Download 5.2. Visualize: 5.6. Model: An object`s acceleration is linearly

5.2. Visualize:
5.6. Model: An object’s acceleration is linearly proportional to the net force.
Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F  a
and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2, four rubber bands will produce an
acceleration of 2.4 m/s2 .
(b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F  ma,
2F  (2m)a  a  F/m  0.6 m/s2
5.10 Solve: Use proportional reasoning. Given that distance traveled is proportional to the square of the
time, d  t 2 , so
d
should be constant. We have
t2
2.0 furlongs
 2.0 s 
2

x
 4.0 s 
2
Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs.
Assess: A longer time should result in a longer distance traveled.
5.14 Solve: Newton’s second law tells us that F  ma. Compute F for each case:
(a) F  (0.200 kg)(5 m/s2 )  1N.
(b) F  (0.200 kg)(10 m/s2 )  2 N .
Assess: To double the acceleration we must double the force, as expected.
5.18. Visualize:
Solve: The object will be in equilibrium if F3 has the same magnitude as F1  F2 but is in the opposite
direction so that the sum of all the three forces is zero.
5.22. Visualize:
Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward.
There is a force labeled FG directed down, a force Fthrust directed up, and a force D directed down. So a possible
description is: “A rocket accelerates upward.”
5.26. Visualize:
Assess: Since the velocity is constant, the acceleration is zero, and the net force is zero.
5.30. Visualize:
Solve: According to Newton’s second law F  ma, the force at any time is found simply by multiplying the
value of the acceleration by the mass of the object.
5.34. Model: Use the particle model for the object.
Solve: (a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the
acceleration of the mass is 10 m/s2 . According to Newton’s second law, F0  m0 (10 m/s2 ). The force then
becomes
1
2
F0 . Newton’s second law gives
1
1
F0  m0a   m0 10 m s 2 
2
2
This means a is 5 m/s2 .
(b) The force is F0 and the mass is now
1
2
m0 . Newton’s second law gives
1
F0  m0a  m0 10 m s 2 
2
This means a  20 m/s2 .
(c) A similar procedure gives a  10 m/s2 .
(d) A similar procedure gives a  2.5 m/s2 .
5.38. Visualize:
Solve: (d) This is an object on a surface because FG  n. It must be moving to the left because the kinetic
friction is to the right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic
block to the left.”
5.42. Visualize:
Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no normal force so
the object is not on a surface. The description could be: “A rocket is fired at an angle to the horizontal and there
is no drag force.”
5.46. Visualize:
The normal force is perpendicular to the hill. The frictional force is parallel to the hill.
5.50. Visualize:
The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. There is a
force of the spring to the right which causes an acceleration.
6.2. Model: We can assume that the ring is a particle.
Visualize:
This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only
three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle  .
Solve: Because the ring is in equilibrium it must obey Fnet  0 N. This is a vector equation, so it has both xand y-components:
 Fnet x  T3 cos  T2  0 N  T3 cos  T2
 Fnet  y  T1  T3 sin  0 N  T3 sin  T1
We have two equations in the two unknowns T3 and  . Divide the y-equation by the x-equation:
T3 sin
T 80 N
 tan  1 
 1.6    tan 1 1.6   58
T3 cos
T2 50 N
Now we can use the x-equation to find
T3 
T2
50 N

 94 N
cos cos58
The tension in the third rope is 94 N directed 58 below the horizontal.
6.6. Visualize: Please refer to Figure EX6.6.
Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice
that the angle between the 3 N force and the –y-axis is the same 20 by which the coordinates are tilted. Applying
Newton’s second law,
ax 
 Fnet  x
m
ay 

 Fnet  y
m
5 N  1 N   3sin 20  N
 1.49 m/s 2
2 kg

2.82 N   3cos 20  N
 0 m/s 2
2 kg
For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15 with the positive xaxis. The other 2-newton force makes an angle of 15 with the negative y-axis. The accelerations are
ax 
ay 
 Fnet  x  2cos15 N   2sin15 N  3 N

