

NgmFN

NgmF

xg

kk

yg

xg

38.9,/42.026.13

27.684.1:2

8.9:1

84.120cos

42.1820cos

7.620sin

2

2,2

11

22

2,22

2,2

T

N

m1

m2

20ºm1g

T

m2g

f

Midterm1_Spring04. The three blocks in the figure below are connected by massless cords and pulleys. Data: m1=5kg, m2=3kg, m3=2kg. Assume that the incline plane is frictionless.

(i) Show all the forces that act on each block.(ii) Calculate the acceleration of m1, m2, m3.

m1

m2

m3 30º

m3g

T2

T1

m2g

T2

T1

N

Fg2x

Fg2y(iii) Calculate the tensions on the cords. (iv) Calculate the normal force acting on m2

Fg2y=m2gcos30ºFg2x=m2gsin30º m1g

Block 1: m1g-T1=m1aBlock 2: m2g(sin30º) +T1-T2=m2aBlock 3: T2-m3g = m3a

(i) Adding (1)+(2)+(3) g(m1+0.5m2-m3)=a(m1+m2+m3) a= 4.41m/s2

(ii) T1=m1(g-a)= 5kg(9.8 m/s2-4.41 m/s2) = 26.95N

(iii) T2=m3(g+a)= 2kg(9.8 m/s2+4.41 m/s2)= 28.42N

(iv) N2= Fg2y= m2gcos30º = 25.46N

1B. (a) What should be the magnitude of F in the figure below if the body of massm=10kg is to slide up along a frictionless incline plane with constant accelerationa=1.98 m/s2? (b) What is the magnitude of the Normal force?

NgamFmamgF 21.7320cos

)5.0(30sin20cos

y

x

30º

FN

Fg

20º

NNFmgN 9.109020sin30cos

2B. Given the system plotted below, where m1=2kg and m2=6kg, calculate theforce F necessary to lift up m2 with a constant acceleration of 0.2m/s2. The pulleysand cords are massless, and the table surface is frictionless.

Fm1

m2

N

m1g

T

m2g

TT

2T

21

212

22

21

22

12

211

/4.0/2.05.05.021

2

21

smasmaatatatadd

tad

11

2222

amTF

amgmT

NsmkggamT 30/)8.92.0)(6(5.0)(5.0 222

Movement

NsmkgNamTF 8.30)/4.0)(2(30 211

Chapter 5 – Force and Motion II

I. Drag forces and terminal speed.

II. Uniform circular motion.

III. Non-Uniform circular motion.

I. Drag force and terminal speed

-Fluid: anything that can flow. Example: gas, liquid.

-Drag force: D

- Appears when there is a relative velocity between a fluid and a body.

- Opposes the relative motion of a body in a fluid.

- Points in the direction in which the fluid flows.

Assumptions:* Fluid = air.* Body is blunt (baseball).* Fast relative motion turbulent air.

)3.6(21 2AvCD

)4.6(2

ACF

v gt

-Terminal speed: vt

0210 2 gg FAvCaifmaFD

C = drag coefficient (0.4-1).

ρ = air density (mass/volume).

A= effective body’s cross sectional area area perpendicular to v

- Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero Fg=D

II. Uniform circular motion

-Centripetal acceleration:

)6.6(2

RvmF

)5.6(2

rva

A centripetal force accelerates a body by changing the direction of thebody’s velocity without changing its speed.

v, a = constant, but direction changes during motion.

-Centripetal force:

a, F are directed toward the center of curvatureof the particle’s path.

III. Non-Uniform circular motion

22tr aaa

- A particle moves with varying speed in a circular path.

- The acceleration has two components: - Radial ar = v2/R

- Tangential at = dv/dt

- at causes the change in the speed of the particle.

- In uniform circular motion, v = constant at = 0 a = ar

rrv

dtvd

aaa rt ˆˆ2

tr FFF

49. A puck of mass m slides on a frictionless table while attached to a hangingcylinder of mass M by a cord through a hole in the table. What speed keepsthe cylinder at rest?

Mg

mg T

TN

0 caMgTMFor

33E. Calculate the drag force on a missile 53cm in diameter cruising with aspeed of 250m/s at low altitude, where the density of air is 1.2kg/m3.Assume C=0.75

kNsmmmkgAvCD 2.6/250)2/53.0()/2.1(75.05.021 2232

32. The terminal speed of a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nose-dive position. Assuming that the diver’s drag coefficient C does not change from one point toanother, find the ratio of the effective cross sectional area A in the slower position to that of thefaster position.

7.32

2

/310/1602

D

E

E

D

D

g

E

g

gt A

AAA

ACFACF

hkmhkm

ACF

v

mMgrv

rvmMg

rvmTmFor

22

11P. A worker wishes to pile a cone of sand onto a circular area in his yard. The radius of the circle isR, and no sand is to spill into the surrounding area. If μs is the static coefficient of friction betweeneach layer of sand along the slope and the sand beneath it (along which it might slip), show thatthe greatest volume of sand that can be stored in this manner is π μs R3/3. (The volume of a coneis Ah/3, where A is the base area and h is the cone’s height).

- To pile the most sand without extending the radius, sand is added to make theheight “h” as great as possible.

- Eventually, the sides become so steep that sand at the surface begins to slip.

