More Applications of Newton’s Laws

113

More Applications ofNewton’s Laws

CHAPTER OUTLINE 5.1 Forces of Friction 5.2 Newton’s Second

Law Applied to a Particle in Uniform Circular Motion

5.3 Nonuniform Circular Motion

5.4 Motion in the Presence of Velocity-Dependent Resistive Forces

5.5 The Fundamental Forces of Nature

5.6 Context ConnectionDrag Coefficients of Automobiles

ANSWERS TO QUESTIONS Q5.1 (a) m m

r r ra R g= + (b) ma T mg= − (c) ma f R= −

r

r

r

r

r

r

r

fr

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 of

the Earth on his foot. In the second case, the action is the force exerted on the girl’s back by the snowball; the reaction is the force exerted on the snowball by the girl’s back. The third action is the force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the force exerted on the window by the air molecules; the reaction is the force on the air molecules exerted 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 the

resultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers were winning with the rope steadily moving in their direction or if the contest was even, then the tension would still be 9 200 N. In all of these cases, the acceleration is zero, and so must be the resultant force on the rope. To win the tug-of-war, a team must exert a larger force on the ground than their opponents do.

Q5.5 If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the

tires 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 the component of the weight down the plane are in the same direction, giving a large acceleration. On the way down, the forces are in opposite directions, giving a relatively smaller acceleration. If the incline 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 the car forward. Drop a stick into a running stream and fluid friction sets the stick into horizontal motion.

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 along

a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around 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 and

around in a circle. Q5.12 Blood pressure cannot supply the force necessary both to balance the gravitational force and to

provide 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,” one

would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space 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 and is in free fall.

Q5.14 From the proportionality of the drag force to the speed squared and from Newton’s second law, we

derive the equation that describes the motion of the skydiver:

mdv

dtmg

D Avy

y= −ρ2

2

where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutist’s

body. At terminal speed,

a

dv

dtyy= = 0 and v

mgD AT =FHGIKJ

21 2

ρ.

When the parachute opens, the coefficient of drag D and the effective area A both increase, thus

reducing the speed of the skydiver. Modern parachutes also add a third term, lift, to change the equation to

mdv

dtmg

D Av

L Avy

y x= − −ρ ρ2 2

2 2

where vy is the vertical velocity, and vx is the horizontal velocity. The effect of lift is clearly seen in

the “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 that

terminal speed is proportional to the square root of radius. When moving with terminal speed, an object 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 be

predicted. Quantum mechanics implies that the future is indeterminate. On the other hand, our sense of free will, of being able to make choices for ourselves that can appear to be random, may be an illusion. It may have nothing to do with the subatomic randomness described by quantum mechanics.

SOLUTIONS TO PROBLEMS Section 5.1 Forces of Friction P5.1 For equilibrium: f F= and n Fg= . Also, f n=µ i.e.,

µ

µ

= =

= =

fn

FFg

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 mgy y

s s s

∑ = + − =≤ =

: 0µ 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 gs=−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 xf i f i2 2 2= + −d i: − = −v gxi s f

2 2µ

(a) xv

gfi=2

2µ

x f =( )

=22 4

2 0 100 9 80256

2.

. .

m s

m s m

2

a fc h

(b) xv

gfi=2

2µ

x f =( )

=22 4

2 0 600 9 8042 7

2.

. ..

m s

m s m

2

a fc h


P5.3 If all the weight is on the rear wheels, (a) F ma mg mas= =: µ

But ∆xat gts= =

2 2

2 2µ

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; or

perhaps the car would flip over.

*P5.4

22.0° 22.0°

+y +y

+x +xf

F = 45 8. lb22.0°

Fg = 170 lb

F2 F1

ntipn Fgground 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 gives

F F F1 2= = . Then, F Fy∑ = °+ − =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 fx∑ = −( ) °=45 8 22 0 0. sin . lb , or

f =17 2. lb .