 0.28 m/s 2
m
 Fnet  y
m
2 kg

1.414 N   2sin15  N   2cos15  N
 0 m/s 2
2 kg
6.10. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless,
the tension in the rope is the only horizontal force.
Visualize:
Solve: (a) Since the box is at rest, ax  0 m/s2 , and the net force on the box must be zero. Therefore, according
to Newton’s first law, the tension in the rope must be zero.
(b) For this situation again, ax  0 m/s2 , so Fnet  T  0 N.
(c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since
ax  5.0 m/s2 ,
Fnet  T  max   50 kg   5.0 m/s 2   250 N
Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is
exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of
about half the weight of a small person) seems reasonable to accelerate a box of this mass
at 5.0 m/s2.
6.14. Model: Use the particle model for the woman.
Solve: (a) The woman’s weight on the earth is
wearth  mgearth   55 kg   9.80 m/s 2   540 N
(b) Since mass is a measure of the amount of matter, the woman’s mass is the same on the moon as on the earth.
Her weight on the moon is
wmoon  mgmoon   55 kg  1.62 m/s 2   89 N
Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to
the moon than to the earth. Thus the woman’s smaller weight on the moon makes sense.
6.18. Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the
farmer’s pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it
will be subject to kinetic friction.
Visualize:
Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that
n  FG  mg. The maximum friction force is
fsmax  s mg   0.8120 kg   9.80 m/s2   940 N
The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not
be able to budge the mule.
Assess: The farmer should have known better.
6.22. Model: We assume that the plane is a particle accelerating in a straight line under the influence of two
forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional
kinematics.
Visualize:
Solve:
We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain:
a
v 82 m/s  0 m/s

 2.34 m/s 2
t
35 s
 Fnet   Fx  Fthrust  fr  ma  Fthrust  fr  ma
For rubber rolling on concrete, r  0.02 (Table 6.1), and since the runway is horizontal, n  FG  mg. Thus:
Fthrust   r FG  ma   r mg  ma  m   r g  a 
  75,000 kg   0.02   9.8 m/s2   2.34 m/s2   190,000 N
Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that
190,000 N is enough to produce the required acceleration.
6.26. Model: We will represent the tennis ball as a particle.
Visualize:
The tennis ball falls straight down toward the earth’s surface. The ball is subject to a net force that is the resultant
of the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively.
Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the
rest of the motion.
Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is
Fnet  FG  D  0 N
Since only the vertical direction matters, one can write:
Fy  0 N  Fnet  D  FG  0 N
When this condition is satisfied, the speed of the ball becomes the constant terminal speed v  vterm . The
magnitudes of the gravitational and drag forces acting on the ball are:
FG  mg  m  9.80 m/s 2 
1
2
2
2
2
Avterm
 0.25  R 2  vterm
  0.25  0.0325 m   26 m/s   0.56 N


4
The condition for dynamic equilibrium becomes:
0.56 N
 57 g
9.80 m/s2  m  0.56 N  0 N  m  9.80
m/s 2
D
Assess: The value of the mass of the tennis ball obtained above seems reasonable.
6.30. Model: The plastic ball is represented as a particle in static equilibrium.
Visualize:
Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the
string’s tension plus two long-range forces. The equilibrium condition is
 Fnet  x  Tx   Felec x  T sin  Felec  0 N
 Fnet  y  Ty   FG  y  T cos  mg  0 N
We can solve the y-equation to get
T
 0.001 kg   9.8 m/s
mg

cos
cos 20
2
  0.0104 N
Substituting this value into the x-equation,
Felec  T sin  1.04 102 N  sin 20  0.0036 N
(b) The tension in the string is 0.0104 N.
6.34. Model: We can assume your body is a particle moving in a straight line under the influence of two
forces: gravity and the support force of the scale.
Visualize:
Solve:
The weight (see Equation 6.10) of an object moving in an elevator is
 a
 w

w  mg 1    a  
 1 g
 g
 mg 
When accelerating upward, the acceleration is
 170 lb 
a 
 1  9.80 m/s 2   1.3 m/s 2
150
lb


When braking, the acceleration is
 120 lb 
a
 1  9.80 m/s 2   2.0 m/s 2
 150 lb 
Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire
State Building must be. Also note that we did not have to convert the units of the weights from pounds to
newtons because the weights appear as a ratio.
6.38. Model: We will represent the bullet as a particle.
Visualize:
Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then
we can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant
to the problem.) The kinematic equation is
v02
 400 m/s   6.67 105 m/s2

