SM PDF Chapter5

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113

More Applications oNewton’s Laws

CHAPTER OUTLINE

5.1 Forces of Friction5.2 Newton’s Second

Law Applied to aParticle in UniformCircular Motion

5.3 Nonuniform CircularMotion

5.4 Motion in the Presenceof Velocity-DependentResistive Forces

5.5 The Fundamental Forcesof Nature

5.6 ContextConnection DragCoefficients of

Automobiles

ANSWERS TO QUESTIONS

Q5.1 (a) m mr r ra R g= + (b) ma T m= − (c) ma R= −

r

r

r

r

r

r

r

f r

FIG. Q5.1

Q5.2 (a) The friction of the road pushing on the tires of a car causes an automobile to move.

(b) The push of the air on the propeller moves the airplane.

(c) The push of the water on the oars causes the rowboat to move.

Q5.3 As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force ofthe Earth on his foot. In the second case, the action is the force exerted on the girl’s back by thesnowball; the reaction is the force exerted on the snowball by the girl’s back. The third action is theforce of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is theforce exerted on the window by the air molecules; the reaction is the force on the air moleculesexerted by the window. We could in each case interchange the terms ‘action’ and ‘reaction.’

Q5.4 The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then theresultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers werewinning with the rope steadily moving in their direction or if the contest was even, then the tensionwould still be 9 200 N. In all of these cases, the acceleration is zero, and so must be the resultant forceon the rope. To win the tug-of-war, a team must exert a larger force on the ground than theiropponents do.

Q5.5 If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between thetires and the road is less than the maximum static friction force. Anti-lock brakes work by “pumping”the brakes (much more rapidly that you can) to minimize skidding of the tires on the road.



114 More Applications of Newton’s Laws

Q5.6 With friction, it takes longer to come down than to go up. On the way up, the frictional force and thecomponent of the weight down the plane are in the same direction, giving a large acceleration. Onthe way down, the forces are in opposite directions, giving a relatively smaller acceleration. If theincline is frictionless, it takes the same amount of time to go up as it does to come down.

Q5.7 As you pull away from a stoplight, friction is the force that accelerates forward a box of tissues on the

level floor of the car. At the same time, friction of the ground on the tires of the car accelerates thecar forward. Drop a stick into a running stream and fluid friction sets the stick into horizontalmotion.

Q5.8 The speed changes. The tangential force component causes tangential acceleration.

Q5.9 A torque is exerted by the thrust force of the water times the distance between the nozzles.

Q5.10 This is the same principle as the centrifuge. All the material inside the cylinder tends to move alonga straight-line path, but the walls of the cylinder exert an inward force to keep everything movingaround in a circular path.

Q5.11 The water has inertia. The water tends to move along a straight line, but the bucket pulls it in andaround in a circle.

Q5.12 Blood pressure cannot supply the force necessary both to balance the gravitational force and toprovide the centripetal acceleration, to keep blood flowing up to the pilot’s brain.

Q5.13 I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” onewould have to be infinitely far away from all other matter. Second, astronauts in orbit are moving ina circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In thespace shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface.Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces andis in free fall.

Q5.14 From the proportionality of the drag force to the speed squared and from Newton’s second law, wederive the equation that describes the motion of the skydiver:

mdv

dtmg

D Av

y y= −

ρ

22

where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutist’s body. At terminal speed,

adv

dt y y= = 0 and v

mg D AT =

F H G

I K J 2

1 2

ρ .

When the parachute opens, the coefficient of drag D and the effective area A both increase, thusreducing the speed of the skydiver. Modern parachutes also add a third term, lift, to change theequation to

mdv

dtmg

D Av

L Av

y y x= − −

ρ ρ

2 22 2

where v y is the vertical velocity, and vx is the horizontal velocity. The effect of lift is clearly seen inthe “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute.



Chapter 5 115

Q5.15 Lower air density reduces air resistance, so a tank-truck-load of fuel takes you farther.

Q5.16 The larger drop has higher terminal speed. In the case of spheres, the text demonstrates thatterminal speed is proportional to the square root of radius. When moving with terminal speed, anobject is in equilibrium and has zero acceleration.

Q5.17 The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot bepredicted. Quantum mechanics implies that the future is indeterminate. On the other hand, oursense of free will, of being able to make choices for ourselves that can appear to be random, may bean illusion. It may have nothing to do with the subatomic randomness described by quantummechanics.

SOLUTIONS TO PROBLEMS

Section 5.1 Forces of Friction

P5.1 For equilibrium: f F= and n F g = . Also, f n= µ i.e.,

µ

µ

= =

= =

f n

FF g

s75 0

25 0 9 800 306.

. ..N

Na f

and µ k = ( )=60 0 0 245. .N

25.0 9.80 N.

r

r

r

r

FIG. P5.1

P5.2 F ma n mg f n mg

y y

s s s

∑ = + − =

≤ =

: 0 µ µ

This maximum magnitude of static friction acts so long as the tires roll without skidding.

F ma f max x s∑ = − =:

The maximum acceleration is a g s= − µ . The initial and final conditions are: xi = 0 ,vi = =50 0 22 4. .mi h m s, v f = 0 , v v a x x f i f i

2 2 2= + −d i: − = −v gxi s f 2 2 µ

(a) x v g f i=2

2 µ

x f =( )

=22 4

2 0 100 9 80256

2.

. .

m s

m s m2

a f

c h

(b) x v g f i=2

2 µ

x f =( )

=22 4

2 0 600 9 8042 7

2.

. ..

m s

m s m2

a fc h




P5.3 If all the weight is on the rear wheels,

(a) F ma mg mas= =: µ

But ∆ x at gts= =2 2

2 2 µ

so µ sx

gt= 2

2

∆: µ s = =

2 0 250 1 609

9 80 4 963 342

.

. ..

mi m mi

m s s2

a fb ge ja f .

(b) Time would increase, as the wheels would skid and only kinetic friction would act; orperhaps the car would flip over.