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

which gives

ntip 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 sf

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 pushes forward on the person with a force of friction. This is the only horizontal force on the person. If the person’s shoe is on the point of slipping the static friction force has its maximum value.

F ma f n maF 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 µ0

0 5 9 8 4 912

3 0 012

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 gtf s= 12

2µ , tx

gf

s= =

( )

( )=

2 2 3

0 8 9 80 875

µ m

m s s

2. ..

c h

P5.6 If the load is on the point of sliding forward on the bed of the

slowing truck, static friction acts backward on the load with its maximum value, to give it the same acceleration as the truck

Σ =F max x : − =f m axload

Σ =F may y : n m g− =load 0

− =µ s xmg ma a gx s= −µ

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

r

r

r

FIG. P5.6

(a) xv

gfxi

s= = =

2 2

2

12

2 0 5 9 814 7

µ m s

m s m

2

b ga fe j. .

.

(b) From the expression xv

gfxi

s=

2

2µ, neither mass affects the answer .

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

µ θk 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 θθ

(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 Fg= − = − ( )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 atf = 12

2 :

2 00

12

1 50

4 00

1 501 78

2

2

. .

.

..

m s

m s2

=

= =

a

a

a f

a f

r r

r

FIG. P5.9

r r r r rF n f g a∑ = + + =m m :

Along :

Along :

x f mg maf m g a

y n mgn mg

0 30 030 0

0 30 0 030 0

− + ° == °−+ − ° == °

sin .sin .

cos .cos .

b g

(b) µ kfn

m g a

mg= =

°−°

sin .

cos .

30 0

30 0a f

, µ ka

g= °−

°=tan .

cos ..30 0

30 00 368

(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 xf i f i

2 2 2= + −c h where x xf i− = 2 00. m

v

v

f

f

2 0 2 1 78 2 00 7 11

7 11 2 67

= + =

= =

. . .

. .

a fa f m s

m s m s

2 2

2 2

P5.10 T f ak− = 5 00. (for 5.00 kg mass) 9 00 9 00. .g 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 rr 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

µµ

(Block #2)(Block #1)

Adding,

68 0

68 01 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

m 1m 2

T F

m 1

n 1

T

m g1 = 118 N

f = n k µ 1 1

m 2

n 2

F

m g 2 = 176 N

f = n k µ 2 2

rrr

r

FIG. P5.11

*P5.12 Let a represent the positive magnitude of the acceleration −a$j of

m1 , of the acceleration −a$i of m2 , and of the acceleration +a$j of m3 .Call T12 the tension in the left rope and T23 the tension in the cord on the right.

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

For m2 , F max x∑ = − + + =−T n T m ak12 23 2µ and F may y∑ = n m g− =2 0

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

we have three simultaneous equations

− + =

+ − − =

+ − =

T a

T T a

T a

12

12 23

23

39 2 4 00

0 350 9 80 1 00

19 6 2 00

. .

. . .

. . .

N kg

N kg

N kg

b ga f b g

b g

n

T12 T23

m g 2

f = n k µ

m g 1

T12

m g 3

T23

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

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

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

(b) Now − + =T12 39 2 4 00 2 31. . . N kg m s2a fc h

T12 30 0= . N

and T23 19 6 2 00 2 31− =. . . N kg m s2a fc h

T23 24 2= . N .


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

F P n

f n f PP P

x

s s s s

∑ = °− == = °

= =

0 50 0 050 0

0 250 0 643 0 161

: cos .: cos .

. . ., , max maxµ µ

a f

Setting

F P P

Py∑ = °− − =

=0 50 0 0 161 3 00 9 80 0

48 6

: sin . . . .

.max

a f N

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

f Ps , . max = 0 161

Setting

F P P

Py∑ = °+ − =

=0 50 0 0 161 3 00 9 80 0

31 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 roof

and when it has gone over the top into free fall. In the first case, we take x and y as parallel and perpendicular to the surface of the roof:

F ma n mgn mg

y y∑ = + − ==

: coscos

θθ

0

then friction is f n mgk k k= =µ µ θcos

r

rr

FIG. P5.14

F ma f mg ma

a g gx x k x

x k

∑ = − − == − − = − °− ° = −

: sincos sin . cos sin . .