2x
2  0.12 m 
2
v12  v02  2ax  a  
Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood
exerts two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the
wood. Second, a retarding frictional force f k that stops the bullet. The only horizontal force is f k , which points
to the left and thus has a negative x-component. The x-component of Newton’s second law is
 Fnet x   fk  ma  fk  ma    0.01 kg   6.67 105 m/s2   6670 N
Notice how the signs worked together to give a positive value of the magnitude of the force.
(b) The time to stop is found from v1  v0  at as follows:
t  
v0
 6.00 104 s  600 μs
a
(c)
Using the above kinematic equation, we can find the velocity as a function of t. For example at t  60 μs,
vx  400 m/s  (6.667 105 m/s2 )(60 106 s)  360 m/s
6.42. Model: We assume Sam is a particle moving in a straight line down the slope under the influence of
gravity, the thrust of his jet skis, and the resisting force of friction on the snow.
Visualize:
Solve:
From the height of the slope and its angle, we can calculate its length:
h
h
50 m
 sin  x1  x0 

 288 m
x1  x0
sin sin10
Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force:
 Fnet  y  Fy  n  FG cos  0 N
 n  FG cos  mg cos  (75 kg)(9.80 m/s2 )(cos10)  724 N
One-dimensional kinematics gives us Sam’s acceleration:
v12  v02  2ax  x  x0   ax 
v12  v02
 40 m/s   0 m2/s 2  2.78 m/s2

2  x1  x2 
2  288 m 
2
Then, from Newton’s second law and the equation f k  k n :
 Fnet  x  Fx  FG sin  Fthrust  f k  max
 k 

mg sin  Fthrust  ma
n
 75 kg   9.80 m/s2   sin10  200 N   75 kg   2.78 m/s2 
724 N
Assess: This coefficient seems a bit high for skis on snow, but not impossible.
 0.165
6.46. Model: We will represent the wood block as a particle, and use the model of kinetic friction and
kinematics. Assume w sin   fs , so it does not hang up at the top.
Visualize:
The block ends where it starts, so x2  x0  0 m. We expect v2 to be negative, because the block will be moving
in the
–x-direction, so we’ll want to take v2 as the final speed. Because of friction, we expect to find v2  v0 .

Solve: (a) The friction force is opposite to v , so f k points down the slope during the first half of the motion
and up the slope during the second half. FG and n are the only other forces. Newton’s second law for the
upward motion is
ax  a0 
 Fnet  x
m
a y  0 m/s 2 

 FG sin  f k mg sin  f k

m
m
 Fnet  y
m

n  FG cos n  mg cos

m
m
The friction model is f k  k n. First solve the y-equation to give n  mg cos . Use this in the friction model to
get f k  k mg cos . Now substitute this result for f k into the x-equation:
a0 
mg sin  k mg cos
  g  sin   k cos     9.8 m/s 2   sin30  0.20cos30   6.60 m/s 2
m
Kinematics now gives
v12  v02  2a0  x1  x0   x1 
2 2
v12  v02 0 m /s  10 m/s 

 7.6 m
2a0
2  6.60 m/s 2 
2
The block’s height is then h  x1 sin  (7.6 m)sin30  3.8 m.
(b) For the return trip, f k points up the slope, so the x-component of the second law is
ax  a1 
 Fnet x
m

 FG sin  f k mg sin  f k

m
m
Note the sign change. The y-equation and the friction model are unchanged, so we have
a1   g  sin  k cos   3.20 m/s2
The kinematics for the return trip are
v22  v12  2a1  x2  x1   v2  2a1x1  2  3.20 m/s2   7.6 m   7.0 m/s
Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction.
The final speed is v2  7.0 m/s.
6.50. Model: The antiques (mass  m) in the back of your pickup (mass  M ) will be treated as a particle.
The antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The
pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the
road.
Visualize:
Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare
that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m  M ,
Newton’s second law is
( Fnet ) x  Fx  N x    FG PA   ( f ) x  0 N  0 N  f  (m  M )ax  (m  M )a
x
( Fnet ) y  Fy  N y    FG PA   ( f ) y  N  (m  M ) g  0 N  0 N
y
The model of static friction is f   N , where  is the coefficient of friction between the tires and the road.
These equations can be combined to yield a   g. Since constant-acceleration kinematics gives
v12  v02  2a( x1  x0 ), we find
v12  v02
v02
 25 m/s 
  min 

