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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
*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 floor 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 floor 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 floor sag” buttons, available to instructors who use this manual and whose classes use the textbook. Estimate the cost of an infinitely 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 floor 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
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 floor 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
= (^) ( + )
= =
. ˆ^. ˆ
kg
m s 2
a i j
F a ii j
F
( + )
∑
∑
N
N N
P5.2 For the same force F , acting on different masses
F = m a 1 1
and F = m a 2 2
(a) m m
a a
1 2
2 1
(b) F = (^) ( m (^) 1 + m (^) 2 ) a = 4 m 1 (^) a = m 1 ( 3 00. m s (^2) )
a = 0 750. m s 2
The Laws of Motion 97
P5.3 m^ =^ 4 00.^ kg,
v (^) i =^ 3 00.^ ˆ i m s,
v (^) 8 = (^) (8 00. i ˆ^ +10 0. j ˆ^ ) m s, t = 8 00. s
a v i j = =
t
. ˆ^. ˆ
m s 2
(^) F = m a = (^) ( 2 50. i ˆ^ +5 00. ˆ j^ )N
F = ( 2 50. ) + ( 2 5 00. ) 2 = 5 59. N
P5.4 (a) Let the x axis be in the original direction of the molecule’s motion.
v (^) f vi at a
a
= + − = + (^) ( × )
= −
m s m s 13 s
..47 × 1015 m s 2
(b) For the molecule,
∑^ F^ = m a^. Its weight is negligible. F wall on molecule = 4 68. × 10 − 26 kg− 4 47. × 1015 mm s 2 N
molecule on wall
( ) = −^ ×
= +
.^10
F 99 × 10 −^10 N
P5.5 (a) (^) ∑ F = ma and v (^) f^2^ = vi^2^ + 2 axf or a x
f i f
v^2 − v^2 2 Therefore,
F m x
F
f i f
∑
∑
( − )
= ×
×
−
v^2 v^2
31
5
kg
m s 22 m s^2 m
⎡⎣ ( ) − (^) ( × )⎤⎦
( )
= ×
2 5 2 3 00 10 2 0 050 0
. 00 −^18 N
(b) The gravitational force exerted by the Earth on the electron is its weight,
Fg = mg = (^) ( 9 11. × 10 − 31 kg)( 9 80. m s (^2) ) = 8 93. × 10 −^30 NN
The accelerating force is 4.08 + 1011 times the weight of the electron.
P5.6 (a) Fg = mg = 120 lb = ( 4 448. N lb)( 120 lb) = 534 N downn
(b) m
F
g
= g = 534 N = 54 5 9.80 m s
- kg
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 g ) p = mgp and ( Fg ) C = mgC give
Δ Fg = m g ( p − gC )
For a person whose mass is 88.7 kg, the change in weight is
Δ Fg = 88 7. kg (^) ( 9 809 5. −9 780 8. (^) ) = 2 55. N
A precise balance scale, as in a doctor’s office, 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.
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. lb up 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 floor.
P5.
∑^ F^ =^ m a^ reads
(^ − 2 00.^ ˆ i^^ +^ 2 00.^ ˆ j^^ +^ 5 00.^ ˆ i^^ −^ 3 00.^ ˆ j^^ −45 0.^ ˆ i^^ ) N^ = m^ ( 3.^7 75 m s^2 ) a ˆ
where ˆ a represents the direction of
a
(^ − 42 0.^ ˆ i^^ −1 00.^ ˆ j )^^ N^ =^^ m ( 3 75.^ m s^2 ) a ˆ ∑^ F =^ (^ 42 0.^ ) + (^2 1 00.^ )^2 N^ at^ tan^
below the – x axis ∑^ F^ =^ 42 0.^ N at 181°^ =^ m (^ 3 75.^ m s^2 ) a ˆ
For the vectors to be equal, their magnitudes and their directions must be equal.