θµ 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 f m 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 01

19 66 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 tension

exceeds the weight corresponding to 25.0 kg, so

T Mgmax . .= = =25 0 9 80 245a f N. When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration,

so Tmv

rv

= =2 23 00

0 800..a f

.

Then vrTm

T T2 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) Fm v

r= =

× ×

×= ×

−

−−

2 31 6 2

108

9 11 10 2 20 10

0 530 108 32 10

. .

..

kg m s

m N inward

e je j

(b) avr

= =×

×= ×−

2 6 2

1022

2 20 10

0 530 109 13 10

.

..

m s

m m s inward2e j

P5.17 n mg= since ay = 0

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

From Newton’s second law f mamv

rc= =2

.

But the friction condition is f ns≤ µ

i.e., mv

rmgs

2

≤ µ

r

r

r

ra c

FIG. P5.17

v rgs≤ =µ 0 600 35 0 9 80. . . m m s2a fe j v ≤ 14 3. m s

P5.18 (a) F may y∑ = , mgmv

rmoon 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) vr

T= 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 mgcos . . .5 00 80 0 9 80° = = kg m s2b 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 of

the circle.

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

r

r

FIG. P5.19

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

sin.

.

θ

θ

=

= °

1 5

48 6

m2 m

r = ° =2 48 6 1 32 m ma fcos . .

F mamv

r

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

48 652 3

N N

N

θ

39.2 N

Ta

Tb

forces

ac v

motion

FIG. P5.20

(a) To solve simultaneously, we add the equations in Ta and Tb :

T T T Ta b a b+ + − = +165 52 3N N.

Ta = =217108

N2

N

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

r

T m gvr

T

c∑ = − = =

= +FHG

IKJ

= +LNMM

OQPP = >

:

. ..

..

2

2

2

85 0 9 808 00

10 01 38 1 000 kg 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 mamv

Ry y∑ = =2

mg nmv

R− =

2

n mgmv

R= −

2

(b) When n = 0 , mgmv

R=

2

Then, v gR= .

P5.23 Fmv

rmg ny∑ = = +

2

But n = 0 at this minimum speed condition, so

mv

rmg v gr

2

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

r

r

FIG. P5.23

P5.24 (a) avrc =2

rvac

= = =2 213 0

2 9 808 62

.

..

m s

m s m

2

b ge j

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

Newton’s law gives

Mg nMv

r+ =

2

rr

FIG. P5.24

n Mvr

g M g g Mg= −FHG

IKJ = − =

2

2b g , downward

continued on next page


(c) avrc =2

ac = =13 0

20 08 45

2.

..

m s

m m s2b g

If the force exerted by the rail is n1

then n MgMv

rMac1

2

+ = =

n M a gc1 = −b g which is < 0 since ac = 8 45. m s2

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

Then a gc > . We need vr

g2

> or v rg> = 20 0 9 80. . m m s2a 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= bg

vT=

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

Thus, vytT = = =1 50

0 300.

.m

5.00 s m s

Then b = = −9 800 300

32 7 1..

. m s

m s s

2

(b) At t = 0, v = 0 and a g= = 9 80. m s2 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 = 0 020 1. m2, R ADv mgT= 1 =2

2ρair

m V= = LNM

OQP =ρ πbead

3 g cm cm kg0 83043

8 00 1 783. . .a f

Assuming a drag coefficient of D = 0 500. for this spherical object, and taking the density of

air 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 ghf i2 2 2 0 2= + = + : h

v

gf= = =2 2

2

53 8

2 9 80148

.