*P5.4

22.0°22.0°

+ y + y

+ x + x f

F = 45 8. lb

22.0°

F g = 170 lb

F2 F1

n tipn F g ground lb= =2 85 0.

Free-Body Diagram of Person Free-Body Diagram of Crutch Tip

FIG. P5.4

From the free-body diagram of the person, F F Fx∑ = ° − ° =1 222 0 22 0 0sin . sin .a f a f , which givesF F F1 2= = . Then, F F y∑ = °+ − =2 22 0 85 0 170 0cos . . lbs lbs yields F = 45 8. lb.

(a) Now consider the free-body diagram of a crutch tip.

F f x∑ = − ( ) °=45 8 22 0 0. sin .lb ,

or

f = 17 2. lb .

F n y∑ = − ( ) °=tip lb45 8 22 0 0. cos . ,

which gives

n tip lb= 42 5. .

For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so

f f ns s= =a fmax µ tip and µ s f

n= = =tip

lb42.5 lb17 2

0 404.

. .

(b) As found above, the compression force in each crutch is

F F F1 2 45 8= = = . lb .



Chapter 5 117

P5.5 (a) The person pushes backward on the floor. The floor pushesforward on the person with a force of friction. This is the onlyhorizontal force on the person. If the person’s shoe is on the pointof slipping the static friction force has its maximum value.

F ma f n ma

F ma n mg ma mg a g

x x v t a t t

t

x x s x

y y

x s x s

f i xi x

∑

∑

= = =

= − == = = =

= + + = + +

=

:

:. . .

.

.

µ

µ µ 0

0 5 9 8 4 912

3 0 0 12

4 9

1 11

2 2

m s m s

m m s

s

2 2

2

e je j

rr

r

FIG. P5.5

(b) x gt f s= 12

2 µ , t

x g f

s= =

( )

( )=

2 2 3

0 8 9 80 875

µ

m

m s s2. .

.c h

P5.6 If the load is on the point of sliding forward on the bed of theslowing truck, static friction acts backward on the load with itsmaximum value, to give it the same acceleration as the truckΣ =F max x : − = f m axload Σ =F ma y y : n m g − =load 0− = µ s xmg ma ax s= − µ

v v a x xxf xi x f i2 2 2= + −d i 0 2 02= + − −v g xxi s f µ b gd i

r

r

r

FIG. P5.6

(a) x v

g f xi

s

= = =2 2

212

2 0 5 9 814 7

µ

m s

m s m2

b ga fe j. .

.

(b) From the expression x v

g f xi

s=

2

2 µ , neither mass affects the answer .

P5.7 − + = f mg sin θ 0 and + − =n mg cosθ 0 with n= µ yield µ θ s c= = ° =tan tan . .36 0 0 727

µ θ k c= = ° =tan tan . .30 0 0 577

P5.8 msuitcase kg = 20 0. , F = 35 0. N

F ma FF ma n F F

x x

y y g

∑∑

= − + =

= + + − =

: . cos: sin

20 0 00

N θ

θ

(a) F cos .cos . .

.

θ θ

θ

=

= =

= °

20 020 0 0 571

55 2

N N

35.0 N

rr

r

r

FIG. P5.8

(b) n F F g = − = − ( )sin . .θ 196 35 0 0 821 N

n = 167 N




P5.9 m = 3 00. kg , θ = °30 0. , x f = 2 00. m, t = 1 50. s , xi = 0, vxi = 0

(a) x at f = 12

2 :

2 0012 1 50

4 001 50

1 78

2

2

. ..

..

m s

m s 2

=

= =

a

a

a f

a f

rr

r

FIG. P5.9

r r r r rF n f g a∑ = + + =m m :

Along :

Along :

x f mg ma f m g a

y n mg n mg

0 30 030 0

0 30 0 030 0

− + ° =

= °−

+ − ° =

= °

sin .sin .

cos .cos .

b g

(b) µ k f n

m g amg

= =°−

°sin .cos .

30 030 0

a f, µ k

a g

= ° −°

=tan .cos .

.30 030 0

0 368

(c) f m g a= ° −sin .30 0a f, f = °− =3 00 9 80 30 0 1 78 9 37. . sin . . .a f N

(d) v v a x x f i f i2 2 2= + −c h where x x f i− = 2 00. m

v

v

f

f

2 0 2 178 2 00 7 11

7 11 2 67

= + =

= =

. . .

. .

a fa f m s

m s m s

2 2

2 2

P5.10 T ak − = 5 00. (for 5.00 kg mass)

9 00 9 00. .T a− = (for 9.00 kg mass)

Adding these two equations gives:

9 00 9 80 0 200 5 00 9 80 14 0

5 605 00 5 60 0 200 5 00 9 8037 8

. . . . . .

.. . . . .

.

a f a fa f

a f a fa f

− =

=

∴ = +

=

a

aT

m s

N

2

r

r

r r

r r r r

r

FIG. P5.10



Chapter 5 119

P5.11 (a)

(b)

See Figure to the right.

68 0 2 2

1 1

. − − =

− =

T m g m aT m g m a

µ

µ (Block #2)(Block #1)

Adding,

68 068 0 1 29

27 2

1 2 1 2

1 2

1 1

.. .

.

− + = +

=+

− =

= + =

µ

µ

µ

m m g m m a

am m

g

T m a m g

b g b g

b g m s

N

2

T

m1m2

T F

m1

n1

T

m g 1 = 118 N

= nk µ 1 1

m2

n2

F

m g 2 = 176 N

f = nk µ 2 2

rrr

r

FIG. P5.11

*P5.12 Let a represent the positive magnitude of the acceleration − a$ j ofm1 , of the acceleration − a$i of m2 , and of the acceleration + a$ j of m3 .Call T 12 the tension in the left rope and T 23 the tension in the cordon the right.

