34
5 The Laws of Motion CHAPTER OUTLINE 5.1 The Concept of Force 5.2 Newton’s First Law and Inertial Frames 5.3 Mass 5.4 Newton’s Second Law 5.5 The Gravitational Force and Weight 5.6 Newton’s Third Law 5.7 Some Applications of Newton’s Laws 5.8 Forces of Friction ANSWERS TO QUESTIONS Q5.1 (a) The force due to gravity of the earth pulling down on the ball—the reaction force is the force due to gravity of the ball pulling up on the earth. The force of the hand pushing up on the ball—reaction force is ball pushing down on the hand. (b) The only force acting on the ball in free-fall is the gravity due to the earth—the reaction force is the gravity due to the ball pulling on the earth. Q5.2 The resultant force is zero, as the acceleration is zero. *Q5.3 Answer (b). An air track or air table is a wonderful thing. It exactly cancels out the force of the Earth’s gravity on the gliding object, to display free motion and to imitate the effect of being far away in space. Q5.4 When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only force on him is the friction between his shoes and the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but the rest of his body experiences almost no accelerating force (only that due to his being attached to his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newton’s first law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls into her lap. Both performers won Academy Awards. *Q5.5 Shake your hands. In particular, move one hand down fast and then stop your hand abruptly. The water drops keep moving down, according to Newton’s first law, and leave your hand. This method is particularly effective for fat, large-mass, high-inertia water drops. Q5.6 First ask, “Was the bus moving forward or backing up?” If it was moving forward, the passenger is lying. A fast stop would make the suitcase fly toward the front of the bus, not toward the rear. If the bus was backing up at any reasonable speed, a sudden stop could not make a suitcase f ly far. Fine her for malicious litigiousness. *Q5.7 (a) The air inside pushes outward on each patch of rubber, exerting a force perpendicular to that section of area. The air outside pushes perpendicularly inward, but not quite so strongly. (b) As the balloon takes off, all of the sections of rubber feel essentially the same outward forces as before, but the now-open hole at the opening on the west side feels no force. The vector sum of the forces on the rubber is to the east. The small-mass balloon moves east with a large acceleration. (c) Hot combustion products in the combustion cham- ber push outward on all the walls of the chamber, but there is nothing for them to push on at the open rocket nozzle. The net force exerted by the gases on the chamber is up if the nozzle is pointing down. This force is larger than the gravitational force on the rocket body, and makes it accelerate upward. 93 13794_05_ch05_p093-126.indd 93 13794_05_ch05_p093-126.indd 93 1/3/07 5:16:39 PM 1/3/07 5:16:39 PM

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Page 1: Sm chapter5

5The Laws of Motion

CHAPTER OUTLINE

5.1 The Concept of Force5.2 Newton’s First Law and Inertial

Frames5.3 Mass5.4 Newton’s Second Law5.5 The Gravitational Force and

Weight5.6 Newton’s Third Law5.7 Some Applications of Newton’s

Laws5.8 Forces of Friction

ANSWERS TO QUESTIONS

Q5.1 (a) The force due to gravity of the earth pulling down on the ball—the reaction force is the force due to gravity of the ball pulling up on the earth. The force of the hand pushing up on the ball—reaction force is ball pushing down on the hand.

(b) The only force acting on the ball in free-fall is the gravity due to the earth—the reaction force is the gravity due to the ball pulling on the earth.

Q5.2 The resultant force is zero, as the acceleration is zero.

*Q5.3 Answer (b). An air track or air table is a wonderful thing. It exactly cancels out the force of the Earth’s gravity on the gliding object, to display free motion and to imitate the effect of being far away in space.

Q5.4 When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only force on him is the friction between his shoes and the fl oor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but the rest of his body experiences almost no accelerating force (only that due to his being attached to his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newton’s fi rst law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls into her lap. Both performers won Academy Awards.

*Q5.5 Shake your hands. In particular, move one hand down fast and then stop your hand abruptly. The water drops keep moving down, according to Newton’s fi rst law, and leave your hand. This method is particularly effective for fat, large-mass, high-inertia water drops.

Q5.6 First ask, “Was the bus moving forward or backing up?” If it was moving forward, the passenger is lying. A fast stop would make the suitcase fl y toward the front of the bus, not toward the rear. If the bus was backing up at any reasonable speed, a sudden stop could not make a suitcase f ly far. Fine her for malicious litigiousness.

*Q5.7 (a) The air inside pushes outward on each patch of rubber, exerting a force perpendicular to that section of area. The air outside pushes perpendicularly inward, but not quite so strongly. (b) As the balloon takes off, all of the sections of rubber feel essentially the same outward forces as before, but the now-open hole at the opening on the west side feels no force. The vector sum of the forces on the rubber is to the east. The small-mass balloon moves east with a large acceleration. (c) Hot combustion products in the combustion cham-ber push outward on all the walls of the chamber, but there is nothing for them to push on at the open rocket nozzle. The net force exerted by the gases on the chamber is up if the nozzle is pointing down. This force is larger than the gravitational force on the rocket body, and makes it accelerate upward.

93

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*Q5.8 A portion of each leaf of grass extends above the metal bar. This portion must accelerate in order for the leaf to bend out of the way. The leaf’s mass is small, but when its acceleration is very large, the force exerted by the bar on the leaf puts the leaf under tension large enough to shear it off.

Q5.9 The molecules of the fl oor resist the ball on impact and push the ball back, upward. The actual force acting is due to the forces between molecules that allow the fl oor to keep its integrity and to prevent the ball from passing through. Notice that for a ball passing through a window, the molecular forces weren’t strong enough.

*Q5.10 The child exerts an upward force on the ball while it is in her hand, but the question is about the ball moving up after it leaves her hand. At those moments, her hand exerts no force on the ball, just as you exert no force on the bathroom scale when you are standing in the shower. (a) If there were a force greater than the weight of the ball, the ball would accelerate upward, not downward. (b) If there were a force equal to the weight of the ball, the ball would move at constant velocity, but really it slows down as it moves up. (c) The “force of the throw” can be described as zero, because it shows up as zero on a force sensor stuck to her palm. (d) The ball moves up at any one moment because it was moving up the previous moment. A limited down-ward acceleration acting over a short time has not taken away all of its upward velocity. We could say it moves up because of ‘history’ or ‘pigheadedness’ or ‘inertia.’

*Q5.11 Since they are on the order of a thousand times denser than the surrounding air, we assume the snowballs are in free fall. The net force on each is the gravitational force exerted by the Earth, which does not depend on their speed or direction of motion but only on the snowball mass. Thus we can rank the missiles just by mass: d > a = e > b > c.

Q5.12 It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass, gravity pulls it downward. A vertical component of the tension must balance the weight for the cable to be in equilibrium. If the cable were completely horizontal, then there would be no verti-cal component of the tension to balance the weight.

Some physics teachers demonstrate this by asking a beefy student to pull on the ends of a cord supporting a can of soup at its center. Some get two burly young men to pull on opposite ends of a strong rope, while the smallest person in class gleefully mashes the center of the rope down to the table. Point out the beauty of sagging suspension-bridge cables. With a laser and an optical lever, demonstrate that the mayor makes the courtroom table sag when he sits on it, and the judge bends the bench. Give them “I make the fl oor sag” buttons, available to instructors who use this manual and whose classes use the textbook. Estimate the cost of an infi nitely strong cable, and the truth will always win.

*Q5.13 The clever boy bends his knees to lower his body, then starts to straighten his knees to push his body up—that is when the branch breaks. When his legs are giving his body upward accelera-tion, the branch is exerting on him a force greater than his weight. He is just then exerting on the branch an equal-size downward force greater than his weight.

*Q5.14 Yes. The table bends down more to exert a larger upward force. The deformation is easy to see for a block of foam plastic. The sag of a table can be displayed with, for example, an optical lever.

Q5.15 As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and the scale reads the larger upward force that the fl oor exerts on them together. Around the top of the weight’s motion, the scale reads less than average. If the iron is moving upward, the lifter can declare that she has thrown it, just by letting go of it for a moment, so our answer applies also to this case.

94 Chapter 5

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The Laws of Motion 95

*Q5.16 (a) Yes, as exerted by a vertical wall on a ladder leaning against it. (b) Yes, as exerted by a ham-mer driving a tent stake into the ground. (c) Yes, as the ball accelerates upward in bouncing from the fl oor. (d) No; the two forces describe the same interaction.

Q5.17 As the sand leaks out, the acceleration increases. With the same driving force, a decrease in the mass causes an increase in the acceleration.

*Q5.18 (a) larger: the tension in A must accelerate two blocks and not just one. (b) equal. Whenever A moves by 1 cm, B moves by 1 cm. The two blocks have equal speeds at every instant and have equal accelerations. (c) yes, backward, equal. The force of cord B on block 1 is the tension in the cord.

Q5.19 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.’

*Q5.20 (a) Smaller. Block 2 is not in free fall, but pulled backward by string tension. (b) The same. Whenever one block moves by 1 cm, the other block moves by 1 cm. The blocks have equal speeds at every instant and have equal accelerations. (c) The same. The light string exerts forces equal in magnitude on both blocks—the tension in the string.