(a) Therefore ˆ a is at 181° counterclockwise from the x axis
(b) m = =
N
3.75 m s 2 kg
(d)
v (^) f = v (^) i + a t = 0 + ( 3 75. m s 2 at 181 °) 10 0. sso
v (^) f = 37 5. m s at 181° v (^) f = 37 5. m s cos 181 ° i ˆ^ +37 5. m s sin 181 °ˆ j so
v (^) f = (^) ( − 37 5. i ˆ^ −0 893. j ˆ^ ) m s
(c)
v (^) f = 37 5. 2 + 0 893. 2 m s = 37 5. m s
FIG. P5.
al, brick on Earth
F normal, pillow on ice
F normal, ice on pillow
F normal, brick on pillow
F normal, pillow on brick
F gravitational, Earth on brick
F gravitational, Earth on pillow
Earth, including ice
Brick
Pillow
F gravitational, brick on Earth
FF gravitatiogravitational, pillow on Earth
Earth, including ice
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 N upward at 35° ahead of the vertical. With the x axis hori- zontally forward, the pole force on the boat is 240 N cos 35 °ˆ j^ + 240 N sin 35 °ˆ i^ = 138 N ˆ i^ + 197 Nˆ j The gravitational force of the whole Earth on boat and worker is Fg = mg = 370 kg 9.8 m s( 2 ) = 3630 N down. The acceleration of the boat is purely horizontal, so ∑^ Fy^ =^ may gives^ +^ B^ +^197 N^ −^3630 N^ =^0.
(a) The buoyant force is B = 3 43. × 10 3 N.
(b) The acceleration is given by (^) ∑ Fx = max : + 138 N − 47 5. N = ( 370 kg) a ; a = =
N
370 kg
m s 2. According to the constant-acceleration model,
v (^) xf = vxi + a tx = 0 857. m s + ( 0 244. m s (^2) )( 0 450. s) = 0 ..
. ˆ
m s
m s
v (^) f = i
P5.16 vx
dx dt
= = 10 t (^) , vy dy dt
= = 9 t 2
a
d dt x =^ v^ x = 10 , a d dt y^ =^ y = t
v 18
At t = 2 00. s, a (^) x = 10 0. m s 2 ,^ ay =36 0. m s^2
∑^ Fx^ =^ max :^ 3 00.^ kg^ 10 0.^ m s^ 30 0. N
2 ( ) =
∑^ Fy^ =^ may :^ 3 00.^ kg^ (36 0^.^ m s^2 ) =^108 N
∑^ F^ =^ Fx^^2^ +^ Fy^2 = 112 N
P5.17 m
mg
tan
kg N m 25.0 m
α
α °° Balance forces,
2 9 80 2
T mg
T
sin . sin
α
α
N
N
FIG. P5.
197 N
138 N 3630 N
47.5 N
B
FIG. P5.
50.0 m α 0.200 m
mg
T T
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: (^) ∑ Fx = max : − T (^) x + T cos θ = 0
Vertical Forces: (^) ∑ Fy = may : − Fg + T sin θ = 0
You need only the equation for the vertical forces to find that the tension in the string is
given by T
Fg
sin θ
. The force the child feels gets smaller, changing from T to T cos θ,
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) T
Fg = =
( )
sin
sin.
θ
kg m s N
2
°
*P5.23 (a) Isolate either mass
T mg ma T mg
The scale reads the tension T , so
T = mg = 5 00. kg ( 9 80. m s 2 ) = 49 0. N
(b) The solution to part (a) is also the solution to (b).
(c) Isolate the pulley T (^) 2 T 1
2 1
T = T = mg =. N.
(d)
∑^ F^ =^ n^ +^ T^ +^ m g^ =^0
Take the component along the incline
n (^) x + T (^) x + mgx = 0
or 0 30 0 0
30 0 2
T mg
T mg mg
sin.
sin.
= 24 5. N.
FIG. P5.