For m1 , F ma y y∑ = + − = −T m g m a12 1 1

For m2 , F max x∑ = − + + = −T n T m ak 12 23 2 µ

and F ma y y∑ = n m g − =2 0

for m3 , F ma y y∑ = T m m a23 3 3− =+

we have three simultaneous equations

− + =

+ − − =

+ − =

T aT T a

T a

12

12 23

23

39 2 4 000 350 9 80 1 00

19 6 2 00

. .. . .

. . .

N kg N kg

N kg

b ga f b gb g

n

T 12 T 23

m g 2

f = nk µ

m g 1

T 12

m g 3

T 23

FIG. P5.12 (a) Add them up:

+ − − =39 2 3 43 19 6 7 00. . . .N N N kg a fa

a m m m= 2 31 1 2 3. ,m s , down for , left for and up for2 .

(b) Now − + =T 12 39 2 4 00 2 31. . .N kg m s 2a fc h

T 12 30 0= . N

and T 23 19 6 2 00 2 31− =. . .N kg m s 2a fc h

T 23 24 2= . N .




P5.13 (Case 1, impending upward motion)Setting

F P n f n f P

P P

x

s s s s

∑ = °− =

= = °

= =

0 50 0 050 0

0 250 0 643 0 161

: cos .: cos .

. . ., ,max max µ µ

a f

Setting

F P PP

y∑ = °− − =

=

0 50 0 0 161 3 00 9 80 048 6

: sin . . . ..max

a f N

(Case 2, impending downward motion)As in Case 1,

f Ps, .max = 0 161

Setting

F P PP

y∑ = °+ − ==

0 50 0 0 161 3 00 9 80 031 7

: sin . . . ..min

a f N

r

rr

r

r r

FIG. P5.13

P5.14 We must consider separately the disk when it is in contact with the roofand when it has gone over the top into free fall. In the first case, we takex and y as parallel and perpendicular to the surface of the roof:

F ma n mg n mg

y y∑ = + − =

=

: coscos

θ

θ 0

then friction is f n mg k k k = = µ µ θ cos

r

rr

FIG. P5.14

F ma f mg maa g g

x x k x

x k

∑ = − − =

= − − = − °− ° = −

: sincos sin . cos sin . .

θ

µ θ θ 0 4 37 37 9 8 9 03a f m s m s2 2

The Frisbee goes ballistic with speed given by

v v a x x

vxf xi x f i

xf

2 2 22 15 2 9 03 10 0 44 4

6 67

= + − = + − − =

=

d i b g e ja fm s m s m m s

m s

2 2 2. .

.

For the free fall, we take x and y horizontal and vertical:

v v a y y

y

y

yf yi y f i

f

f

2 2

2

2

2

0 6 67 37 2 9 8 10 37

6 024 0119 6

6 84

= + −

= ° + − − °

= + =

d ib g e jd i

b g. sin . sin

...

.

m s m s m

m m s m s

m

2

2



Chapter 5 121

Section 5.2 Newton’s Second Law Applied to a Particle in Uniform Circular Motion

P5.15 m = 3 00. kg , r = 0 800. m. The string will break if the tensionexceeds the weight corresponding to 25.0 kg, so

T Mg max . .= = =25 0 9 80 245a f N.

When the 3.00 kg mass rotates in a horizontal circle, the tensioncauses the centripetal acceleration,

so T mvr

v= =

2 23 000 800..

a f.

Then v rT m

T T 2 0 8003 00

0 8003 00

0 800 2453 00

65 3= = ≤ = =.

..

..

..maxa f a f a f

m s2 2

and 0 65 3≤ ≤v .

or 0 8 08≤ ≤v . m s .

r

r

r

FIG. P5.15

P5.16 (a) F m v

r= =

× ×

×= ×

−

−

−2 31 6 2

1089 11 10 2 20 10

0 530 108 32 10

. .

..

kg m s

m N inward

e je j

(b) a vr

= =×

×= ×

−

2 6 2

10222 20 10

0 530 109 13 10

.

..

m s

m m s inward2e j

P5.17 n mg = since a y = 0

The force causing the centripetal acceleration is the frictional force f .

From Newton’s second law f ma mvrc= =

2.

But the friction condition is f ns≤ µ

i.e., mvr

mg s

2≤ µ

r

r

r

ra c

FIG. P5.17

v rg s≤ = µ 0 600 35 0 9 80. . .m m s 2a fe j v ≤ 14 3. m s

P5.18 (a) F ma y y∑ = , mg mvrmoon down down=

2

v g r= = × + × = ×moon2 m s m m m s1 52 1 7 10 100 10 1 65 106 3 3. . .e je j

(b) v rT

= 2π , T =

×

×= × =

2 1 8 10

1 65 106 84 10 1 90

6

33

π .

.. .

m

m s s h

e j




P5.19 T mg cos . . .5 00 80 0 9 80° = = kg m s 2b ge j

(a) T = 787 N:rT i j= +68 6 784. $ $ N Na f a f

(b) T macsin .5 00° = : ac = 0 857. m s2

toward the center ofthe circle.

The length of the wire is unnecessary information. Wecould, on the other hand, use it to find the radius of thecircle, the speed of the bob, and the period of the motion.

r

r

FIG. P5.19

*P5.20 F mg g = = =4 9 8 39 2kg m s N2b ge j. .

sin .

.

θ

θ

=

= °

1 5

48 6

m2 m

r = ° =2 48 6 1 32m ma fcos . .

F ma mvr

T T

T T

x x

a b

a b

∑ = =

°+ ° =

+ =°

=

2

2

48 6 48 64 6

1 32109

48 6165

cos . cos ..

cos .

kg m s m

N N

b gb g

F ma

T T

T T

y y

a b

a b

∑ =

+ °− °− =

− =°

=

sin . sin . ..

sin . .

48 6 48 6 39 2 039 2

48 6 52 3

N N

N

θ

39.2 N

T a

T bforces

ac v

motion

FIG. P5.20

(a) To solve simultaneously, we add the equations in T a and T b:

T T T T a b a b+ + − = +165 52 3N N.