.

m s

m s m

2

b ge j

Chapter 5 125

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

∴ = =×

×= ⋅

−

−bmgvT

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

or at time tmb

= −FHGIKJ = × −ln . .0 368 2 04 10 3a f s

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

P5.28 vmgb

btm

= FHGIKJ − −F

HGIKJ

LNM

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

At t → ∞ , v vmgbT→ =

At t = 5 54. s 0 500 15 54

9 00. exp

..

v vb

T T= −−FHG

IKJ

LNMM

OQPP

s kga f

exp.

.. ;

.

.ln . . ;

. .

..

−FHG

IKJ =

−= = −

= =

b

b

b

554

9 000500

554

9 000500 0 693

9 00 0 693

554113

s

kg

s

kg

kg

s kg s

b g

b g

b gb g

(a) vmgbT = vT = =

9 00 9 80

1 1378 3

. .

..

kg m s

kg s m s

2b ge j

(b) 0 750 11 13

9 00. exp

..

v vt

T T= − −FHG

IKJ

LNM

OQP s exp

..

.−FHG

IKJ =

1 139 00

0 250t

s

t =−

=9 00 0 250

1 1311 1

. ln ..

.a f

s s

(c) dxdt

mgb

btm

= FHGIKJ − −FHG

IKJ

LNM

OQP1 exp ; dx

mgb

btm

dtx

x t

0

10

z z= FHG IKJ − −FHGIKJ

LNM

OQPexp

x xmgt

bm gb

btm

mgtb

m gb

btm

t

− = +FHGIKJ

−FHGIKJ = +

FHGIKJ

−FHGIKJ −

LNM

OQP0

2

20

2

2 1exp exp

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

IKJJ − −9 00

554 9 00 9 80

1130 693 1

2

2.

. . .

.exp . kg 9.80 m s

s

1.13 kg s

kg m s

kg s

22

e jb g e jb g b g

x = + − =434 626 0 500 121 m 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 − = FHGIKJ20 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

dvdt

cv e cvict= = − = −−

P5.30 F ma∑ =

− =

− =

− =

− − =−

= − +

= + =+

=+

z z −

−

kmv mdvdt

kdtdvv

k dt v dv

k tv

v v

v vkt

v ktv

vvv 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 FGm m

r= =

×= ×

−−1 2

2

11

29

6 672 10 2 2

0 302 97 10

.

..

e ja fa fa f N

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

FGm m

rg = =× ⋅−

−1 22

11

27

6 67 10 70 70

210

.~

N m kg kg kg

m N

2 2e jb gb ga f

P5.33 aMG

RE

= = =4

9 816

0 6132b g.

. m s

m s2

2 toward the earth

P5.34 F kq q

re= = ×

+ −= − × = ×1 2

122

92

6 68 99 1040 40

20003 60 10 3 60 10

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

Chapter 5 127

Section 5.6 Context ConnectionDrag Coefficients of Automobiles

*P5.35 The resistive force is

R D Av

R

aRm

= =

=

= − = − = −

12

12

0 250 1 20 2 20 27 8

255255

0 212

2 2ρ . . . .

.

a fe je jb g kg 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 2ρ . . . .a fe je jb g kg m m m s N3 2 .

Now Newton’s second law is F max x∑ =

+ − =

= ×

f

f

s

s

53 0 1 300

3 95 103

.

.

N kg 3 m s

N forward

2e j

(b) Newton’s second law changes to

+ − =

=−

=

3 95012

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 95012

0 34 1 2 2 6 0

3 9500 530

86 3

2 Nkgm

kg m m

s kg m s2

. . .