Q5.21 The tension in the rope when pulling the car is twice that in the tug-of-war. One could consider the car as behaving like another team of twenty more people.

*Q5.22 (b) Newton’s 3rd law describes all objects, breaking or whole. The force that the locomotive exerted on the wall is the same as that exerted by the wall on the locomotive. The framing around the wall could not exert so strong a force on the section of the wall that broke out.

Q5.23 The sack of sand moves up with the athlete, regardless of how quickly the athlete climbs. Since the athlete and the sack of sand have the same weight, the acceleration of the system must be zero.

*Q5.24 (i) b. In this case the compressional force on the bug’s back only has to be large enough to accelerate the smaller block. (ii) d. and (iii) d. These two forces are equal, as described by Newton’s third law.

Q5.25 An object cannot exert a force on itself. If it could, then objects would be able to accelerate themselves, without interacting with the environment. You cannot lift yourself by tugging on your bootstraps.

*Q5.26 answer (b) 200 N must be greater than the force of friction for the box’s acceleration to be forward.

*Q5.27 Static friction exerted by the road is the force making the car accelerate forward. Burning gasoline can provide energy for the motion, but only external forces—forces exerted by objects outside—can accelerate the car. If the road surface were icy, the engine would make the tires spin. The rubber contacting the ice would be moving toward the rear of the car. When the road is not icy, static friction opposes this relative sliding motion by exerting a force on the rubber toward the front of the car. If the car is under control and not skidding, the relative speed is zero along the lines where the rubber meets the road, and static friction acts rather than kinetic friction.

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96 Chapter 5

*Q5.28 (i) answer d. The stopping distance will be the same if the mass of the truck is doubled. The normal force and the frictional force both double, so the backward acceleration remains the same as without the load. (ii) answer g. The stopping distance will decrease by a factor of four if the initial speed is cut in half.

*Q5.29 Answer (d). Formulas a, b, f, and g have the wrong units for speed. Formula c would give an imaginary answer. Formula e would imply that a more slippery table, with smaller µ, would require a larger original speed, when really it would require a smaller original speed.

*Q5.30 Answer (e). All the other possibilities would make the total force on the crate be different from zero.

Q5.31 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 “pump-ing” the brakes (much more rapidly that you can) to minimize skidding of the tires on the road.

Q5.32 As you pull away from a stoplight, friction is the force that accelerates forward a box of tissues on the level fl oor of the car. At the same time, friction exerted by the ground on the tires of the car accelerates the car forward. When you take a step forward, friction exerted by the fl oor on your shoes causes your acceleration.

*Q5.33 (a) B (b) B (c) B Note that the mass of the woman is more than one-half that of the man. A free-body diagram of the pulley is the best guide for explanation. (d) A.

SOLUTIONS TO PROBLEMS

Section 5.1 The Concept of Force

Section 5.2 Newton’s First Law and Inertial Frames

Section 5.3 Mass

Section 5.4 Newton’s Second Law

Section 5.5 The Gravitational Force and Weight

Section 5.6 Newton’s Third Law

P5.1 m

m

=

= +( )= =

3 00

2 00 5 00

6 00

.

. ˆ . ˆ

. ˆ

kg

m s2�

� �

a i j

F a ii j

F

+( )= ( ) + ( ) =

15 0

6 00 15 0 16 22 2

. ˆ

. . .

N

N N�

P5.2 For the same force F, acting on different masses

F m a= 1 1

andF m a= 2 2

(a) m

m

a

a1

2

2

1

1

3= =

(b) F m m a m m= +( ) = = ( )1 2 1 14 3 00a . m s2

a = 0 750. m s2

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The Laws of Motion 97

P5.3 m = 4 00. kg, �v ii = 3 00. ˆ m s,

�v i j8 8 00 10 0= +( ). ˆ . ˆ m s, t = 8 00. s

��

av i j= = +Δt

5 00 10 0

8 00

. ˆ . ˆ

.m s2

� �F a i j= = +( )m 2 50 5 00. ˆ . ˆ N

F = ( ) + ( ) =2 50 5 00 5 592 2. . . N

P5.4 (a) Let the x axis be in the original direction of the molecule’s motion.

v vf i at a

a

= + − = + ×( )= −

−: .670 670 3 00 10

4

13m s m s s

..47 1015× m s2

(b) For the molecule, � �F a=∑ m . Its weight is negligible.

�Fwall on molecule kg= × − ×−4 68 10 4 47 1026 15. . mm s N2

molecule on wall

( ) = − ×

= +

−2 09 10

2 0

10.

.�F 99 10 10× − N

P5.5 (a) F ma∑ = and v vf i fax2 2 2= + or ax

f i

f

=−v v2 2

2

Therefore,

F mx

F

f i

f

=−( )

= ××

v v2 2

31

5

2

9 11 107 00 10

..

kgm s22 2m s

m

( ) − ×( )⎡⎣

⎤⎦

( ) = ×

2 5 23 00 10

2 0 050 03 64 1

.

.. 00 18− N

(b) The gravitational force exerted by the Earth on the electron is its weight,

F mgg = = ×( )( ) = ×− −9 11 10 9 80 8 93 1031 30. . .kg m s2 NN

The accelerating force is 4.08 + 1011 times the weight of the electron .

P5.6 (a) F mgg = = = ( )( ) =120 4 448 120 534lb N lb lb N dow. nn

(b) mF

gg= = =534

54 5N

9.80 m skg2 .

P5.7 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just

canceled out by a glass of tomato juice. By subtraction, F mgg p p( ) = and F mgg C C( ) = give

ΔF m g gg p C= −( ) For a person whose mass is 88.7 kg, the change in weight is

ΔFg = −( ) =88 7 9 809 5 9 780 8 2 55. . . .kg N

A precise balance scale, as in a doctor’s offi ce, reads the same in different locations because it compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.

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98 Chapter 5

P5.8 We fi nd acceleration:

� � � �

r r v a

i j a

f i it t− = +

1

2

4 20

2

. ˆ ˆm 3.30 m = 0+1

211 20 0 720

5 83 4 58

2. .

. ˆ . ˆ

s s

m s

2

2

( ) =

= −( )

a

a i j ..

Now � �F a∑ = m becomes

� � �

�F F a

F i j

g m+ =

= −( ) +

2

2 2 80 2. ˆ ˆkg 5.83 4.58 m s2 .. . ˆ

. ˆ . ˆ .

80 9 80

16 3 14 62

kg m s

N

2( )( )= +( )

j

F i j�

P5.9 (a) � � �F F F i j∑ = + = +( )1 2 20 0 15 0. ˆ . ˆ N

� � �

�F a i j a

a i

∑ = + =

= +

m : . ˆ . ˆ .

. ˆ . ˆ

20 0 15 0 5 00

4 00 3 00 jj( ) m s2

or

a = =5 00 36 9. .m s at2 θ °

(b) F

Fx

y

2

2

15 0 60 0 7 50

15 0 60 0 13

= == =

. cos . .

. sin . .

°

°

N

00

7 50 13 0

27 5

2

1 2

N

N�

� � �F i j

F F F i

= +( )= + =∑

. ˆ . ˆ

. ˆ ++( ) = =

= +( )13 0 5 00

5 50 2 60

. ˆ .

. ˆ . ˆ

j a a

a i j

N

m

m� �

�ss m s at 25.32 2= 6 08. °

*P5.10 (a) force exerted by spring on hand, to the left; force exerted by spring on wall, to the right. (b) force exerted by wagon on handle, downward to the left. Force exerted by wagon on planet, upward. Force exerted by wagon on ground, downward. (c) Force exerted by football on player, downward to the right. Force exerted by football on planet, upward. (d) Force exerted by small-mass object on large-mass object, to the left. (e) Force exerted by negative charge on positive charge, to the left. (f ) Force exerted by iron on magnet, to the left.

P5.11 (a) You and the earth exert equal forces on each other: m g M ay e e= . If your mass is 70.0 kg,

ae =( )( )

×= −70 0

5 98 101024

22.

.~

kg 9.80 m s

kgm

2

ss2

(b) You and the planet move for equal time intervals according to x at= 1

22. If the seat is

50.0 cm high,

2 2x

a

x

ay

y

e

e

=

xa

ax

m

mxe

e

yy

y

ey= = = ( )

×70 0 0 500

5 98 1024

. .

.

kg m

kkgm~ 10 23−

FIG. P5.9

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The Laws of Motion 99

*P5.12 The free-body diagrams (a) and (b) are included in the following diagram. The action–reaction pairs (c) are shown joined by the dashed lines.

P5.13 (a) 15 0. b upl to counterbalance

the Earth’s force on the block

(b) 5 00. lb up The forces on the

block are now the Earth pulling down with 15 lb and the rope pulling up with 10 lb.

(c) 0 The block now accelerates

up away from the fl oor.

P5.14 � �F a∑ = m reads

− + + − −( ) =2 00 2 00 5 00 3 00 45 0 3. ˆ . ˆ . ˆ . ˆ . ˆ .i j i j i N m 775 m s2( ) a

where a represents the direction of �a

− −( ) = ( )42 0 1 00 3 75. ˆ . ˆ . ˆi j aN m s2m

�F∑ = ( ) + ( )42 0 1 002 2. . N at tan

.