T a = =217 108N2

N

(b) T T b a= − = − =165 165 108 56 2N N N N.



Chapter 5 123

Section 5.3 Nonuniform Circular Motion

P5.21 Let the tension at the lowest point be T .

F ma T mg ma mvr

T m g vr

T

c∑ = − = =

= +F H G I K J

= +LNMM

OQPP = >

:

. ..

..

2

2

2

85 0 9 808 00

10 01 38 1 000kg m s

m s m

kN N2b g b g

He doesn’t make it across the river because the vine breaks.

r

r

rr

FIG. P5.21

P5.22 (a) F ma mvR y y∑ = =

2

mg n mvR

− =2

n mg mvR

= −2

(b) When n = 0, mg mvR

=2

Then, v gR= .

P5.23 F mvr

mg n y∑ = = +2

But n = 0 at this minimum speed condition, so

mvr

mg v gr2

9 80 1 00 3 13= ⇒ = = =. . .m s m m s2e ja f .

r

r

FIG. P5.23

P5.24 (a) a vrc =2

r vac

= = =2 2

13 0

2 9 808 62

.

..

m s

m s m2

b ge j

(b) Let n be the force exerted by the rail.

Newton’s law gives

Mg n Mv

r+ =

2

rr

FIG. P5.24

n M vr

g M g g Mg = −F H G

I K J = − =

22b g , downward

continued on next page




(c) a vrc =2

ac = =13 0

20 08 45

2.

..

m s m

m s 2b g

If the force exerted by the rail is n1

then n Mg Mv

r

Mac1

2+ = =

n M a g c1 = −b g which is < 0 since ac = 8 45. m s 2

Thus, the normal force would have to point away from the center of the curve. Unless theyhave belts, the riders will fall from the cars. To be safe we must require n1 to be positive.

Then a g c > . We need vr

g 2

> or v rg > = 20 0 9 80. .m m s 2a fe j, v > 14 0. m s .

Section 5.4 Motion in the Presence of Velocity-Dependent Resistive Forces

P5.25 (a) a g bv= −

When v vT = , a = 0 and g bvT = b g

vT

=

The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.

Thus, vtT = = =1 50 0 300. .m

5.00 s m s

Then b = = −9 800 300

32 7 1..

.m s m s

s2

(b) At t = 0, v = 0 and a g = = 9 80. m s 2 down

(c) When v = 0 150. m s, a g bv= − = − =−9 80 32 7 0 150 4 901. . . .m s s m s m s2 2e jb g down

P5.26 (a) ρ = mV

, A = 00201. m2 , R ADv mg T = 1

=2

2 ρ air

m V = = LNM

OQP

= ρ π bead3 g cm cm kg 0 830 4

38 00 1 783. . .a f

Assuming a drag coefficient of D = 0 500. for this spherical object, and taking the density ofair at 20°C from the endpapers, we have

vT = =2 1 78 9 80

0 500 1 20 0 020 153 8

. .

. . ..

kg m s

kg m m m s

2

3 2

b ge je je j

(b) v v gh gh f i2 2 2 0 2= + = + : h

v g f = = =2 2

253 8

2 9 80148

.

.

m s

m s m2

b ge j



Chapter 5 125

P5.27 (a) At terminal velocity, R v b mg T = =

∴ = =×

×= ⋅

−

−b mg

vT

3 00 10 9 80

2 00 101 47

3

2

. .

..

kg m s

m s N s m

2e je j

(b) In the equation describing the time variation of the velocity, we have

v v eT bt m= − −1e j v vT = 0 632. when e bt m− = 0 368.

or at time t mb

= − F H G I

K J = × −ln . .0 368 2 04 10 3a f s

(c) At terminal velocity, R v b mg T = = = × −2 94 10 2. N

P5.28 v mg

bbt

m= F H G I

K J − −F H G I

K J LNM

OQP1 exp where exp x exa f= is the exponential function.

At t→ ∞

, v v m

bT → =

At t = 5 54. s 0 500 15 54

9 00. exp

..

v v b

T T = − −F H G

I K J

LNMM

OQPP

s kg

a f

exp.

.. ;

.

.ln . . ;

. .

..

−F H G

I K J =

−= = −

= =

b

b

b

554

9 000500

554

9 000 500 0 693

9 00 0 693

554113

s

kg

s

kg

kg

s kg s

b g

b g

b gb g

(a) v mg

bT = vT = =9 00 9 80

1 1378 3

. .

..

kg m s

kg s m s

2b ge j

(b) 0 750 1 1 139 00

. exp ..

v v tT T = −

−F H G I

K J LNM

OQP s exp .

..

−F H G I

K J =1 139 00

0 250t s

t =−

=9 00 0 250

1 1311 1

. ln ..

.a f

s s

(c) dxdt

mg b

btm

= F H G I

K J − −F H G I

K J LNM

OQP1 exp ; dx

mg b

btm

dtx

x t

0

10

z z = F H G I

K J − −F H G I

K J LNM

OQPexp

x x mgt

bm g b

btm

mgtb

m g b

btm

t− = +

F H G

I K J

−F H G I

K J = +F H G

I K J

−F H G I

K J −LNM

OQP0

2

20

2

2 1exp exp

At t = 5 54. s , x = +F

H GG

I

K J J − −9 00

554 9 00 9 80

1130 693 1

2

2.. . .

.exp .kg 9.80 m s

s1.13 kg s

kg m s

kg s

22

e j b g e j

b g b g

x = + − =434 626 0 500 121m m m.a f




P5.29 (a) v t v eicta f= − v v ei

c20 0 5 00 20 0. . . sa f= = − , vi = 10 0. m s .

So 5 00 10 0 20 0. . .= −e c and − = F H G I

K J 20 0 12

. lnc c = − = × − −ln.