- Goal: find the greatest height (greatest slope) for which the sand does not slide.

Cross section of sand’s cone

R

h

θ

N

mg

x

yf

Fgx

Fgy

Static friction grain does not move

sin

cos

mgFf

mgFN

gx

gy

tancossin max, ssssgx mgNfmgF

If grain does not slide

The surface of the cone has the greatest slope and the height of the cone ismaximum if :

33)(

3

tan

32 RRRhAV

RhRh

sscone

ss

21. Block B weighs 711N. The coefficient of static friction between the block and the table is 0.25;assume that the cord between B and the knot is horizontal. Find the maximum weight of block Afor which the system will be stationary.

FgA

N

FgB

T1

f

T3

T1

T2

T3

NfstationarySystem ss max,

NNTfT

gmNBBlock

s

B

75.17771125.00 1max,1

322

2221

30sin

25.20530cos75.17730cos

TTT

NNTTTTKnot

y

x

NNTgmTABlock A 62.10225.2055.030sin23

23P. Two blocks of weights 3.6N and 7.2N, are connected by a massless string and slide down a 30ºinclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10;that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find(a) the magnitude of the acceleration of the blocks and (b) the tension in the string. (c) Describethe motion if, instead, the heavier block leads.

A

B

FgA

FgB

NA

NB

T

T

fk,B

fk,A

Block A Block BNA NB

fkAfkB

TFgxA

FgyA

FgxB

FgyB

T

Light block A leads

Light block A leads

Heavy block B leads

aTaNTNamfTF

NNNf

NgmFNBBlock

aTaTNNamTfF

NNNf

NgmFNABlock

BkBgxB

BkBkB

BgyBB

AkAgxA

AkAkA

AgyAA

73.035.273.025.130sin)2.7(

25.1)23.6)(2.0(

23.630cos

37.049.137.0312.030sin)6.3(

312.0)12.3)(1.0(

12.330cos

NTsma

2.0/49.3 2

NWW

WWT kAkBBA

BA 2.0cos

The above set of equations is not valid in this circumstance aA≠ aBThe blocks move independentlyfrom each other.

Reversing the blocks is equivalent to switching the labels. This would give T~(μkA-μkB)<0 impossible!!!

NNNfNNNf

kk

ss

8.22)60(38.033)60(55.0max,

74. A block weighing 22N is held against a vertical wall by a horizontal force F of magnitude 60N. Thecoefficient of static friction between the wall and the block is 0.55 and the coefficient of kineticfriction between them is 0.38. A second P acting parallel to the wall is applied to the block. For thefollowing magnitudes and directions of P, determine whether the block moves, the direction ofmotion, and the magnitude and direction of the frictional force acting on the block: (a) 34N up(b) 12N up, (c) 48N up, (d) 62N up, (e) 10N down, (f) 18N down.

F=60N(a) P=34N, up

mg=22N

N

N=F=60N

P

22Nf

Without P, the block is at rest

movenotdoesBlockNff

downNffNN

affassumeweIfmafmgP

s

s

33

122234

0

max,

(b) P=12N, up

P

22N

f

movingNotNff

upNNNfmamgfP

s

33

1012220

max,

(c) P=48N, up

P

22Nf movingNotNff

downNNNfmamgfP

s

33

2622480

max,

(d) P=62N, up P

22Nf

downNffwithmamgfP

wrongAssumptionupmovesBlockNff

upNNNfmgfP

k

s

8.22

(*)33

402262(*)0

max,

(e) P=10N, down

P

22N

f

movingNotNff

upNNNfmamgPf

s

33

3212220

max,

(f) P=18N, down

P22N

f

upNffmovesNff

upNNNfmamgPf

k

s

8.2233

4022180

max,

28. Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight ofblock C to keep A from sliding if μs between A and the table is 0.2. (b) Block C suddenly is lifted ofA. What is the acceleration of block A if μk between A and the table is 0.15?

T

T

WB=22N

WA=44N

f

Wc

N)1(00 max,

max,

NTfTaABlockNff

ss

ss

)2(220 NTgmTBBlock B

NNTNs

1102.0

22)2()1(

NNNWWWNBABlocks CCA 6644110,

amTgmamNT

NgmNdisappearsC

BB

Ak

A

44

(a)

(b)

NTaTsmaaT

172.222/3.25.46.6 2

29. The two blocks (with m=16kg and m=88kg) shown in the figure below are not attached. The coefficientof static friction between the blocks is: μs=0.38 but the surface beneath the larger block is frictionless.What is the minimum value of the horizontal force F required to keep the smaller block from slippingdown the larger block?

MmFaamF total

)2(0'0

)1('

mgFmgfMm

FmmaFFblockSmall

ss

NM

MmmgFmgMm

FMs

s 488)2()1(

Mg

N

F’

mg

fF’

Movement

Fmin required to keep m from sliding down?

Treat both blocks as a single system sliding across a frictionless floor

44. An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom ofnegligible mass. The combined weigh of the car and riders is 5kN, and the radius of the circle is 10m.What are the magnitude and the direction of the force of the boom on the car at the top of the circle ifthe car’s speed is (a) 5m/s (b) 12m/s?

RgvWF

RvmWF BB

221

The force of the boom on the car is capable of pointing any directionFB

W

y

downFsmvbupNFsmva BB 3.2/12)(7.3/5)(