.− ⎛

⎝⎞⎠

1 1 00

42 0 below the –x axis

�F a∑ = = ( )42 0 3 75. . ˆN at 181 m s2° m

For the vectors to be equal, their magnitudes and their directions must be equal.

(a) Therefore a is at 181° counterclockwise from the x axis

(b) m = =42 011 2

..

N

3.75 m skg2

(d) � � �v v af i t= + = + ( )0 3 75 10 0. .m s at 181 s2 ° so

�v f = 37 5. m s at 181°

�v i jf = +37 5 181 37 5 181. cos ˆ . sin ˆm s m s° ° so

�v i jf = − −( )37 5 0 893. ˆ . ˆ m s

(c) �v f = + =37 5 0 893 37 52 2. . .m s m s

FIG. P5.12

al, brick on Earth

Fnormal, pillow on ice

Fnormal, ice on pillow

Fnormal, brick on pillow

Fnormal, pillow on brick

Fgravitational, Earth on brick

Fgravitational, Earth on pillow

Earth, includingice

Brick

Pillow

Fgravitational, brick on Earth

FgravitatioFgravitational, pillow on Earth

Earth, includingice

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100 Chapter 5

Section 5.7 Some Applications of Newton’s Laws

*P5.15 As the worker through the pole exerts on the lake bottom a force of 240 N downward at 35° behind the vertical, the lake bottom through the pole exerts a force of 240 Nupward at 35° ahead of the vertical. With the x axis hori-zontally forward, the pole force on the boat is

240 35 240 35 138 197N N N Ncos ˆ sin ˆ ˆ ˆ° °j i i j+ = +

The gravitational force of the whole Earth on boat and worker is F mgg = = ( ) =370 3630kg 9.8 m s N down2 . The acceleration of the boat is purely horizontal, so

F may y∑ = gives + + − =B 197 3630 0N N .

(a) The buoyant force is B = ×3 43 103. N .

(b) The acceleration is given by F max x∑ = : + − = ( )138 47 5 370N N kg. a;

a = =90 20 244

..

N

370 kgm s2. According to the constant-acceleration model,

v vxf xi xa t= + = + ( )( ) =0 857 0 244 0 450 0. . .m s m s s2 ..

. ˆ

967

0 967

m s

m s�v if =

P5.16 vx

dx

dtt= = 10 , vy

dy

dtt= = 9 2

ad

dtxx= =v

10, ad

dtty

y= =v

18

At t = 2 00. s, a ax y= =10 0 36 0. , .m s m s2 2

F max x∑ = : 3 00 10 0 30 0. . .kg m s N2( ) =

F may y∑ = : 3 00 36 0 108. .kg m s N2( ) =

F F Fx y∑ = + =2 2 112 N

P5.17 m

mg

==

=

=

1 00

9 80

0 200

0 458

.

.

tan.

.

kg

N

m

25.0 mα

α °°

Balance forces,

2

9 80

2613

T mg

T

sin

.

sin

α

α

=

= =NN

FIG. P5.15

197 N

138 N 3630 N

47.5 N

B

FIG. P5.17

50.0 m

0.200 mα

mg

TT

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The Laws of Motion 101

*P5.18 (a)

FIG. P5.18

T sin 38.3°

28.0 cm

35.7 cmθ

T

mg

Fmagnetic

T cos 38.3°

0.637 N

Fmagneticθ

The fi rst diagram shows the geometry of the situation and lets us fi nd the angle of the string with the horizontal: cosθ = 28�35.7 = 0.784 θ = 38.3°. The second diagram is the free body diagram, and the third diagram is the same free body diagram with some calculated results already shown, including 0.065 kg (9.8 m�s2) = 0.637 N.

(b) ∑Fx = ma

x: −T cos 38.3° + F

magnetic = 0

∑Fy = ma

y: +T sin 38.3° − 0.637 N = 0

from the second equation, T = 0.637 N� sin 38.3° = 1.03 N

(c) Now Fmagnetic

= 1.03 N cos 38.3° = 0.805 N to the right .

*P5.19 (a) P cos 40°− n = 0 and P sin 40°− 220 N = 0 P = 342 N and n = 262 N

(b) P − n cos 40°− 220 N sin 40° = 0 and n sin 40 − 220 N cos 40° = 0 n = 262 N and P = 342 N.

(c) The results agree. The methods are basically of the same level of diffi culty. Each involves one equation on one unknown and one equation in two unknowns. If we are interested in fi nding n without fi nding P, method (b) is simpler.

P5.20 From equilibrium of the sack: T Fg3 = (1)

From ∑Fy = 0 for the knot: T T Fg1 1 2 2sin sinθ θ+ = (2)

From ∑Fx = 0 for the knot: T T1 1 2 2cos cosθ θ= (3)

Eliminate T2 = T T2 1 1 2= cos / cosθ θ and solve for T1

TF Fg g

12

1 2 1 2

2=+( ) =

cos

sin cos cos sin

cos

si

θθ θ θ θ

θnn

cos .

sin .

θ θ1 2

3

1

325

25 0

85 0

+( )= =

= ⎛⎝

T F

T F

g

g

N

°

°⎞⎞⎠ =

=⎛⎝⎜

⎞⎠⎟

=

296

29660 0

2 11

2

N

NT Tcos

cos

cos .θθ

°°

°cos .25 0163⎛

⎝⎞⎠ = N

P5.21 See the solution for T1 in Problem 5.20. The equation indicates that the tension is directly pro-portional to Fg. As θ θ1 2+ approaches zero (as the angle between the two upper ropes approaches 180°) the tension goes to infi nity. Making the right-hand rope horizontal maximizes the tension in the left-hand rope, according to the proportionality of T1 to cosθ2.

FIG. P5.20

θ1

θ1 θ2

F

θ2

3

T21

g

CE

ME

NT

T

T

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Page 10: Sm chapter5

102 Chapter 5

P5.22 (a) An explanation proceeding from fundamental physical principles will be best for the parents and for you. Consider forces on the bit of string touching the weight hanger as shown in the free-body diagram:

Horizontal Forces: F max x∑ = : − + =T Tx cosθ 0

Vertical Forces: F may y∑ = : − + =F Tg sinθ 0

You need only the equation for the vertical forces to fi nd that the tension in the string is

given by TFg=

sinθ. The force the child feels gets smaller, changing from T to T cosθ,

while the counterweight hangs on the string. On the other hand, the kite does not notice what you are doing and the tension in the main part of the string stays constant. You do not need a level, since you learned in physics lab to sight to a horizontal line in a building. Share with the parents your estimate of the experimental uncertainty, which you make by thinking critically about the measurement, by repeating trials, practicing in advance and looking for variations and improvements in technique, including using other observers. You will then be glad to have the parents themselves repeat your measurements.

(b) TFg= =

( )=

sin

. .

sin ..

θ0 132 9 80

46 31 79

kg m sN

2

°

*P5.23 (a) Isolate either mass

T mg ma

T mg

+ = =

=

0

.

The scale reads the tension T, so

T mg= = ( ) =5 00 9 80 49 0. . .kg m s N2

(b) The solution to part (a) is also the solution to (b).

(c) Isolate the pulley

� �T T2 1

2 1

2 0

2 2 98 0

+ =

= = =T T mg . .N

(d) � � � �F n T g∑ = + + =m 0

Take the component along the incline

n T mgx x x+ + = 0

or

0 30 0 0

30 02

5 00 9 80

+ − =

= = = ( )T mg

T mgmg

sin .

sin .. .

°

°22

24 5= . .N

FIG. P5.22

FIG. P5.23(a) and (b)

FIG. P5.23(c)

FIG. P5.23(d)

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Page 11: Sm chapter5

The Laws of Motion 103

P5.24 The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the fi gure to the right. The angle θ is the angle of inclination of the plane. Applying Newton’s second law for the accel-erating system (and taking the direction up the plane as the positive x direction) we have

F n mgy∑ = − =cosθ 0: n mg= cosθ

F mg max∑ = − =sinθ : a g= − sinθ

(a) When θ = 15 0. °

a = −2 54. m s2

(b) Starting from rest

v v

v

f i f i f

f f

a x x ax

ax

2 2 2 2

2 2 2 54

= + −( ) =

= = −( ) −. m s2 22 00 3 18. .m m s( ) =

P5.25 Choose a coordinate system with i East and j North.

� �F a∑ = = ( )m 1 00 10 0. .kg m s2 at 30 0. °

5 00 10 0 30 0 5 00 81. ˆ . . . ˆ .N N N( ) + = ( )∠ = ( ) +j F j�

° 666 N( ) i

∴ = ( )F1 8 66. N East

P5.26 First, consider the block moving along the horizontal. The only force in the direction of movement is T. Thus,

F max∑ =

T a= ( )5 kg (1)

Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N.