.12 2 1

20 03 47 10

c h s

(b) At t = 40 0. s v e c= = =−10 0 10 0 0 250 2 5040 0. . . .. m s m s m sb g b ga f

(c) v v eict= − a dv

dtcv e cvi

ct= = − = −−

P5.30 F ma∑ =

− =

− =

− =

− − =−

= − +

= + = +

=+

z z −

−

kmv m dvdt

kdt dvv

k dt v dv

k t vv v

v vkt

v ktv

v v

v kt

t

v

v

v

v

2

2

0

2

1

0

0

0

0

0

0

0

0

01

1 1

1 1 1

1

a f

Section 5.5 The Fundamental Forces of Nature

P5.31 F Gm mr

= = × = ×

−

−1 22

11

296 672 10 2 2

0 302 97 10.

..e ja fa fa f N

P5.32 For two 70–kg persons, modeled as spheres,

F Gm m

r g = =× ⋅

−

−1 22

11

276 67 10 70 70

210

.~

N m kg kg kg

m N

2 2e jb gb ga f

P5.33 a MGRE

= = =4

9 816

0 6132b g.

.m s

m s2

2 toward the earth

P5.34 F k q qr

e= = × + − = − × = ×1 2

122

92

6 68 99 10 40 402000

3 60 10 3 60 10b g e j a fa fa f. . .N (attractive) N downward



Chapter 5 127

Section 5.6 Context Connection Drag Coefficients of Automobiles

*P5.35 The resistive force is

R D Av

R

a Rm

= =

=

= − = − = −

12

12

0 250 1 20 2 20 27 8

255255 0 212

2 2 ρ . . . .

.

a fe je jb gkg m m m s

N N

1 200 kg m s

3 2

2

*P5.36 (a) The drag force on this car body at this speed is

R D Av= = =12

12

0 34 1 2 2 6 10 53 02 2 ρ . . . .a fe je jb gkg m m m s N3 2 .

Now Newton’s second law is F max x∑ =

+ − =

= × f

f

s

s

53 0 1 3003 95 103

..

N kg 3 m s N forward

2

e j

(b) Newton’s second law changes to

+ − =

= −

=

3 950 12

0 2 1 2 2 6 10 1 300

3 950 31 23 02

2 N N kg

N1 300 kg

m s forward2

. . .

..

a fa fa fa f b ga

a

Streamlining to reduce drag makes little difference to the acceleration at this low speed.

(c) F max x∑ = now reads

+ − =

= ⋅ ⋅

⋅

=

3 950 12

0 34 1 2 2 6 0

3 9500 530

86 3

2 Nkg m

kg m m

s kg m s2

. . .

..

a fa fa f v

v

(d) Still again, + − =3 950 12

0 2 1 2 2 6 02 Nkg m

. . .a fa fa f v

v = =3 9500 312

113.

ms

m s2

2

According to this model, drag reduction significantly affects maximum speed.




Additional Problems

P5.37 Applying Newton’s second law to each object gives:

(1) T f m g a1 1 2= + +sin θ b g

(2) T T f m g a2 1 2− = + +sinθ b g (3) T M g a2 = −b g

(a), (b) Equilibrium a = 0a f and frictionless incline f f 1 2 0= =b g

Under these conditions, the equations reduce to

(1’) T mg 1 2= sin θ

(2’) T T mg 2 1− = sin θ

(3’) T Mg 2 =

r rr

r

r

r

rrr

r

r

r

FIG. P5.37

Substituting (1’) and (3’) into equation (2’) then gives M m= 3 sinθ

so equation (3’) becomes T mg 2 3= sin θ

(c), (d) M m= 6 sinθ (double the value found above), and f f 1 2 0= = . With these conditions present,the equations become T m g a1 2= +sin θ b g, T T m g a2 1− = +sin θ b g and T m g a2 6= −sin θ b g.Solved simultaneously, these yield

a g =

+

sinsin

θ θ 1 2

, T mg 1 41

1 2=

+

+

F H G

I K J sin

sinsin

θ θ θ

and T mg 2 61

1 2=

+

+

F H G

I K J sin

sinsin

θ θ θ

(e) Equilibrium a = 0a f and impending motion up the incline so M M= max while f mg s1 2= µ θ cos and mg s2 = µ θ cos , both directed down the incline. Under theseconditions, the equations become T mg s1 2= +sin cosθ µ θ b g, T T mg s2 1− = +sin cosθ µ θ b g, and

T M2 = max , which yield M m smax sin cos= +3 θ µ θ b g

(f) Equilibrium a = 0a f and impending motion down the incline so M M= min , while f mg s1 2= µ θ cos and mg s2 = µ θ cos , both directed up the incline. Under these conditions,the equations are T mg s1 2= −sin cosθ µ θ b g, T T mg s2 1− = −sin cosθ µ θ b g, and T M g 2 = min ,

which yield M m smin sin cos= −3 θ µ θ b g. When this expression gives a negative value, itcorresponds physically to a mass M hanging from a cord over a pulley at the bottom end ofthe incline.

(g) T T M g M g mg s2 2 6,max ,min max min cos− = − = µ θ



Chapter 5 129

P5.38 For the system to start to move when released, theforce tending to move m2 down the incline,m g 2 sin θ , must exceed the maximum friction forcewhich can retard the motion:

f f f n n f m g m g

s s

s s

max max max= + = +

= +1 2 1 1 2 2

1 1 2 2

, , , ,

max , , cos µ µ

µ µ θ

From Table 5.1, µ s, .1 0 610= (aluminum on steel) and µ s, .2 0 530= (copper on steel). With

m m1 22 00 6 00 30 0= = = °. . . ,kg, kg, θ

rr

FIG. P5.38

the maximum friction force is found to be f max N= 38 9. . This exceeds the force tending to cause thesystem to move, m g 2 6 00 9 80 29 4sin . . sin .θ = ° =kg m s 30 N2e j . Hence,

the system will not start to move when released .

The friction forces increase in magnitude until the total friction force retarding the motion, f f f = +1 2 , equals the force tending to set the system in motion. That is, until

f m g = =2 29 4sin .θ N .