..

a fa fa f v

v

(d) Still again, + − =3 95012

0 2 1 2 2 6 02 Nkgm

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

(a), (b) Equilibrium a = 0a f

and frictionless incline f f1 2 0= =b g Under these conditions, the equations reduce to (1’) T mg1 2= sinθ

(2’) T T mg2 1− = sinθ (3’) T Mg2 =

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 mg2 3= sinθ

(c), (d) M m= 6 sinθ (double the value found above), and f f1 2 0= = . With these conditions present,

the equations become T m g a1 2= +sinθb g, T T m g a2 1− = +sinθb g and T m g a2 6= −sinθb g . Solved simultaneously, these yield

ag

=+

sinsin

θθ1 2

, T mg1 41

1 2=

++FHG

IKJsin

sinsin

θθθ

and T mg2 61

1 2=

++FHG

IKJsin

sinsin

θθθ

(e) Equilibrium a = 0a f and impending motion up the incline so M M= max while

f mgs1 2= µ θcos and f mgs2 = µ θcos , both directed down the incline. Under these conditions, the equations become T mg s1 2= +sin cosθ µ θb g, T T mg s2 1− = +sin cosθ µ θb g , and

T M g2 = max , which yield M m smax sin cos= +3 θ µ θb g

(f) Equilibrium a = 0a f and impending motion down the incline so M M= min , while

f mgs1 2= µ θcos and f mgs2 = µ θcos , both directed up the incline. Under these conditions, the equations are T mg s1 2= −sin cosθ µ θb g, T T mg s2 1− = −sin cosθ µ θb g , and T M g2 = min ,

which yield M m smin sin cos= −3 θ µ θb g . When this expression gives a negative value, it

corresponds physically to a mass M hanging from a cord over a pulley at the bottom end of the incline.

(g) T T M g M g mgs2 2 6,max ,min max min cos− = − = µ θ

Chapter 5 129

P5.38 For the system to start to move when released, the force tending to move m2 down the incline, m g2 sinθ , must exceed the maximum friction force which can retard the motion:

f f f n n

f m g m gs s

s s

max max max= + = += +

1 2 1 1 2 2

1 1 2 2

, , , ,

max , , cos

µ µµ 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 fmax N= 38 9. . This exceeds the force tending to cause the

system to move, m g2 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 normal

force acting on it be n and the friction force, fs . Resolving vertically: n F Pg= + sinθ . Horizontally: P fscosθ = . But, f ns s≤ µ i.e.,

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

P Fs s g1− ≤µ θ µ θtan seca f . Then

PFs g

sminimum = −

µ θµ θ

sec

tan1.

rr

r

r

FIG. P5.39

(b) P=( )

−0 400 100

1 0 400. sec

. tan N θ

θ

θ degb g 0.00 15.0 30.0 45.0 60.0

P Na 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 Ts

+ − =

= −

sin

sin

θµ θ0

b g and

T mg Ts scos sinθ µ µ θ− + = 0 so

r

r

r

r

FIG. P5.40

Tmgs

s=

+µ

θ µ θcos sin.

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

d

d s sθθ µ θ θ µ θcos sin sin cos+ = = − +b g 0 .

(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 f T m ak1 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 fµ s T (1) Σ =F m a2 2 : + °− − =m g f T m ak2 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 fµ 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 C

C

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 rr

r

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

F ma n C

ny y B

B

∑ = + − ==

: cos

.

θ 0

357 N

(d) For the foot with motion impending,

F ma f C

n CC

n

x x s s

s B s

ss

B

s ss

∑ = + − ==

= = =

: sinsinsin cos sin

tan .

θµ θ

µ θ θ θθ

0

357

357

N

Nb g

(e) The maximum coefficient is

µ θs s= = °=tan tan . .50 2 1 20 .


P5.43 (a) First, draw a free-body diagram, (top figure) of the top block. Since ay = 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 s2 (for top block). Now draw a free-

body diagram (middle figure) of the bottom block and observe that Σ =F Max B gives f aB= =5 88 8 00. . N kgb g

or aB = 0 735. m s2 (for the bottom block). In time t, the distance 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

12

2 0612

0 735 3 002 2. . . m s m s m2 2e 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 1 and appear in both diagrams as action-reaction pairs

(b) 5.00 kg: Σ =F max : n m g1 1 5 00 9 80 49 0= = =. . .a f N f T1 0− =

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

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

Σ =Fy 0: n n2 1 98 0 0− − =.