We have F may∑ =

88 2 9. N kg− = ( )T a (2)

Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be added to give 88 2 14. N kg= ( )a . Then

a T= =6 30 31 5. .m s and N2

*P5.27 (a) and (b) The slope of the graph of upward velocity versus time is the acceleration of the person’s body. At both time 0 and time 0.5 s, this slope is (18 cm�s) � 0.6 s = 30 cm�s2.

For the person’s body, ∑Fy = ma

y: + F

bar − 64 kg(9.8 m�s2) = 64 kg (0.3 m�s2)

Note that there is no fl oor touching the person to exert a normal force. Note that he does not exert any extra force ‘on himself .’ Solving, F

bar = 646 N up.

FIG. P5.24

FIG. P5.25

FIG. P5.26

5 kg

n

T

49 N

+x

9 kg

T +y

Fg= 88.2 N

continued on next page

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Page 12: Sm chapter5

104 Chapter 5

(c) ay = slope of v

y versus t graph = 0 at t = 1.1 s. The person is moving with maximum speed

and is momentarily in equilibrium:

+ Fbar

− 64 kg (9.8 m�s2) = 0 Fbar

= 627 N up.

(d) ay = slope of v

y versus t graph = (0 − 24 cm�s)�(1.7 s − 1.3 s) = − 60 cm�s2

+ Fbar

− 64 kg (9.8 m�s2) = 64 kg (− 0.6 m � s2) Fbar

= 589 N up.

P5.28 m1 2 00= . kg, m2 6 00= . kg, θ = 55 0. °

(a) F m g T m ax∑ = − =2 2sinθ

and

T m g m a

am g m g

m m

− =

= −+

=

1 1

2 1

1 2

3 57sin

m s2

(b) T m a g= +( ) =1 26 7. N

(c) Since vi = 0, v f at= = ( )( ) =3 57 2 00 7 14. . .m s s m s2 .

P5.29 After it leaves your hand, the block’s speed changes only because of one component of its weight:

F ma mg ma

a x x

x x

f i f i

∑ = − =

= + −( )sin .20 0

22 2

°

v v

Taking v f = 0, vi = 5 00. m s, and a g= − ( )sin .20 0° gives

0 5 00 2 9 80 20 0 02= ( ) − ( ) ( ) −( ). . sin . ° x f

or

x f =( ) ( ) =25 0

2 9 80 20 03 73

.

. sin ..

°m

P5.30 As the man rises steadily the pulley turns steadily and the tension in the rope is the same on both sides of the pulley. Choose man-pulley-and-platform as the system:

F ma

T

T

y y∑ =

+ − ==

950 0

950

N

N

The worker must pull on the rope with force 950 N .

FIG. P5.28

FIG. P5.29

FIG. P5.30

T

950 N

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Page 13: Sm chapter5

The Laws of Motion 105

P5.31 Forces acting on 2.00 kg block:

T m g m a− =1 1 (1)

Forces acting on 8.00 kg block:

F T m ax − = 2 (2)

(a) Eliminate T and solve for a:

aF m g

m mx= −

+1

1 2

a F m gx> > =0 19 61for N.

(b) Eliminate a and solve for T:

Tm

m mF m gx=

++( )1

1 22

T F m gx= ≤ − = −0 78 42for N.

(c) Fx , N −100 −78.4 −50.0 0 50.0 100

ax , m s2 −12.5 −9.80 −6.96 −1.96 3.04 8.04

P5.32 (a) Pulley P1 has acceleration a2.

Since m1 moves twice the distance P1 moves in the same time, m1 has twice the acceleration of P1, i.e., a a1 22= .

(b) From the fi gure, and using

F ma m g T m a

T m a m a

T T

∑ = − = ( )= = ( )

− =

: 2 2 2 2

1 1 1 1 2

2 1

1

2 2

2 00 3( ) Equation (1) becomes m g T m a2 1 2 22− = . This equation com-

bined with Equation (2) yields

T

mm

mm g1

11

222

2+⎛

⎝⎞⎠ =

Tm m

m mg1

1 2

112 22

=+

and Tm m

m mg2

1 2

114 2

=+

(c) From the values of T1 and T2 we fi nd that

aT

m

m g

m m11

1

2

112 22

= =+

and a am g

m m2 12

1 2

1

2 4= =

+

FIG. P5.31

T

FIG. P5.32

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Page 14: Sm chapter5

106 Chapter 5

P5.33 First, we will compute the needed accelerations:

1 0

2

( ) =

( )

Before it starts to move:

During t

ay

hhe first 0.800 s:m s

aty

yf yi=−

= −v v 1 20 0

0 80

.

. 001 50

3

sm s

While moving at constant ve

2=( )

.

llocity:

During the last 1.50 s:

a

a

y

yy

=

( ) =

0

4v ff yi

t

−= −

= −

v 0 1 20

1 500 800

.

..

m s

sm s2

Newton’s second law is: F may y∑ =

+ − ( )( ) = ( )

= +

S a

S

y72 0 9 80 72 0

706 7

. . .kg m s kg

N

2

22 0. kg( )ay

(a) When ay = 0, S = 706 N .

(b) When ay = 1 50. m s2, S = 814 N .

(c) When ay = 0, S = 706 N .

(d) When ay = −0 800. m s2, S = 648 N .

P5.34 Both blocks move with acceleration am m

m mg= −

+⎛⎝⎜

⎞⎠⎟

2 1

2 1

:

a = −+

⎛⎝⎜

⎞⎠⎟

=7 29 8 5 44

kg kg

7 kg 2 kgm s m s2 2. .

(a) Take the upward direction as positive for m1.

v vxf xi x f ia x x2 2 2 22 0 2 4 2 5 44= + −( ) = −( ) +: . .m s m s(( ) −( )= − ( ) = −

x

x

x

f

f

0

5 76

2 5 440 529

.

..

m s

m sm

2 2

2

ff = 0 529. m below its initial level

(b) v v vxf xi x xfa t= + = − + ( )( ): . . .2 40 5 44 1 80m s m s s2

vvxf = 7 40. m s upward

FIG. P5.33

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Page 15: Sm chapter5

The Laws of Motion 107

Section 5.8 Forces of Friction

P5.35 F ma n mg

f n mgy 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 gs= −μ

The initial and fi nal conditions are: xi = 0, vi = =50 0 22 4. .mi h m s, v f = 0

v v vf i f i i s fa x x gx2 2 22 2= + −( ) − = −: μ

(a) xgf

i= v2

x f =( )

( )( ) =22 4

2 0 100 9 80256

2.

. .

m s

m sm2

(b) xgf

i= v2

x f =( )

( )( ) =22 4

2 0 600 9 8042 7

2.

. ..

m s

m sm2

P5.36 For equilibrium: f F= and n Fg= . Also, f n= μ i.e.,

μ

μ

= =

=( ) =

f

n

F

Fg

s

75 0

25 0 9 800 306

.

. ..

N

N

and

μk =( ) =60 0

0 245.

.N

25.0 9.80 N

*P5.37 (a) The car’s acceleration in stopping is given by v xf

2 = v

xi 2 + 2 a

x(x

f − x

i) 0 =

(20 m �s)2 + 2 ax(45 m − 0) a

x = −4.44 m �s2.

For the book not to slide on the horizontal seat we need

F ma f max x s∑ = − =: = 3.8 kg(−4.44 m �s2) fs = 16.9 N

F ma n mgy y∑ = − =: 0 n = 3.8 kg(9.8 m �s2) = 37.2 N

To test whether the book starts to slide, we see if this required static friction force is avail-able in the allowed range

fs ≤ µ

s n = 0.65(37.2 N) = 24.2 N

Because 16.9 N is less than 24.2 N, the book does not start to slide .

(b) The actual friction force is 16.9 N backwards, and the whole force exerted by the seat on the book is 16.9 N backward + 37.2 N upward = 40.9 N upward and backward at 65.6° with the horizontal.

FIG. P5.36

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Page 16: Sm chapter5

108 Chapter 5

P5.38 If all the weight is on the rear wheels,

(a) F ma mg mas= =: μ But

Δx = =at gts2 2

2 2

μ

so μs

x

gt= 2

2

Δ:

μs =( )( )( )( )

2 0 250 1 609

9 80 4 96 2

.

. .

mi m mi

m s s2 == 3 34.

(b) Time would increase, as the wheels would skid and only kinetic friction would act; or perhaps the car would f lip over.

P5.39 m = 3 00. kg, θ = 30 0. °, x = 2 00. m, t = 1 50. s

(a) x = 1

22at :

2 001

21 50

4 00

1 501 78

2

2

. .

.

..

m s

m s2

= ( )

=( ) =

a

a

� � � � �F n f g a∑ = + + =m m :

Along :

Al

x f mg ma

f m g a

0 30 0

30 0

− + == −( )

sin .

sin .

°

°

oong :y n mg

n mg

+ − ==

0 30 0 0

30 0

cos .

cos .

°

°

(b) μk

f

n

m g a

mg= =

−( )sin .

cos .

30 0

30 0

°

°, μk

a

g= − =tan .

cos ..30 0

30 00 368°

°

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

(d) v vf i f ia x x2 2 2= + −( ) where

x xf i− = 2 00. m

v

v

f

f

2 0 2 1 78 2 00 7 11

7 11 2

= + ( )( ) =

= =

. . .