 0.58
2  x1  x0 
2 g  x1  x0   2   9.8 m/s 2   55 m 
2
a
The truck cannot stop if  is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and
0.80 respectively (see Table 6.1), are larger. So the truck can stop.
(b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for
 min . This value is smaller than the given coefficient of static friction (s  0.60) between the antiques and the
truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m.
Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According
to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle
traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to
stop the truck.
6.54. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity.
Visualize:
Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus
the drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot
straight up,
 F 
y
1
1
2


 4mg 
 ma    FG  D     mg  Av 2   mg  A  2vterm   mg  A 
  5mg
4
4


 A 
Thus a  5 g , where the minus sign indicates the downward direction. We have used Equations 6.16 for the
drag force and 6.19 for the terminal velocity.
(b) When the ball is shot straight down,
 F 
y
 ma  D  FG 
1
2
A 2vterm   mg  3mg
4
Thus a  3 g , this time directed upward.
(c)
The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the
terminal velocity because the drag force decreases as the ball slows.
6.58. Model: The ball hanging from the ceiling of the truck by a string is represented as a particle.
Visualize:
Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between
“at rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight
down.
(b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle.
(c) The ball moves with the truck, so its acceleration is 5 m/s2 in the forward direction.
(d) The free-body diagram shows that the horizontal component of T provides a net force in the forward
direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck.
(e) Newton’s second law for the ball is
ax 
 Fnet x
m

Tx T sin10

m
m
a y  0 m2 /s2 
 Fnet  y
m

Ty   FG  y
m

T cos10  mg
m
We can solve the second equation for the magnitude of the tension:
T
1.0 kg  9.8 m/s
mg

cos10
cos10
2
  9.95 N
Then the first equation gives the acceleration of the ball and truck:
ax 
T sin10  9.95 N  sin10

=1.73 m/s 2
m
m
The truck’s velocity cannot be determined.
6.62. Solve: (a) The terminal velocity for a falling object is reached when the downward gravitational force
is balanced by the upward drag force.
FG  D
mg  bvterm  6 Rvterm
 vterm 
mg
6 R
4

(b) The mass of the spherical sand grain of density p  2400 kg/m3 is m     R3  .
3

3
2
4
2  gR 2 2  2400 kg/m  9.80 m/s  5.0  10 m 

 1.3 m/s
9
9

3 Ns 
1.0  10

m2 

50 m
 38 s.
The time required for the sand grain to fall 50 m at this speed is t 
1.3 m/s
Assess: The speed of 1.3 m/s for a sand grain falling through water seems about right.
2
Thus
vterm 
6.66. Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How
far does the auto slide before coming to rest?
(b)
(c) Newton’s second law is
Fy  ny   FG  y  n  mg  may  0 N
Fx  0.80n  max
The y-component equation gives n  mg  1500 kg   9.8 m/s2 . Substituting this into the x-component equation
yields
1500 kg  ax  0.80 1500 kg  9.8 m/s2   ax   0.80 9.8 m/s2   7.8 m/s2
Using the constant-acceleration kinematic equation v12  v02  2ax, we find
v02
 40 m/s   102 m