P5.39 (a) The crate is in equilibrium, just before it starts to move. Let the normalforce acting on it be n and the friction force, s . Resolving vertically:n F P g = + sinθ . Horizontally: P scosθ = . But, f ns s≤ µ i.e.,

P F Ps g cos sinθ µ θ ≤ +c h or P Fs s g cos sinθ µ θ µ − ≤a f . Divide by cos θ :

P Fs s g 1− ≤ µ θ µ θ tan seca f . Then

PFs g

sminimum =

−

µ θ

µ θ

sectan1

.

rr

r

r

FIG. P5.39

(b) P = ( )

−0 400 100

1 0 400. sec

. tan N θ

θ

θ deg b g 0.00 15.0 30.0 45.0 60.0P N a f 40.0 46.4 60.1 94.3 260

If the angle were 68 2. ° or more, the expression for P would go to infinity and motion would become impossible.




P5.40 With motion impending,

n T mg

f mg T s

+ − =

= −

sin

sin

θ

µ θ

0

b g

and

T mg T s scos sinθ µ µ θ − + = 0

so

r

r

r

r

FIG. P5.40

T ms

s=

+ µ

θ µ θ cos sin.

To minimize T , we maximize cos sinθ µ θ + s

dd s sθ

θ µ θ θ µ θ cos sin sin cos+ = = − +b g0 .

(a) θ µ = = = °− −tan tan . .1 1 0 350 19 3s

(b) T =°+ °

=0 350 1 30 9 80

19 3 0 350 19 34 21

. . .

cos . . sin ..

kg m s N

2a fc h

P5.41 Σ =F m a1 1 : − °− + =m g T m ak 1 1 135 0sin . ,

− °− °+ =3 50 9 80 35 0 3 50 9 80 35 0 3 50 1 50. . sin . . . cos . . .a fa f a fa f a f µ s T (1)

Σ =F m a2 2 : + °− − =m g T m ak 2 2 235 0sin . ,

+ °− °− =8 00 9 80 35 0 8 00 9 80 35 0 8 00 1 50. . sin . . . cos . . .a fa f a fa f a f µ s T (2)

Solving equations (1) and (2) simultaneously gives

(a) µ k = 0 0871.

(b) T = 27 4. N

FIG. P5.41



Chapter 5 131

P5.42 (a) See Figure (a) to the right.

(b) See Figure (b) to the right.

(c) For the pin,

F ma CC

y y∑ = − =

=

: cos

cos.

θ

θ

357 0357

N N

For the foot,

mrg = =36 4 9 8 357. .kg m s N2b ge j

r

r r

r

r

FIG. P5.42(a) FIG. P5.42(b)

F ma n Cn

y y B

B

∑ = + − =

=

: cos.

θ 0357 N

(d) For the foot with motion impending,

F ma f Cn C

Cn

x x s ss B s

ss

B

s ss

∑ = + − =

=

= = =

: sinsinsin cos sin

tan .

θ

µ θ

µ θ θ θ

θ

0

357357

N N

b g

(e) The maximum coefficient is

µ θ s s= = °=tan tan . .50 2 1 20 .




P5.43 (a) First, draw a free-body diagram, (top figure) of the top block. Since a y = 0 , n1 19 6= . N . And f nk k = = = µ 1 0 300 19 6 5 88. . .N Na f . Σ =F max T

10 0 5 88 2 00. . .N N kg − = b gaT

or aT = 2 06. m s 2 (for top block). Now draw a free- body diagram (middle figure) of the bottom block andobserve that Σ =F Max B gives f aB= =5 88 8 00. .N kg b g or aB = 0 735. m s 2 (for the bottom block). In time t, thedistance each block moves (starting from rest) is

d a tT T = 12

2 and d a tB B= 12

2 . For the top block to reach

the right edge of the bottom block, (see bottom figure)it is necessary that d d LT B= + or

1

22 06 1

20 735 3 002 2. . .m s m s m2 2

e j e jt t= +

which gives: t = 2 13. s .

(b) From above, dB = =12

0 735 2 13 1 672. . .m s s m2e ja f .

r

r

r

r

r

rr

FIG. P5.43

P5.44 (a)

rr

r

r

r

r

r r

FIG. P5.44

f n1 1and appear in both diagrams as action-reaction pairs

(b) 5.00 kg: Σ =F max : n m g 1 1 5 00 9 80 49 0= = =. . .a f N f T 1 0− =

T f mg = = = =

1 0 200 5 00 9 80 9 80 µ . . . .a fa f N

10.0 kg: Σ =F max : 45 0 10 01 2. .− − = f f a

Σ =F y 0: n n2 1 98 0 0− − =.







*P5.46 (a) Directly n = ° =63 7 13 62 1. cos .N N

f k = =0 36 62 1 22 3. . .N Na f .

Now adding + + − =T a14 3 22 3 6 5. . .N N kg b g and − + =T a37 2 3 8. .N kg b g gives37 2 8 01 10 3. . .N N kg − = b ga

a = 2 84. m s 2 .

Then T = − =37 2 3 8 2 84 26 5. . . .N kg m s N2e j .

(b) We recognize the equations are describing a 6.5-kg block on an incline at 13° with thehorizontal. It has coefficient of friction 0.36 with the incline. It is pulled forward, which isdown the incline, by the tension in a cord running to a hanging 3.8-kg object.

f k n

T

13° 63.7 N

T

37.2 N

3.8 kg

6.5 kg

13°

FIG. P5.46

*P5.47 When the cloth is at a lower angle θ , the radialcomponent of F ma∑ = reads

n mg mv

r+ =sin θ

2

.

At θ = °68 0. , the normal force drops to zero and

g vr

sin682

° = .