.

m s

m s

2 2

2 2 ..67 m s

P5.40 msuitcase kg= 20 0. , F = 35 0. N

F ma F

F ma n F Fx x

y y g

∑∑

= − + == + + − =

: . cos

: sin

20 0 0N θθ 00

(a) F cos .

cos.

.

.

θ

θ

θ

=

= =

=

20 0

20 00 571

55 2

N

N

35.0 N

°

(b) n F Fg= − = − ( )[ ]sin . .θ 196 35 0 0 821 N

n = 167 N

FIG. P5.39

FIG. P5.40

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Page 17: Sm chapter5

The Laws of Motion 109

P5.41 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 60

. . . . . .

.

( ) − ( )( ) =

=

a

a m s22

N

∴ = ( ) + ( )( )=

T 5 00 5 60 0 200 5 00 9 80

37 8

. . . . .

.

P5.42 Let a represent the positive magnitude of the acceleration −aj of m1, of the acceleration −ai of m2, and of the acceleration +aj 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

12

12 2

39 2 4 00

0 350 9 80

. .

. .

N kg

N 33

23

1 00

19 6 2 00

= ( )+ − = ( )

.

. .

kg

N kg

a

T a

(a) Add them up:

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

a m m= 2 31 1 2. ,m s , down for , left for and u2 pp for m3

(b) Now − + = ( )( )T12 39 2 4 00 2 31. . .N kg m s2

T12 30 0= . N

and T23 19 6 2 00 2 31− = ( )( ). . .N kg m s2

T23 24 2= . N

FIG. P5.41

FIG. P5.42

n

T12 T23

m g 2

f = n k μ

m g 1

T12

m g 3

T23

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Page 18: Sm chapter5

110 Chapter 5

P5.43 (a) See the fi gure adjoining

(b) 68 0 2 2

1 1

. − − =− =

T m g m a

T m g m a

μμ

(Block #2)

(Block #11)

Adding,

68 0

68 01 2

1 2 1 2

1 2

.

..

− +( ) = +( )=

+( ) − =

μ

μ

m m g m m a

am m

g 99

27 21 1

m s

N

2

T m a m g= + =μ .

P5.44 (a) To fi nd the maximum possible value of P, imagine impend-ing upward motion as case 1. Setting

F P n

f n f Px

s s s s

∑ = − == =

0 50 0 0: cos .

: c, ,

°

max maxμ μ oos .

. . .

50 0

0 250 0 643 0 161

°

= ( ) =P P Setting

F P P

P

y∑ = − − ( ) =

=

0 50 0 0 161 3 00 9 80 0

48

: sin . . . .

.max

°

66 N

To fi nd the minimum possible value of P, consider impend-ing downward motion. As in case 1,

f Ps , .max = 0 161

Setting

F P P

P

y∑ = + − ( ) =

=

0 50 0 0 161 3 00 9 80 0

31

: sin . . . .

.min

°

77 N

(b) If P > 48.6 N, the block slides up the wall. If P < 31.7 N, the block slides down the wall.

(c) We repeat the calculation as in part (a) with the new angle. Consider impending upward motion as case 1. Setting

F P n

f n f Px

s s s s

∑ = − == =

0 13 0: cos

: cos, ,

°

max maxμ μ 113

0 250 0 974 0 244

°

= ( ) =. . .P P Setting

F P P

Py∑ = − − ( ) =

= −0 13 0 244 3 00 9 80 0

1580

: sin . . .

max

°

N

The push cannot really be negative. However large or small it is, it cannot produce upward motion. To fi nd the minimum possible value of P, consider impending downward motion. As in case 1,

f Ps , .max = 0 244

Setting

F P P

P

y∑ = + − ( ) =

=

0 13 0 244 3 00 9 80 0

62 7

: sin . . .

.min

°

NN

P ≥ 62.7 N. The block cannot slide up the wall. If P < 62.7 N, the block slides down the wall.

FIG. P5.43

T

m 1m 2

TF

m 1

n 1

T

m g 1 = 118 N

f1 = mkn1

m 2

n 2

F

m g 2 = 176 N

f2 = mkn2

FIG. P5.44

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Page 19: Sm chapter5

The Laws of Motion 111

*P5.45 (a) If P is too small, static friction will prevent the block from moving. We will fi nd the value of P when motion is just ready to begin:

∑Fy = ma

y: −P sin 37° − 0.42 kg 9.8 m �s2 + n = 0 n = 4.12 N + P sin 37°

with motion impending we read the equality sign in

fs = µ

sn = 0.72(4.12 N + P sin 37°) = 2.96 N + 0.433 P

∑Fx = ma

x: P cos 37° − f = 0 P cos 37° − 2.96 N − 0.433 P = 0 P = 8.11 N

Thus a = 0 if P ≤ 8.11 N . If P > 8.11 N, the block starts moving. Immediately kinetic fric-tion acts, so it controls the acceleration we measure. We have again n = 4.12 N + P sin 37°

fk = µ

kn = 0.34(4.12 N + P sin 37°) = 1.40 N + 0.205 P

∑Fx = ma

x: P cos 37° − f = 0.42 kg a a = (P cos 37° − 1.40 N − 0.205 P)�0.42 kg

a = 1.41 P − 3.33 where a is in m �s2 when P is in N, to the right if P > 8.11 N

(b) Since 5 N is less than 8.11 N, a = 0 .

(c) fs ≤ µ

sn does not tell us the value of the friction force. We know that it must counterbalance

5 N cos 37° = 3.99 N, to hold the block at rest. The friction force here is

3.99 N horizontally backward .

(d) a = 1.41(10) − 3.33 = 10.8 m �s2 to the right

(e) From part (a), f = 1.40 N + 0.205(10) = 3.45 N to the left .

(f ) The acceleration is zero for all values of P less than 8.11 N. When P passes this threshold, the acceleration jumps to its minimum nonzero value of 8.14 m �s2. From there it increases linearly with P toward arbitrarily high values.

P5.46 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 fi rst case, we take x and y as parallel and perpendicular to the surface of the roof:

F ma n mg

n mgy y∑ = + − =

=: cos

cos

θθ

0

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

F ma f mg ma

a g gx x k x

x k

∑ = − − == − − = −

: sin

cos sin .

θμ θ θ 0 44 37 37 9 8 9 03cos sin . .° °−( ) = −m s m s2 2

The Frisbee goes ballistic with speed given by

v vxf xi x f ia x x2 2 2

2 15 2 9 03 1= + −( ) = ( ) + −( )m s m s2. 00 0 44 4

6 67

m m s

m s

2 2−( ) =

=

.

.vxf

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

v vyf yi y f ia y y2 2

2

2

0 6 67 37 2 9

= + −( )= ( ) + −. sinm s ° .. sin

..

8 10 37

6 024 01

m s m

mm s

2( ) −( )= +

(y

y

f

f

°

)) =2

19 66 84

..

m sm2

FIG. P5.46

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Page 20: Sm chapter5

112 Chapter 5

P5.47 Since the board is in equilibrium, Fx∑ = 0 and we see that the normal forces must be the same on both sides of the board. Also, if the mini-mum normal forces (compression forces) are being applied, the board is on the verge of slipping and the friction force on each side is

f f ns s= ( ) =max

μ

The board is also in equilibrium in the vertical direction, so

F f Fy g∑ = − =2 0, or f

Fg=2

The minimum compression force needed is then

. nf F

s

g

s

= = =( ) =

μ μ2

95 572 0

..

N

2 0.663N

P5.48 Take +x in the direction of motion of the tablecloth. For the mug:

F ma a

ax x x

x

∑ = ==

0 1 0 2

0 5

. .

.

N kg

m s2

Relative to the tablecloth, the acceleration of the mug is 0 5 3 2 5. .m s m s m s2 2 2− = − . The mug reaches the edge of the tablecloth after time given by

Δ x t a t

t

t

xi x= +

− = + −( )=

v1

2

0 3 01

22 5

0 49

2

2. .

.

m m s2

00 s. The motion of the mug relative to tabletop is over distance

1

2

1

20 5 0 490 0 060 02 2a tx = ( )( ) =. . .m s s m2

The tablecloth slides 36 cm over the table in this process.

*P5.49 (a) When the truck has the greatest acceleration it can without the box sliding, the force of friction on the box is forward and is described by f

s = µ

s n.

We also have ∑Fx = ma

x: + f

s = ma ∑F

y = ma

y: + n − mg = 0

Combining by substitution gives µs mg = ma a = 0.3 (9.8 m�s2) = 2.94 m�s2 forward.

(b) The truck is accelerating forward rapidly and exerting a forward force of kinetic friction on the box, making the box accelerate forward more slowly;

n = mg fk = µ

k mg = ma a = µ

k g = 0.25 (9.8 m�s2) = 2.45 m�s2 forward.

(c) Now take the x axis along the direction of motion and the y axis perpendicular to the slope.