Chapter 5 133

f n n2 2 1 98 0 0 20 49 0 98 0 29 4= = + = + =µ µ . . . . .b g a f N

45 0 9 80 29 4 10 0. . . .− − = a a = 0 580. m s2

*P5.45 (a) If the car is about to slip down the incline, f is directed up

the incline.

F n f mgy∑ = + − =cos sinθ θ 0 where f ns= µ gives

nmg

s=

+cos tanθ µ θ1b g and fmgs

s=

+µ

θ µ θcos tan1b g .

Then, F n f mv

Rx∑ = − =sin cos minθ θ2

yields

vRg s

smin

tantan

=−

+θ µ

µ θb g

1.

When the car is about to slip up the incline, f is directed

down the incline. Then, F n f mgy∑ = − − =cos sinθ θ 0

with f ns= µ yields

nmg

s=

−cos tanθ µ θ1b g and fmgs

s=

−µ

θ µ θcos tan1b g .

In this case, F n f mv

Rx∑ = + =sin cos maxθ θ2

, which gives

vRg s

smax

tantan

=+

−θ µ

µ θb g

1.

(b) If vRg s

smin

tantan

=−

+=

θ µµ θb g

10 , then µ θs = tan .

(c) vmin

. tan . .

. tan ..=

°−

+ °=

100 9 80 10 0 0 100

1 0 100 10 08 57

m m s m s

2a fe ja fa f

vmax

. tan . .

. tan ..=

°+

− °=

100 9 80 10 0 0 100

1 0 100 10 016 6

m m s m s

2a fe ja fa f

θ

θ

m gr

fr

nr

mg

f sin θ

f cosθ

ncosθ

nsin θ

θ

θ

m gr

fr

nr

mg

f sin θ

f cosθ

ncosθ

nsin θ

FIG. 5.45


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

fk = =0 36 62 1 22 3. . . N Na f . Now adding + + − =T a14 3 22 3 6 5. . . N N kgb g and − + =T a37 2 3 8. . N kgb g gives

37 2 8 01 10 3. . . N N kg− = b ga

a = 2 84. m s2 .

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 the

horizontal. It has coefficient of friction 0.36 with the incline. It is pulled forward, which is down the incline, by the tension in a cord running to a hanging 3.8-kg object.

fk

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 radial component of F ma∑ = reads

n mgmv

r+ =sinθ

2

.

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

gvr

sin682

° = .

R68°

p

mg

p

mg cos68°

mg sin68°

FIG. P5.47

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

The rate of revolution is

angular speed = FHGIKJFHG

IKJ = =1 73

12

22 0 33

0 835 50 1..

. . m s rev

m rev s rev minb g a fπ

ππr

r.

Chapter 5 135

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

Horizontally: Tmv

rsinθ =

2

Vertically: T mgcosθ = where r is the radius of the curve, and v is the speed of the car.

By division, tanθ = vrg

2

Then avr

gc = =2

tanθ : ac = °9 80 15 0. tan . m s2e j

ac = 2 63. m s2

r

r

r

FIG. P5.48

(b) rvac

=2

r = =23 0

2 63201

2.

.

m s

m s m2

b g

(c) v rg2 201 9 80 9 00= = °tan . tan .θ m m s2a fe j v = 17 7. m s

P5.49 (a) Since the centripetal acceleration of a person is downward (toward

the axis of the earth), it is equivalent to the effect of a falling elevator. Therefore,

′ = −F Fmv

rg g

2

or F Fg g> ′

(b) At the poles v = 0 and ′ = = = =F F mgg g 75 0 9 80 735. .a f N down.

rr

FIG. P5.49

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

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

(c) From Fm v

Rc = 12

we have vRFm

mm

gRc= =FHGIKJ1

2

1.


*P5.51 vr

T= = =

2 2 9 0015 0

3 77π π .

..

m s

m sa fa f

(a) avrr = =2

1 58. m s2

(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 15agr inward .