FIG. P5.23(a) and (b)
FIG. P5.23(c)
FIG. P5.23(d)
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 figure 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
∑^ Fy^ =^ n^ −^ mg cos^ θ^ =^0 :^ n^ =^ mg cos^ θ
∑^ Fx^ = −^ mg^ sin^ θ^ = ma :^ a^ = − g^ sin^ θ
(a) When θ = 15 0. °
a = −2 54. m s 2
(b) Starting from rest v v
v
f i f i f
f f
a x x ax
ax
= = (^) ( −. m s (^2) ) −( 2 2 00. m ) = 3 18. m s
P5.25 Choose a coordinate system with ˆ i East and ˆ j North.
(^) ∑^ F^ =^ m a^ =^ 1 00.^ kg(^ 10 0. m s^2 )at^ 30 0. °
( 5 00. N ) ˆ j^ + F 1 = (10 0. N )∠30 0. = (5 00. N) ˆ j^ + 8.
° ( 6 66 N) ˆ i
∴ F 1 = 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, ∑^ Fx^ = ma
T = ( 5 kg) a (1)
Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N.
We have (^) ∑ Fy = ma
88 2. N − T = ( 9 kg) a (2)
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be added to give 88 2. N = ( 14 kg) a. Then
a = 6 30. m s and^2 T =31 5. N
*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 cms) 0.6 s = 30 cms^2.
For the person’s body, ∑ Fy = may : + F bar − 64 kg(9.8 ms 2 ) = 64 kg (0.3 ms 2 )
Note that there is no floor 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.
FIG. P5.
FIG. P5.
5 kg
n T
49 N
9 kg
T + y
F g = 88.2 N
continued on next page
The Laws of Motion 105
P5.31 Forces acting on 2.00 kg block:
T − m g 1 = m a 1 (1)
Forces acting on 8.00 kg block:
Fx − T = m a 2 (2)
(a) Eliminate T and solve for a :
a
F m g m m
= x −
1 1 2
a > 0 for Fx > m g 1 =19 6. N
(b) Eliminate a and solve for T :
T
m m m
= Fx m g
(^1) ( + ) 1 2
2
T = 0 for Fx ≤ − m g 2 = −78 4. N
(c) Fx , N − 100 −78.4 −50.0 0 50.0 100
a (^) x , m s (^2) −12.5 −9.80 −6.96 −1.96 3.04 8.
P5.32 (a) Pulley P 1 has acceleration a 2.
Since m 1 moves twice the distance P 1 moves in the same time, m 1 has twice the acceleration of P 1 , i.e., a 1^ =^2 a 2.
(b) From the figure, and using
F ma m g T m a T m a m a T T
∑ =^ −^ =^ ( )
1 1 1 1 2 2 1
Equation (1) becomes m g 2 − 2 T 1 (^) = m a 2 2. This equation com- bined with Equation (2) yields T m
m
m (^1) m g 1
1 2 2 + 2 2
T
m m m m 1 1 2 g 1 (^12) (^22)
and T^
m m m m 2 1 2 g 1 1 4 2
(c) From the values of T 1 and T 2 we find that
a
T
m
m g (^1) m m
1 1
2 1 1 (^22 )
and a a
m g (^2 1) m m 2 1 2
FIG. P5.
T
FIG. P5.
106 Chapter 5
P5.33 First, we will compute the needed accelerations:
1 0
2
( )^ =
Before it starts to move:
During t
a (^) y
hhe first 0.800 s: m s a y t = v^ yf^ −^ vyi^ = 1 20^ −^0 0 80
s m s While moving at constant ve
=^2
llocity:
During the last 1.50 s:
a
a
y
y
y
( )^ =
v (^) ff yi t
v (^) 0 1 20 1 50 0 800
m s s m s 2
Newton’s second law is: (^) ∑ Fy = may
S a S
72 0 9 80 72 0 y 706 7
. kg. m s. kg N
2
( 2 2 0.^ kg) a^ y
(a) When a (^) y = 0 , S = 706 N.
(b) When a (^) y = 1 50. m s 2 , S = 814 N.
(c) When a (^) y = 0 , S = 706 N.