We have ∑Fy = ma

y: + n − mg cos 10° = 0 + n = mg cos 10° f

s = µ

s mg cos 10°

∑Fx = ma

x: + f

s − mg sin 10° = ma

a = µs g cos 10° − g sin 10° = 0.3(9.8 m �s2) cos 10° − (9.8 m �s2) sin 10° =

1.19 m �s2 up the incline

(d) This time kinetic friction acts:

∑Fy = ma

y: + n − mg cos 10° = 0 + n = mg cos 10° f

k = µ

k mg cos 10°

∑Fx = ma

x: + f

k − mg sin 10° = ma

a = µk g cos 10° − g sin 10° = [0.25 cos 10° − sin 10°]9.8 m �s2 = 0.711 m �s2 up the incline

FIG. P5.47

fn

F = 95.5 N

fn

g

continued on next page

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Page 21: Sm chapter5

The Laws of Motion 113

(e) Model the box as in equilibrium (total force equals zero) with motion impending (static friction equals coeffi cient times normal force).

∑Fy = ma

y: + n − mg cosθ = 0 + n = mg cosθ f

s = µ

s mg cosθ

∑Fx = ma

x: + f

s − mg sinθ = ma = 0 µ

s mg cosθ − mg sinθ = 0

µs = sinθ � cosθ = tan θ θ = tan−1 0.3 = 16.7°

(f ) The mass makes no difference. Mathematically, the mass has divided out in each determi-nation of acceleration and angle. Physically, if several packages of dishes were placed in the truck, they would all slide together, whether they were tied to one another or not.

Additional Problems

*P5.50 (a) Directly n = =63 7 13 62 1. cos .N N°

fk = ( ) =0 36 62 1 22 3. . .N N

Now adding + + − = ( )T a14 3 22 3 6 5. . .N N kg and − + = ( )T a37 2 3 8. .N kg gives

37 2 8 01 10 3. . .N N kg− = ( )a

a = 2 84. m s2

Then T = − ( ) =37 2 3 8 2 84 26 5. . . .N kg m s N2 .

(b) We recognize the equations are describing a 6.5-kg block on an incline at 13° with the horizontal. It has coeffi cient 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.

FIG. P5.50

fkn

T

13º 63.7 N

T

37.2 N

3.8 kg

6.5 kg

13º

fkn

T

13º 63.7 N

T

37.2 N

3.8 kg

6.5 kg

13º

P5.51 (a) see fi gure to the right

(b) First consider Pat and the chair as the system. Note that two ropes support the sys-tem, and T = 250 N in each rope. Applying

F ma∑ = 2 480T ma− = , where

m = =480

9 8049 0

.. kg

Solving for a gives

a = − =500 480

49 00 408

.. m s2

(c) F ma∑ = on Pat:

F n T ma∑ = + − =320 , where m = =320

9 8032 7

.. kg

n ma T= + − = ( ) + − =320 32 7 0 408 320 250 83 3. . . N .

FIG. P5.51

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Page 22: Sm chapter5

114 Chapter 5

*P5.52 (a) As soon as Pat passes the rope to the other child,

Pat and the seat, with total weight480 N, wwill accelerate down andthe other child, oonly 440 N, willaccelerate up.

We have + − =480480

NN

9.8 m s2T a and + − =T a440440

NN

9.8 m s2

Adding,

+ − + − = +( )

=

480 440 49 0 44 9

40

N N kg kg

N

93

T T a

a

. .

..9 kgm s2= =0 426. a

The rope tension is T = + ( ) =440 44 9 459N kg 0.426 m s N2. .

(b) In problem 51, a rope tension of 250 N does not make the rope break. In part (a), the rope is strong enough to support tension 459 N. But now the tension everywhere in the rope is 480 N, so it can exceed the breaking strength of the rope.

The tension in the chain supporting the pulley is 480 480 960+ =N N, so that chain may break fi rst.

P5.53 � �F a∑ = m gives the object’s acceleration

ai j

a

= =−( )

=

∑ F

m

t8 00 4 00

4 00

. ˆ . ˆ

.

N

2.00 kg

m s2(( ) − ( ) =ˆ . ˆi jv

2 00 m s3 td

dt

Its velocity is

d dt

i

i

t� � � � �

v v v v a

v i

v

v

∫ ∫= − = − =

= ( ) −

0

4 00

0

. ˆm s2 22 00

4 00 1 0

0

. ˆ

. ˆ .

m s

m s

3

2

( )⎡⎣ ⎤⎦

= ( ) −

∫ t dt

t

t

j

v i�

00 2t m s3( ) ˆ .j

FIG. 5.52

480 N 440 N

aa

T

T

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Page 23: Sm chapter5

The Laws of Motion 115

(a) We require �v = 15 0. m s, �

v 2 225= m s2 2

16 0 1 00 225

1 00 16 0

2 4

4

. .

. .

t t

t

m s m s m s2 4 2 6 2 2+ =

+ s s2 4t

t

2

22

225 0

16 0 16 0 4 225

2 00

− =

=− ± ( ) − −( ). .

.==

=

9 00

3 00

.

. .

s

s

2

t

Take �ri = 0 at t = 0. The position is

� �r v i j= = ( ) − ( )⎡⎣ ⎤⎦∫ dt t t

t

0

24 00 1 00. ˆ . ˆm s m s2 3 ddt

t t

t

0

2 3

4 002

1 003

= ( ) − ( )�r i j. ˆ . ˆm s m s2 3

at t = 3 s we evaluate.

(b) �r i j= −( )18 0 9 00. ˆ . ˆ m

(c) So �r = ( ) + ( ) =18 0 9 00 20 12 2. . .m m

P5.54 (a) We write ∑Fx = ma

x for each object.

18 2

3

4

N kg

kg

kg

− = ( )− = ( )

= ( )

P a

P Q a

Q a

Adding gives 18 9N kg= ( )a so

. a = 2 00 2. m s

(b) Q = ( ) =4 2 8 00kg m s N net force on the 4 kg2 .

P − = ( ) =8 3 2 6 00N kg m s N net force on the 32 . kg and P = 14 N

18 14 2 2 4 00N N kg m s N net force on t2− = ( ) = . hhe 2 kg

FIG. P5.54

continued on next page

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Page 24: Sm chapter5

116 Chapter 5

(c) From above, Q = 8 00. N and P = 14 0. N

(d) The 3-kg block models the heavy block of wood. The contact force on your back is represented by Q, which is much less than the force F. The difference between F and Q is the net force causing acceleration of the 5-kg pair of objects. The acceleration is real and nonzero, but lasts for so short a time that it never is associated with a large velocity. The frame of the building and your legs exert forces, small relative to the hammer blow, to bring the partition, block, and you to rest again over a time large relative to the ham-mer blow. This problem lends itself to interesting lecture demonstrations. One person can hold a lead brick in one hand while another hits the brick with a hammer.

*P5.55 (a) Take the rope and block together as the system:

∑Fx = ma

x: + 12 N = (4 kg + m

1)a a = 12N�(4 kg + m

1) forward

(b) The rope tension varies along the massive rope, with the value 12 N at the front end and a value we call T

back at the back end. Take the block alone as the system:

∑Fx = ma

x: + T

back = (4 kg)a = 4 kg(12N�(4 kg + m

1) = 12 N�(1 + m

1�4 kg) forward

(c) We substitute m1 = 0.8 kg: a = 12N�(4 kg + 0.8 kg) = 2.50 m �s2 forward

Tback

= 12 N�(1 + 0.8�4) = 10.0 N forward

(d) As m1 → ∞, 12 N�(1 + m

1�4) goes to zero

(e) As m1 → 0, 12 N�(1 + m

1�4) goes to 12 N

(f ) A cord of negligible mass has constant tension along its length.

*P5.56 (a) Choose the black glider plus magnet as the system.

∑Fx = ma

x: + 0.823 N = 0.24 kg a a = 3.43 m �s2 toward the scrap iron

(b) The analysis in part (a) applies here with no change. ablack

=

3.43 m �s2 toward the scrap iron .

For the green glider with the scrap iron,

∑Fx = ma

x: + 0.823 N = 0.12 kg a a = 6.86 m �s2 toward the magnet

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Page 25: Sm chapter5

The Laws of Motion 117

P5.57 (a) First, we note that F T= 1. Next, we focus on the massM and write T Mg5 = . Next, we focus on the bottom pulley and write T T T5 2 3= + . Finally, we focus on the top pulley and write T T T T4 1 2 3= + + .

Since the pulleys are not starting to rotate and are frictionless, T T1 3= , and T T2 3= . From this

information, we have T T5 22= , so TMg

2 2= .

Then T T TMg

1 2 3 2= = = , and T

Mg4

3

2= ,

and T Mg5 = .

(b) Since F T= 1 , we have FMg=2

.

*P5.58 (a) The cord makes angle θ with the horizontal where tan θ = 0.1�0.4 θ = 14.0°.

∑Fy = ma

y: + 10 N sin 14.0° − 2.2 kg 9.8 m�s2 +

n = 0 n = 19.1 N

fk = μ

kn = 0.4(19.1 N) = 7.65 N

∑Fx = ma

x: + 10 N cos 14.0° − 7.65 N = 2.2 kg a a = 0.931 m�s2

(b) When x is large we have n = 21.6 N, f

k = 8.62 N and a = (10 N − 8.62 N) � 2.2 kg = 0.625 m �s2.