R

68°p

mg

p

mg cos68°

mg sin68°

FIG. P5.47

v rg = ° = ° =sin . . sin .68 0 33 9 8 68 1 73m m s m s2a fe j

The rate of revolution is

angular speed= F

H G I

K J F H G

I K J

= =1 73

12

22 0 33 0 835 50 1. . . .m s

rev m rev s rev minb g a fπ

π π r

r.



Chapter 5 135

P5.48 (a) While the car negotiates the curve, the accelerometer is at the angle θ .

Horizontally: T mvr

sin θ =2

Vertically: T mg cosθ =

where r is the radius of the curve, and v is the speed of the car.

By division, tan θ = vrg

2

Then a vr

g c = =2

tan θ : ac = °9 80 15 0. tan .m s 2e j

ac = 2 63. m s 2

r

r

r

FIG. P5.48

(b) r v

ac

=2

r = =23 0

2 63201

2.

.

m s

m s m

2

b g

(c) v rg 2 201 9 80 9 00= = °tan . tan .θ m m s 2a fe j v = 17 7. m s

P5.49 (a) Since the centripetal acceleration of a person is downward (towardthe axis of the earth), it is equivalent to the effect of a fallingelevator. Therefore,

′ = −F F mvr g g

2 or F F g g > ′

(b) At the poles v =

0 and′ = = = =

F F mg g g 75 0 9 80 735. .a f N down.

rr

FIG. P5.49

At the equator, ′ = − = − =F F ma g g c 735 75 0 0 033 7 732N N N. .b g down.

P5.50 (a) Since the object of mass m2 is in equilibrium, F T m g y∑ = − =2 0

or T m g = 2 .

(b) The tension in the string provides the required centripetal acceleration of the puck.

Thus, F T m g c = = 2 .

(c) From F m v

Rc = 12

we have v RFm

mm

gRc= = F

H G I K J

1

2

1.




*P5.51 v r

T = = =2 2 9 00

15 03 77

π π ..

.m

s m s

a fa f

(a) a vrr = =2

1 58. m s 2

(b) F m g arlow N= + =

b g 455(c) F m g arhigh N= − =b g 329

(d) F m g armid N upward and= + =2 2 397 at θ = = = °− −tan tan ..

.1 1 1 589 8

9 15a g

r inward .

*P5.52 (a) The mass at the end of the chain is in verticalequilibrium. Thus T mg cosθ = .

Horizontally T ma mvrrsin θ = =

2

rr

= += °+ =

2 50 4 002 50 28 0 4 00 5 17. sin .. sin . . .

θ a fa f m m m

Then a vr =

2

5 17. m.

By division tan.

θ = =a g

v g

r2

5 17

v g

v

2 5 17 5 17 9 80 28 0

5 19

= = °

=

. tan . . tan .

.

θ a fa fa f m s

m s

2 2

(b) T mg cosθ =

T mg = =

°=

cos

. .

cos .θ

50 0 9 80

28 0555

kg m s N

2b ge j

TR = 4.00 m

θ l = 2.50 m

r

mg

FIG. P5.52

P5.53 Standing on the inner surface of the rim, and moving with it, each person will feel a normal forceexerted by the rim. This inward force causes the 3 00 . m s 2 centripetal acceleration:

a vrc =2

: v a rc= = =3 00 60 0 13 4. . .m s m m s2e ja f

The period of rotation comes from v rT

= 2π : T r

v= = =2 2 60 0

13 428 1

π π ..

.m

m s s

a f

so the frequency of rotation is f T

= = = F H G I K J =1 128 1

128 1

60 2 14. .

. s s

s1 min

rev min .



Chapter 5 137

*P5.54 The volume of the grain is V r= = × = ×− −43

43

5 10 5 24 103 7 3 19π π m m 3e j . . Its weight is

F mg Vg g = = = × ×

= ×

−

−

ρ 19 3 10 5 24 10 9 8

9 90 10

3 19

14

. . .

.

kg m m m s

N down

3 3 2e je je j

(a) As it settles at terminal speed, F ma y y∑ =

− × + ⋅

× =

= ×

− −

−

9 90 10 0 018 8 5 10 0

1 05 10

14 7

5

. .

.

N N sm

m

m s

2 e jvv

(b) ∆ v t y=

t =×

= × =−

0 087 59 10 2 115

3.. .

m

1.05 10 m s s h

(c) The speed of the middle of the tube is

3 0002 0 09

11 314 0 09 28 3rev min

m rev

min60 s

radian s m m sπ .

. .a f a fF

H G I K J F H G I

K J = = .

Now a vrc = = = ×2 2

328 30 09

8 88 10..

. m s

m m s 2b g

.

(d) In place of part (a) we have F max x∑ =

0 018 8 5 10 1 01 10

9 55 10

0 08 8 38

7 14

3

3

. .

.

. .

N s m m kg 8.88 10 m s

m s

m s9.55 10 m

s

2 3 2⋅ × = × ×

= × =

= ⋅

×=

− −

−

−

e j e jv

v xt

t

∆




P5.55 (a) n mvR

=2

f mg − = 0

ns= µ v RT

= 2π

T R

g s

=4 2π µ

(b) T = 2 54. s

# .revmin

rev2.54 s

smin

revmin

= F H G I

K J =1 60 23 6

rf s

r

n

mrg

FIG. P5.55

P5.56 (a) The bead moves in a circle with radius v R= sin θ at aspeed of

v rT

RT

= =2 2π π θ sin

The normal force has

an inward radial component of n sin θ

and an upward component of n cosθ

F ma n mg y y∑ = − =: cosθ 0

or

n mg =cosθ

r

r

FIG. P5.56

Then F n m vrx∑ = =sinθ 2

becomesmg m

RRT cos

sinsin

sinθ

θ θ

π θ F H G I

K J = F H G I

K J 2 2




Chapter 5 139

which reduces to g R

T sin

cossinθ

θ π θ = 4 2

2

This has two solutions: sin θ θ = ⇒ = °0 0 (1)

and cos θ π = gT

R

2

24 (2)

If R = 15 0. cm and T = 0 450. s, the second solution yields

cos. .