(d) When a (^) y = −0 800. m s 2 , S = 648 N.
P5.34 Both blocks move with acceleration a
m m m m
= g
2 1 2 1
a =
⎝⎜^
⎠⎟^
kg kg 7 kg 2 kg
. m s 2. m s^2
(a) Take the upward direction as positive for m 1.
v (^) xf^2 vxi^2 ax x (^) f xi (^2 )
= + 2 ( − ) : 0 = ( −2 4. m s ) +2 5 44((. m s) ( − )
( )
x
x
x
f
f
m s m s
m
2 2 2
ff =^ 0 529.^ m below its initial level
(b) v (^) xf = v (^) xi + a tx : vxf = − 2 40. m s + ( 5 44. m s (^2) )( 1 80. s)
vvxf = 7 40. m s upward
FIG. P5.
108 Chapter 5
P5.38 If all the weight is on the rear wheels,
(a) F = ma : μ smg = ma But Δ x = =
at^2 sgt^2 2 2
μ
so μ s x gt
2
μ s =
( )( ) ( )(^ )
mi m mi m s 2 s
(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 =
at^2 :
2 00
2
2
m s
m s 2
a
a (^) (^) ∑^ F^ =^ n^ +^ f^ +^ m g^^ = m a^ : Along :
Al
x f mg ma f m g a
= ( − )
sin. sin.
oong y : n mg n mg
cos. cos.
(b)^ μ k
f n
m g a mg
( sin. − ) cos.
μ k
a g
= tan. − = cos.
(c) f = m g ( sin 30 0°. − a ), f = 3 00 9 80. (. sin 30 0. ° −1 78. ) = 9 37. N
(d) v^2 f^ = vi^2^ + 2 a x ( f − xi )
where x (^) f − xi = 2 00. m
v
v
f
f
m s
m s
2 2
(^2 2) ..67 m s
P5.40 m suitcase = 20 0. kg, F^ =^ 35 0.^ N
F ma F F ma n F F
x x y y g
∑ ∑
:. cos : sin
20 0 N θ 0 θ 00
(a) F^ cos^.
cos
θ
θ
θ
N
N
35.0 N
(b) n = Fg − F sin θ = [ 196 − 35 0 0 821. (. ) ]N
n = 167 N
FIG. P5.
FIG. P5.
The Laws of Motion 109
P5.41 T^ −^ f^ k =^ 5 00.^ a (for 5.00 kg mass)
9 00. g − T = 9 00. a (for 9.00 kg mass)
Adding these two equations gives:
a a m s 22
N
T 5 00 5 60 0 200 5 00 9 80
P5.42 Let a represent the positive magnitude of the acceleration − a ˆ j of m 1 , of the acceleration − a ˆ i of m 2 , and of the acceleration
- a ˆ j of m 3. Call T 12 the tension in the left rope and T 23 the tension in the cord on the right.
For m 1 , (^) ∑ Fy = may + T 12 (^) − m g 1 = − m a 1
For m 2 , (^) ∑ Fx = max − T 12 (^) + μ kn + T (^) 23 = − m a 2
and (^) ∑ Fy = may n − m g 2 = 0
for m 3 , (^) ∑ Fy = may T (^) 23 − m g 3 = + m a 3 we have three simultaneous equations
− + = ( )
T a T T
12 12 2
N kg N (^33)
23
= ( )
kg N kg
a T a
(a) Add them up:
- 39 2. N − 3 43. N − 19 6. N = ( 7 00. kg ) a
a = 2 31. m s , down for^2 m 1 (^) , left for m 2 ,and upp for m 3
(b) Now −^ T 12 +^ 39 2.^ N^ = ( 4 00.^ kg)(^ 2 31. m s^2 )
T 12 = 30 0. N
and T 23 − 19 6. N = (2 00. kg)( 2 31. m s 2 )
T 23 = 24 2. N
FIG. P5.