As x decreases, the acceleration increases gradually, passes through a maximum, and then drops more rapidly, becoming negative. At x = 0 it reaches the value a = [0 − 0.4(21.6 N− 10 N)] � 2.2 kg = −2.10 m �s2.

(c) We carry through the same calculations as in part (a) for a variable angle, for which cosθ = x[x2 + (.1m)2]−1�2 and sinθ = 0.1 m[x2 + (.1m)2]−1�2 We fi nd

ax x x

=+⎡⎣ ⎤⎦ − −

−10 0 1 0 4 21 6 102 2 1 2

N N N 0.1. . ./ 22 2 1 2

2 2

0 1

2 2

4 55 0 1

+⎡⎣ ⎤⎦( )

= +⎡⎣ ⎤⎦

.

.

. .

/

kg

a x x11 2 2 2 1 2

3 92 0 182 0 1/ /

. . .− + +⎡⎣ ⎤⎦−

x

Now to maximize a we take its derivative with respect to x and set it equal to zero:

da

dxx x x x= +( ) + −( ) +(−

4 55 0 1 4 55 2 0 12 2 1 2 12

2 2. . . ./ )) + −( ) +( ) =

+

− −3 2 12

2 2 3 2

2

0 182 2 0 1 0

4 55 0

/ /. .

. .

x x

x 11 4 55 0 182 0 0 252 2( ) − − = =. . .x x x 00 m

At this point a = ( ) +⎡⎣ ⎤⎦ − +−

4 55 0 25 0 1 3 92 0 182 02 2 1 2. . . . .

/0.25 .. . .

/25 0 1 0 9762 2 1 2

+⎡⎣ ⎤⎦ =−

m/s2

(d) We solve

0 4 55 0 1 3 92 0 182 0 12 2 1 2 2 2= +⎡⎣ ⎤⎦ − + +⎡⎣−

. . . . ./

x x x ⎤⎤⎦

+⎡⎣ ⎤⎦ = +

−1 2

2 2 1 2

2

3 92 0 1 4 55 0 182

15 4

/

/. . . .

.

x x

x ++⎡⎣ ⎤⎦ = + +

+ −

0 1 20 7 1 65 0 0331

5 29 1 65

2 2

2

. . . .

. .

x x

x x 00 121 0. = only the positive root is directlly meaningful: = 0.0610 mx

FIG. P5.57

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Page 26: Sm chapter5

118 Chapter 5

*P5.59 (a) The cable does not stretch: Whenever one car moves 1 cm, the other moves 1 cm.

At any instant they have the same velocity and at all instants they have the same acceleration.

(b) Consider the BMW as object. ∑F

y = ma

y: + T − (1461 kg)(9.8 m �s2) = 1461 kg (1.25 m �s2) T = 1.61 × 104 N

(c) Consider both cars as object. ∑F

y = ma

y: + Tabove − (1461 kg + 1207 kg)(9.8 m �s2) = (1461 kg + 1207 kg) (1.25 m �s2)

Tabove = 2.95 × 104 N

(d) The Ferrari pulls up on the middle section of cable with 16.1 kN. The BMW pulls down on the middle section of cable with 16.1 kN. The net force on the middle section of cable is 0 .

The current velocity is 3.50 m �s up. After 0.01 s the acceleration of 1.25 m �s2 gives the cable additional velocity 0.0125 m �s, for a total of 3.51 m �s upward .

The 3.50 m �s in 3.51 m �s needs no dynamic cause; the motion of the cable continues on its own, as described by the law of ‘inertia’ or ‘pigheadedness.’ The 0.01 m �s of extra upward speed must be caused by some total upward force on the section of cable. But because the cable’s mass is very small compared to a thousand kilograms, the force is very small compared to 1.61 × 104 N, the nearly uniform tension of this section of cable.

*P5.60 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 ms s

s

max max max= + = +=

1 2 1 1 2 2

1 1

, , , ,

max ,

μ μμ gg m gs+ μ θ, cos2 2

From the table of coeffi cients of friction in the text, we take μs , .1 0 610= (aluminum on steel) and μs , .2 0 530= (copper on steel). With

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

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 N2 . 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

FIG. P5.60

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Page 27: Sm chapter5

The Laws of Motion 119

P5.61 (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≤ +( )

or

P Fs s gcos sinθ μ θ μ−( ) ≤ Divide by cosθ :

P Fs s g1−( ) ≤μ θ μ θtan sec

Then

PFs g

sminimum =

−μ θ

μ θsec

tan1

(b) P = ( )−

0 400 100

1 0 400

. sec

. tan

N θθ

θ deg( ) 0.00 15.0 30.0 45.0 60.0

P N( ) 40.0 46.4 60.1 94.3 260

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

P5.62 (a) Following the in-chapter example about a block on a frictionless incline, we have

a g= = ( ) °sin . sin .θ 9 80 30 0m s2

a = 4 90. m s2

(b) The block slides distance x on the incline, with sin ..

30 00 500

° = m

x

x = 1 00. m: v vf i f ia x x2 2 2 0 2 4 90 1 00= + −( ) = + ( )( ). .m s m2

v f = 3 13. m s after time tx

sf

f

= = ( )=

2 2 1 00

3 130 639

v.

..

m

m ss

FIG. P5.61

continued on next page

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Page 28: Sm chapter5

120 Chapter 5

(c) Now in free fall y y t a tf i yi y− = +v1

22:

− = −( ) − ( )2 00 3 13 30 01

29 80

4 9

2. . sin . .

.

m s m s2°t t

00 1 56 2 00 0

1 56 1

2m s m s m

m s

2( ) + ( ) − =

=− ±

t t

t

. .

. .556 4 4 90 2 00

9 80

2m s m s m

m s

2

2

( ) − ( ) −( ). .

.

Only the positive root is physical

t

x tf x

=

= = ( )⎡⎣ ⎤⎦

0 499

3 13 30 0 0 499

.

. cos . .

s

m sv ° ss m( ) = 1 35.

(d) total time = + = + =t ts 0 639 0 499 1 14. . .s s s

(e) The mass of the block makes no difference.

*P5.63 (a) The net force on the cushion is in a fi xed direction, downward and forward making angle tan−1(F� mg) with the vertical. Starting from rest, it will move along this line with (b) increasing speed. Its velocity changes in magnitude.

(c) Since the line of motion is in the direction of the net force, they both make the same angle

with the horizontal: x� 8 m = F� mg = 2.40 N�(1.2 kg)(9.8 m �s2) so x = 1.63 m

(d) The cushion will move along a parabola. The axis of the parabola is parallel to the dashed line in the problem fi gure. If the cushion is thrown in direction above the dashed line, its path will be concave downward, to make its velocity become more and more nearly parallel to the dashed line over time. If the cushion is thrown down more steeply, its path will be concave upward, again making its velocity turn toward the fi xed direction of its acceleration.

P5.64 t t xs s m2( ) ( ) ( )2

0 0 0

1 02 1 04 0 0 100

1 53 2 34 1 0 2

. . .

. . . 000

2 01 4 04 0 0 350

2 64 6 97 0 0 500

3 30 10 89 0 7

. . .

. . .

. . . 550

3 75 14 06 1 00. . .

From x at= 1

22 the slope of a graph of x versus t 2 is

1

2a, and

a = × = ( ) =2 2 0 071 4 0 143slope m s m s2 2. .

From ′ =a g sinθ ,

′ = ⎛⎝

⎞⎠ =a 9 80

1 77 4

127 10 137.

.

..m s m s2 2, different by 4%.

The difference is accounted for by the uncertainty in the data, which we may estimate from the third point as

0 350 0 071 4 4 04

0 35018

. . .

.%

− ( )( )=

Thus the acceleration values agree.

FIG. P5.64

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Page 29: Sm chapter5

The Laws of Motion 121

P5.65 With motion impending,

n T mg

f mg Ts

+ − =

= −( )sin

sin

θμ θ0

and

T mg Ts scos sinθ μ μ θ− + = 0

so

Tmgs

s

=+

μθ 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+( ) = = − +0

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

(b) T =( )( )

+0 350 1 30 9 80

19 3 0 350

. . .

cos . . sin

kg m s2

° 119 34 21

..

°= N

*P5.66 (a) When block 2 moves down 1 cm, block 1 moves 2 cm forward, so block 1 always has twice

the speed of block 2, and a1 = 2a

2 relates the magnitudes of the accelerations.

(b) Let T represent the uniform tension in the cord. For block 1 as object,

∑Fx = m

1a

1: T = m

1(2a

2)

For block 2 as object, ∑Fy = m

1a

1: T + T −(1.3 kg)(9.8 m�s2) = (1.3 kg)(−a

2)

To solve simultaneously we substitute for T: 4 m1a

2 + (1.3 kg)a

2 = 12.7 N

a2 = 12.7 N (1.30 kg + 4m

1)−1 down

(c) a2 = 12.7 N (1.30 kg + 4[0.55 kg])−1 down = 3.64 m�s2 down

(d) a2 approaches 12.7 N�1.3 kg = 9.80 m�s2 down

(e) a2 approaches zero .