..θ

π = =

9 80 0 450

4 0 1500 335

2

2

m s s

m

2e ja fa f and θ = °70 4.

Thus, in this case, the bead can ride at two positions θ = °70 4. and θ = °0 .

(b) At this slower rotation, solution (2) above becomes

cos. .

..θ

π = =

9 80 0 850

4 0 1501 20

2

2

m s s

m

2e ja fa f , which is impossible.

In this case, the bead can ride only at the bottom of the loop, θ = °0 . The loop’s rotationmust be faster than a certain threshold value in order for the bead to move away from thelowest position.

P5.57 At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg arv br v= + 2 2

(a) mg v v= × + ×− −3 10 10 0 870 109 10 2. .e j e j

For water,m V

v v

= = LNM

OQP

× = × + ×

−

− − −

ρ π 1 000 43

10

4 11 10 3 10 10 0 870 10

5 3

11 9 10 2

kg m m3 e je j e j. . .

Assuming v is small, ignore the second term on the right hand side: v = 0 013 2. m s .

(b) mg v v= × + ×− −3 10 10 0 870 108 8 2. .e j e j Here we cannot ignore the second term because thecoefficients are of nearly equal magnitude.

4 11 10 3 10 10 0 870 103 10 3 10 4 0 870 4 11

2 0 8701 03

8 8 8 2

2. . .

. . . ..

.

× = × + ×

=− ± +

=

− − −

e j e ja f a fa f

a f

v v

v m s





(c) mg v v= × + ×− −3 10 10 0 870 107 6 2. .e j e j

Assuming v > 1 m s , and ignoring the first term: 411 10 0 870 105 6 2. .× = ×− −e jv

v = 6 87. m s

*P5.58 (a) t ds m........

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

a f a f1 002 003 004 005 006 007 008 00

9 0010 011 012 013 014 015 016 017 018 0

19 020 0

4 8818 942 173 8

112154199246

296347399452505558611664717770

823876

(b)

(s)

(m)

900

800

700

600

500

400

300

200

100

00 2 4 6 8 10 12 14 16 18 20

t

d

FIG. P5.58(b)

(c) A straight line fits the points from t = 11 0. s to 20.0 s quite precisely. Its slope is the terminalspeed.

vT = = −

−=slope m m

20.0 s s m s876 399

11 053 0

..



Chapter 5 141

P5.59 F L T mg L T ma

F L T L T m vr

m vr

L T L T

L T

L T

T

y y y y

x x x

∑

∑

= − − = °− °− = =

= + = °+ ° =

=°

=

∴ °+ ° =

°− ° =

+ °

°=

°

− °

°=

°

cos . sin . .

sin . cos .

..

. cos ..

sin . cos . .cos . sin . .

cos .sin .

.sin .

sin .cos .

.

cot

20 0 20 0 7 35 0

20 0 20 0

0 75035 0

60 0 20 016 3

20 0 20 0 16 320 0 20 0 7 35

20 020 0

16 320 0

20 020 0

7 35

2

2 2

N

kg m s

m N

N N

N

Ncos20.0

b ga f

20 0 20 0 16 320 0

7 3520 0

3 11 39 8

12 8

. tan . .sin .

.cos .

. .

.

°+ ° =°

−°

=

=

a fa f

N N

N

N

T

T

r

r

rr

FIG. P5.59

P5.60 v v kxi= − implies the acceleration is a dvdt

k dxdt

kv= = − = −0

Then the total force is F ma m kv∑ = = −a f The resistive force is opposite to the velocity:

r rF v∑ = − km .

ANSWERS TO EVEN PROBLEMS

P5.2 (a) 256 m; (b) 42.7 m

P5.4 (a) 0.404; (b) 45.8 lb

P5.6 (a) 14.7 m; (b) neither mass is necessary

P5.8 see the solution (a) 55.2 ° ; (b) 167 N

P5.10 37.8 N

P5.12 (a) 2 31 1. m s , down for2 m , left for m2 ,and up for m3 ; (b) 30.0 N and 24.2 N

P5.14 6.84 m

P5.16 (a) 8 32 10 8. × − N inward ;(b) 9 13 1022. × m s inward2

P5.18 (a) 1 65 103. × m s ; (b) 6 84 103. × s

P5.20 (a) 108 N; (b) 56.2 N

P5.22 (a) mg mvR

−2

upward; (b) gR

P5.24 (a) 8.62 m; (b) Mg downward(c) 8 45. m s 2 unless they have belts, theriders will fall from the cars

P5.26 (a) 53 8. m s ; (b) 148 m

P5.28 (a) 78 3. m s ; (b) 11.1 s; (c) 121 m

P5.30 see the solution

P5.32 ~10 7− N toward you

P5.34 3 60 106. × N downward




P5.36 (a) 3 95 103. × N forward ;(b) 3 02. m s forward2 , (c) 86.3 m/s(d) 113 m/s

P5.38 they do not, 29.4 N

P5.40 (a) 19.3° ; (b) 4.21 N

P5.42 (a) see the solution; (b) see the solution;(c) 357 N; (d) see the solution; (e) 1.20

P5.44 (a) see the solution;(b) 9 80. N , 0 580. m s 2

P5.46 (a) a = 2 84. m s 2 , 26.5 N;(b) see the solution

P5.48 (a) 2 63. m s 2 ; (b) 201 m; (c) 17 7. m s

P5.50 (a) m g 2 ; (b) m g 2 ; (c) mm

gR2

1

F H G I

K J

P5.52 (a) 5 19. m s ; (b) see the solution, 555 N

P5.54 (a) 1 05 10 5. × − m s ; (b) 7 59 103. × s ;(c) 8 88 103. × m s 2 ; (d) 8.38 s

P5.56 (a) either 70.4 ° or 0° ; (b) 0°

P5.58 (a) see the solution; (b) see the solution;(c) 53 0. m s

P5.60 r rF v∑ = − km

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