2a
2  7.8 m/s 2 
2
x  
7.2. Visualize:
Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational
force on them. The interaction forces between the two are equal and opposite.
(b) The system consists of the soccer ball and bowling ball, as indicated in the figure.
(c)
Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on the
other are part of an action/reaction pair, and therefore have equal magnitudes. Each ball’s acceleration due to the
forces on it is determined by Newton’s second law, a  Fnet / m, which depends on the mass. Since the masses of
the balls are different, their accelerations are different.
7.6. Visualize: Please refer to Figure EX7.6.
Solve: (a) For each block, there is a gravitational force with the earth, a normal force and kinetic friction with
the surface, and a tension force due to the rope.
(b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates
down the incline with the same acceleration that Block B has up the incline. The system consists of the two
blocks, as indicated in the figure.
(c)
Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is
convenient since then one component of a is zero, simplifying the mathematical expression of Newton’s second
law.
7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless
and along with the earth it is a part of the environment. The three blocks are our three systems of interest.
Visualize:
The force applied on block 1 is FA on 1  12 N. The acceleration for all the blocks is the same and is denoted by a.
Solve: (a) Newton’s second law for the three blocks along the x-direction is
 F 
on 1 x
 FA on 1  F2 on 1  m1a
 F 
on 2 x
 F1on 2  F3 on 2  m2a
 F 
on 3 x
 F2 on 3  m3a
Adding these three equations and using Newton’s third law ( F2 on 1  F1 on 2 and F3 on 2  F2 on 3 ), we get
FA on 1   m1  m2  m3  a  12 N   1 kg  2 kg  3 kg  a  a  2 m/s2
Using this value of a, the force equation on block 3 gives
F2 on 3  m3a   3 kg   2 m/s2   6 N
(b) Substituting into the force equation on block 1,
12 N  F2 on 1  1 kg   2 m/s 2   F2 on 1  10 N
Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the
2 kg block exerts on the 1 kg block is reasonable.
7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must
therefore consider both the block of ice and the rope as objects in the system.
Visualize:
Solve:
The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same
acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts
on the rear end of the rope.
Newton’s second law along the x-axis for the ice block and the rope is
 F 
 FR on I  mI a  10 kg   2.0 m/s2   20 N
 F 
 Fext  FI on R  mR a  Fext  FR on I  mR a
on I x
on R x
 Fext  FR on I  mR a  20 N   0.500 kg   2.0 m/s2   21 N
7.18. Visualize: Please refer to Figure P7.18.
Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton’s third law force
pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these
massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs
when the static friction on B is at its maximum value. Applying Newton’s first law to the vertical forces on block
B gives nB   FG B  mB g. The static friction force on B is thus
 fs B  snB  smB g.
 fs B  TA on B , and the same
TB on A   FG A  mA g. Since TA on B  TB on A , we have  fs B  mA g , so
Applying Newton’s first law to the horizontal forces on B gives
vertical forces on A gives
s mB g  mA g
 mA  s mB   0.60 20 kg   12.0 kg
analysis of the
7.22. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight
of this segment of rope is a downward force. It is balanced by the tension force at height y.
Solve: The mass m of this segment of rope is the same fraction of the total mass M  2.2 kg as length y is a
fraction of the total length L  3.0 m. That is, m/M  y/L, from which we can write the mass of the rope
segment
m
M
y
L
This segment of rope is in static equilibrium, so the tension force pulling up on it is
T  FG  mg 
Mg
(2.2 kg)(9.8 m /s2 )
y
y  7.19 y N
L
3.0 m
where y is in m. The tension increases linearly from 0 N at the bottom ( y  0 m) to 21.6 N at the top ( y  3 m).
This is shown in the graph.
7.26. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be
treated as particles. We will also use the constant-acceleration kinematic equations.
Visualize:
Solve: (a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by
Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the
starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in).
However, the very different masses of the two crafts means that the distances they each move will also be very
different. The pictorial representation shows that they meet at time t1 when xM1  xm1. There’s only one force on
each craft, so Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair,
FM on m  Fm on M  Ftractor beam  4.0 104 N
The accelerations of the two craft are
aM 
Fm on M
M

FM on m 4.0  104 N
4.0  104 N
2

0.020
m/s
a


 2.0 m/s 2
m
2.0 106 kg
m
2.0  104 kg
Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. The
accelerations are very different even though the forces are the same. Now we have a constant-acceleration
problem in kinematics. At a later time t1 the positions of the crafts are
xM1  xM0  vM0  t1  t0   12 aM  t1  t0   12 aMt12
2
xm1  xm0  vm0  t1  t0   12 am t1  t0   xm0  12 amt12
2
The craft meet when xM1  xm1, so
1
2
aMt12  xm0  12 amt12  t1 
2 10,000 m 
2 xm0
2 xm0


 99.5 s
aM  am
aM  am
2.02 m/s 2
Knowing t1 , we can now find the starship’s position as it meets the shuttlecraft:
xM1  12 aMt12  99 m
The starship moves 99 m as it pulls in the shuttlecraft from 10 km away.
7.30. Model: The two blocks form a system of interacting objects.
Visualize: Please refer to Figure P7.30.
Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand
block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or
negative answer for the tension in the string.
Newton’s first law applied to the y-direction on Block L yields
F 
L y
 0  nL   FG L cos20  nL  mL g cos20
Therefore
 fk L   k L mL g cos20   0.201.0 kg  9.80 m/s2  cos20  1.84 N
A similar analysis of the vertical forces on Block R gives  f k R  1.84 N as well. Using Newton’s second law in
the x-direction for Block L,
  FL x  mLa  TR on L   fk L   FG L sin 20  mLa  TR on L  1.84 N  mL g sin 20.
For Block R,
F 
R x
 mR a   FG R sin 20  1.84 N  TL on R  mR a  mR g sin 20  1.84 N  TL on R
These are two equations in the two unknowns a and TL on R  TR on L  T . Solving them, we obtain a  2.12 m/s2 and
T  0.61 N.
Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks,
which is reasonable.
7.34. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a
massless rope and frictionless pulley.
Visualize:
Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration
constraint is a2  a  a1, where a will have a positive value. There are two real action/reaction pairs. The two
tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a
frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the freebody diagrams, especially force f1 exerted on block 2 by block 1. We have quite a few pieces of information to
include. First, Newton’s second law for blocks 1 and 2:
F
net on 1