(f ) From 2T = 12.74 N + 0, T = 6.37 N

(g) Yes. As m1 approaches zero, block 2 is essentially in free fall. As m

2 becomes negligible

compared to m1, the system is nearly in equilibrium.

P5.67 F ma∑ =

For m1: T m a= 1

For m2: T m g− =2 0

Eliminating T, a

m g

m= 2

1

For all 3 blocks:

F M m m a M m mm g

m= + +( ) = + +( )⎛

⎝⎜⎞⎠⎟1 2 1 2

2

1

FIG. P5.67

FIG. P5.65

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Page 30: Sm chapter5

122 Chapter 5

P5.68 Throughout its up and down motion after release the block has

F ma n mg

n mgy y∑ = + − =

=: cos

cos

θθ

0

Let �R i j= +R Rx y

ˆ ˆ represent the force of table on incline. We have

F ma R n

R mg

F ma Mg

x x x

x

y y

= + − ==

= −

: sin

cos sin

:

θθ θ

0

−− + == +

n R

R Mg mgy

y

cos

cos

θθ

02

�R = + +( )mg M m gcos sin cosθ θ θto the right upw2 aard

P5.69 Choose the x axis pointing down the slope.

v vf i at a

a

= + = + ( )=

: . .

. .

30 0 0 6 00

5 00

m s s

m s2

Consider forces on the toy.

F ma mg m

F ma m

x x

y y

= = ( )=

= −

: sin .

.

:

θ

θ

5 00

30 7

m s2

°

gg T

T mg

T

cos

cos . . cos .

θθ

+ == = ( )( ) °

=

0

0 100 9 80 30 7

0..843 N

FIG. P5.68

FIG. P5.69

25.00 m s=a

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Page 31: Sm chapter5

The Laws of Motion 123

P5.70 F may y∑ = : n mg− =cosθ 0

or n

n

= ( )= ( )

8 40 9 80

82 3

. . cos

. cos

θθN

F max x∑ = : mg masinθ = or a g

a

=

= ( )sin

. sin

θ

θ9 80 m s2

θ , deg , N , m s

0.00

5.00

10.0

15.0

20.0

25.0

3

2n a

00.0

35.0

40.0

45.0

50.0

55.0

60.0

65.0

70.0

75.0

80.00

85.0

90.0

82.3

82.0

81.1

79.5

77.4

74.6

71.3

67.4

633.1

58.2

52.9

47.2

41.2

34.8

28.2

21.3

14.3

7.17

0..00

0.00

0.854

1.70

2.54

3.35

4.14

4.90

5.62

6.30

6.993

7.51

8.03

8.49

8.88

9.21

9.47

9.65

9.76

9.80

At 0°, the normal force is the full weight and the acceleration is zero. At 90°, the mass is in free fall next to the vertical incline.

FIG. P5.70

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Page 32: Sm chapter5

124 Chapter 5

P5.71 (a) Apply Newton’s second law to two points where butterfl ies are attached on either half of mobile (other half the same, by symmetry)

(1) T T2 2 1 1 0cos cosθ θ− = (2) T T mg1 1 2 2 0sin sinθ θ− − = (3) T T2 2 3 0cosθ − = (4) T mg2 2 0sinθ − = Substituting (4) into (2) for T2 2sinθ ,

T mg mg1 1 0sinθ − − =

Then

Tmg

11

2=sinθ

Substitute (3) into (1) for T2 2cosθ :

T T3 1 1 0− =cosθ , T T3 1 1= cosθ

Substitute value of T1:

T mgmg

T31

1 132

2= = =cos

sin tan

θθ θ

From Equation (4),

Tmg

22

=sinθ

(b) Divide (4) by (3):

T

T

mg

T2 2

2 2 3

sin

cos

θθ

=

Substitute value of T3:

tantanθ θ

21

2= mg

mg, θ θ

21 1

2= ⎛

⎝⎞⎠

−tantan

Then we can fi nish answering part (a):

Tmg

2 1 12 1

= ( )⎡⎣ ⎤⎦−sin tan tanθ

(c) D is the horizontal distance between the points at which the two ends of the string are attached to the ceiling.

D = + +2 21 2� � �cos cosθ θ and L = 5�

DL= + ⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

+⎧⎨⎩

⎫−

52 2

1

211

11cos cos tan tanθ θ ⎬⎬

FIG. P5.71

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Page 33: Sm chapter5

The Laws of Motion 125

ANSWERS TO EVEN PROBLEMS

P5.2 (a) 1�3 (b) 0.750 m�s2

P5.4 (a) 4 47 1015. × m s2 away from the wall (b) 2 09 10 10. × − N toward the wall

P5.6 (a) 534 N down (b) 54.5 kg

P5.8 16 3 14 6. ˆ . ˆi j+( ) N

P5.10 (a) Force exerted by spring on hand, to the left; force exerted by spring on wall, to the right.(b) Force exerted by wagon on handle, downward to the left. Force exerted by wagon on planet, upward. Force exerted by wagon on ground, downward. (c) Force exerted by football on player, downward to the right. Force exerted by football on planet, upward. (d) Force exerted by small-mass object on large-mass object, to the left. (e) Force exerted by negative charge on positive charge, to the left. (f ) Force exerted by iron on magnet, to the left.

P5.12 see the solution

P5.14 (a) 181° (b) 11.2 kg (c) 37 5. m s (d) − −( )37 5 0 893. ˆ . ˆi j m s

P5.16 112 N

P5.18 (a) see the solution (b) 1.03 N (c) 0.805 N to the right

P5.20 T1 296= N; T2 163= N; T3 325= N

P5.22 (a) see the solution (b) 1.79 N

P5.24 see the solution (a) 2 54. m s2 down the incline (b) 3 18. m s

P5.26 see the solution 6 30. m s2; 31.5 N

P5.28 see the solution (a) 3 57. m s2 (b) 26.7 N (c) 7 14. m s

P5.30 950 N

P5.32 (a) a a1 22= (b) Tm m g

m m11 2

1 22 2=

+ /; T

m m g

m m21 2

1 2 4=

+ / (c) a

m g

m m12

1 22 2=

+ /; a

m g

m m22

1 24=

+P5.34 (a) 0.529 m (b) 7 40. m s upward

P5.36 μs = 0 306. ; μk = 0 245.

P5.38 (a) 3.34 (b) time would increase

P5.40 see the solution (a) 55.2° (b) 167 N

P5.42 (a) 2 31. m s2 down for m1, left for m2 and up for m3 (b) 30.0 N and 24.2 N

P5.44 (a) Any value between 31.7 N and 48.6 N (b) If P > 48.6 N, the block slides up the wall. If P < 31.7 N, the block slides down the wall. (c) P ≥ 62.7 N. The block cannot slide up the wall. If P < 62.7 N, the block slides down the wall.

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Page 34: Sm chapter5

126 Chapter 5

P5.46 6.84 m

P5.48 0 060 0. m

P5.50 (a) a = 2.84 m �s2; T = 26.5 N (b) A 3.80-kg object and a 6.50-kg object are joined by a light string passing over a light frictionless pulley. The 3.80-kg object is hanging and tows the heavier object down a ramp inclined at 13.0°, with which it has coeffi cient of kinetic friction 0.360. See the solution.

P5.52 (a) Pat and the seat accelerate down at 0.426 m �s2. The other child accelerates up off the ground at the same rate. (b) The tension throughout the rope becomes 480 N, larger than 250 N.

P5.54 (a) 2 00 2. m s to the right (b) 8.00 N right on 4 kg; 6.00 N right on 3 kg; 4.00 N right on 2 kg (c) 8.00 N between 4 kg and 3 kg; 14.0 N between 2 kg and 3 kg (d) see the solution

P5.56 (a) 3.43 m �s2 toward the endstop (b) 3.43 m �s2 toward the green glider; 6.86 m �s2 toward the black glider

P5.58 (a) 0.931 m �s2 (b) From a value of 0.625 m �s2 for large x, the acceleration gradually increases, passes through a maximum, and then drops more rapidly, becoming negative and reaching −2.10 m �s2 at x = 0. (c) 0.976 m �s2 at x = 25.0 cm (d) 6.10 cm.

P5.60 They do not; 29.4 N

P5.62 (a) 4 90. m s2 (b) 3 13. m s at 30.0° below the horizontal (c) 1.35 m (d) 1.14 s (e) No

P5.64 see the solution; 0 143. m s2 agrees with 0 137. m s2

P5.66 (a) When block 2 moves down 1 cm, block 1 moves 2 cm forward, so block 1 always has twice the speed of block 2, and a

1 = 2 a

2. (b) a

2 = 12.7 N (1.30 kg + 4m

1)−1 down (c) 3.64 m �s2

down (d) a2 approaches 9.80 m �s2 down (e) a

2 approaches zero. (f ) 6.37 N (g) Yes. As m

1

approaches zero, block 2 is essentially in free fall. As m2 becomes negligible compared to m

1, the

system is nearly in equilibrium.

P5.68 mg cos sinθ θ to the right + +( )M m gcos2 θ upward

P5.70 see the solution

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