*P5.52 (a) The mass at the end of the chain is in vertical

equilibrium. Thus T mgcosθ = .

Horizontally T mamv

rrsinθ = =2

r

r

= += °+ =

2 50 4 00

2 50 28 0 4 00 5 17

. sin .

. sin . . .

θa fa f

m

m m

Then av

r =2

5 17. m.

By division tan.

θ = =ag

vg

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 mgcosθ =

Tmg

= =°

=cos

. .

cos .θ

50 0 9 80

28 0555

kg m s N

2b ge j

T

R = 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 force

exerted by the rim. This inward force causes the 3 00. m s2 centripetal acceleration:

avrc =2

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

The period of rotation comes from vr

T= 2π

: Tr

v= = =2 2 60 0

13 428 1

π π ..

. m

m s s

a f

so the frequency of rotation is fT

= = = FHGIKJ =

1 128 1

128 1

602 14

. ..

s ss

1 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 m3e j . . Its weight is

F mg Vgg = = = × ×

= ×

−

−

ρ 19 3 10 524 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 may 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) ∆y v ty=

t =×

= × =−0 08

7 59 10 2115

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 3 rev min m

rev min60 s

radian s m m sπ .

. .a f a fF

HGIKJFHGIKJ = = .

Now avrc = = = ×2 2

328 3

0 098 88 10

.

..

m s

m m s2b g

.

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

0 018 8 5 10 1 01 10

9 55 10

0 088 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

vxt

t

∆


P5.55 (a) nmv

R=

2

f mg− = 0

f ns= µ vR

T= 2π

TRg

s=4 2π µ

(b) T = 2 54. s

# . revmin

rev2.54 s

smin

revmin

= FHGIKJ =

1 6023 6

rfs

rn

mrg

FIG. P5.55

P5.56 (a) The bead moves in a circle with radius v R= sinθ at a

speed of

vr

TRT

= =2 2π π θsin

The normal force has

an inward radial component of n sinθ

and an upward component of n cosθ

F ma n mgy y∑ = − =: cosθ 0

or

nmg

=cosθ

r

r

FIG. P5.56

Then F n mvrx∑ = =sinθ2

becomes mg m

RRTcos

sinsin

sinθ

θθ

π θFHGIKJ = F

HGIKJ

2 2


Chapter 5 139

which reduces to g R

Tsin

cossinθ

θπ θ= 4 2

2

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

and cosθπ

=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 rotation

must be faster than a certain threshold value in order for the bead to move away from the lowest 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 π1 00043

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 the

coefficients are of nearly equal magnitude.

4 11 10 3 10 10 0 870 10

3 10 3 10 4 0 870 4 11

2 0 8701 03

8 8 8 2

2

. . .

. . . .

..

× = × + ×

=− ± +

=

− − −e j e ja f a fa fa 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: 4 11 10 0 870 105 6 2. .× = ×− −e jv

v = 6 87. m s

*P5.58 (a) t ds m

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

a f a f1 002 003 004 005 006 007 008 009 00

10 011 012 013 014 015 016 017 018 019 020 0

4 8818 942 173 8

112154199246296347399452505558611664717770823876

(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 terminal

speed.

vT = = −−

=slopem m

20.0 s s m s

876 39911 0

53 0.

.

Chapter 5 141

P5.59 F L T mg L T ma

F L T L T mvr

mvr

L TL 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 20 016 3

20 07 35

20 03 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 advdt

kdxdt

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 for 2 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) mgmv

R−

2

upward; (b) gR

P5.24 (a) 8.62 m; (b) Mg downward (c) 8 45. m s2 unless they have belts, the

riders 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 s2 P5.46 (a) a = 2 84. m s2 , 26.5 N; (b) see the solution P5.48 (a) 2 63. m s2 ; (b) 201 m; (c) 17 7. m s

P5.50 (a) m g2 ; (b) m g2 ; (c) mm

gR2

1

FHGIKJ

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 s2 ; (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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