FIG. P5.
n
T 12 T 23
m g 2
f = μ kn
m g 1
T 12
m g 3
T 23
The Laws of Motion 111
*P5.45 (a) If P is too small, static friction will prevent the block from moving. We will find the value of P when motion is just ready to begin: ∑ Fy = may : − P sin 37° − 0.42 kg 9.8 ms 2 + n = 0 n = 4.12 N + P sin 37° with motion impending we read the equality sign in fs = μsn = 0.72(4.12 N + P sin 37°) = 2.96 N + 0.433 P ∑ Fx = max : 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 = μkn = 0.34(4.12 N + P sin 37°) = 1.40 N + 0.205 P ∑ Fx = max : 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 ms 2 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 ≤ μsn 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 ms^2 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 ms^2. 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 first case, we take x and y as parallel and perpendicular to the surface of the roof: F ma n mg n mg
∑ y =^ y +^ −^ =
: cos cos
θ θ
then friction is f (^) k = μ (^) k n =μ (^) kmg cosθ
F ma f mg ma a g g
x x k x x k
∑ =^ −^ −^ = = − − = −
: sin cos sin.
θ
μ θ θ ( 0 4 4 cos 37 ° −sin 37 °)9 8. m s 2 = −9 03. m s^2
The Frisbee goes ballistic with speed given by v (^) xf^2 vxi^2 ax x (^) f xi 2
= + 2 ( − ) = ( 15 m s ) + 2 ( −9 03. m s 2 ) 100 0 44 4
m m s m s
( − ) =^2
vxf. For the free fall, we take x and y horizontal and vertical: v (^) yf^2 vyi^2 ay y (^) f yi 2
= (. m s sin °) + − .. sin
m s m
m
m s
(
y
y
f
f
))
2
19 6
m s 2 m
FIG. P5.
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 (^) s ) (^) max= μ sn
The board is also in equilibrium in the vertical direction, so
∑^ Fy^ =^2 f^ −^ Fg =^0 , or^ f^
Fg
2 The minimum compression force needed is then
. n f F s
g s
μ 2 μ
N
N
P5.48 Take + x in the direction of motion of the tablecloth. For the mug:
F ma a a
x x x x
∑ =^ =
N kg m s 2 Relative to the tablecloth, the acceleration of the mug is (^) 0 5. m s 2 − 3 m s 2 = −2 5. m s^2. The mug reaches the edge of the tablecloth after time given by
Δ x t a t
t
t
= (^) xi + x
− = + (^) ( − )
=
v
2
..^2
m m s 2
0 0 s. The motion of the mug relative to tabletop is over distance 1 2
a tx^2 = (^) (0 5. m s (^2) )( 0 490. s) 2 = 0 060 0. m
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 fs = μs n. We also have ∑ Fx = max : + fs = ma ∑ Fy = may : + n − mg = 0 Combining by substitution gives μs mg = ma a = 0.3 (9.8 ms^2 ) = (^) 2.94 ms 2 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 = μk mg = ma a = μk g = 0.25 (9.8 ms^2 ) = 2.45 ms^2 forward.
(c) Now take the x axis along the direction of motion and the y axis perpendicular to the slope. We have ∑ Fy =^ may :^ +^ n^ −^ mg^ cos 10°^ =^0 +^ n^ =^ mg^ cos 10°^ fs =^ μs mg^ cos 10° ∑ Fx = max : + fs − mg sin 10° = ma a = μs g cos 10° − g sin 10° = 0.3(9.8 ms^2 ) cos 10° − (9.8 ms^2 ) sin 10° = 1.19 ms^2 up the incline
(d) This time kinetic friction acts: ∑ Fy = may : + n − mg cos 10° = 0 + n = mg cos 10° fk = μk mg cos 10° ∑ Fx = max : + fk − mg sin 10° = ma a = μk g cos 10° − g sin 10° = [0.25 cos 10° − sin 10°]9.8 ms^2 = (^) 0.711 ms^2 up the incline
FIG. P5.
f n
F = 95.5 N
f n
g
continued on next page