x
 f1  T1  k n1  T1  m1a1  m1a
F
F

net on 1 y
 n1  m1g  0 N  n1  m1g
  T  f  f  T  T  f   n  T  m a
 F   n  n  m g  0 N  n  n  m g
net on 2 x
pull
net on 2 y
1
2
2
2
1
pull
2
1
k 2
2
2
1
2 2
 m2a
2
We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law:
n1  n1  m1g f1  f1  k n1  k m1g T1  T2  T
Knowing n1, we can now use the y-equation of block 2 to find n2 . Substitute all these pieces into the two xequations, and we end up with two equations in two unknowns:
k m1g  T  m1a Tpull  T  k m1g  k  m1  m2  g  m2a
Subtract the first equation from the second to get
Tpull  k  3m1  m2  g   m1  m2  a  a 
Tpull   k  3m1  m2  g
m1  m2
 1.77 m/s 2
7.38. Model: Assume the particle model for m1 , m2 , and m3 , and the model of kinetic friction. Assume the
ropes to be massless, and the pulleys to be frictionless and massless.
Visualize:
Solve: Newton’s second law for m1 is T1  ( FG )1  m1a1. Newton’s second law for m2 is
 F 
on m2
(F
y
 n2   FG 2  0 N  n2   2.0 kg   9.8 m/s2   19.6 N
)  T2  f k2  T  m2a2  T2  k n2  T1  (2.0 kg)a2
on m2 x
Newton’s second law for m3 is
 F 
on m3
y
 T2   FG 3  m3a3
Since m1, m2 , and m3 move together, a1  a2  a3  a. The equations for the three masses thus become
T1   FG 1  m1a  1.0 kg  a T2  k n2  T1  m2a   2.0 kg  a T2   FG 3  m3a    3.0 kg  a
Subtracting the third equation from the sum of the first two equations yields:
  FG 1  k n2   FG 3   6.0 kg  a
  1.0 kg   9.8 m/s2    0.3019.6 N    3.0 kg   9.8 m/s2    6.0 kg  a  a  2.3 m/s 2
7.42. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable.
Visualize:
Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2).
Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1
to allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet  0 N
for both systems. Newton’s second law for the cable car is
F

net on 1 x
Newton’s second law for the counterweight is
F
F
 T1  FB  m1g sin1  0 N

net on 2 x
 m2 g sin 2  T2  0 N

net on 1 y
F

net on 2 y
 n1  m1g cos1  0 N
 n2  m2 g cos 2  0 N
From the x-equation for the counterweight, T2  m2 g sin 2 . By Newton’s third law, T1  T2 . Thus the x-equation
for the cable car becomes
FB  m1g sin1  T1  m1g sin1  m2 g sin2  3770 N
(b) If the brakes fail, then FB  0 N. The car will accelerate down the hill on one side while the counterweight
accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the
acceleration vectors. The constraint is a1x  a2 x  a, where a will have a negative value. Using T1  T2  T , the
two x-equations are
 Fnet on 1 x  T  m1g sin1  m1a1x  m1a
F

net on 2 x
 m2 g sin 2  T  m2a2x  m2a
Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T:
m sin1  m2 sin 2
m1g sin1  m2 g sin   m1  m2  a  a   1
g  0.991 m/s 2
m1  m2
Now we have a problem in kinematics. The speed at the bottom is calculated as follows:
v12  v02  2a  x1  x0   2ax1  v1  2ax1  2  0.991 m/s2   400 m   28.2 m/s
Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless
slope of 30 is reasonable.
7.46. Model: Use the particle model for the wedge and the block.
Visualize:
The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note:
n1 on 2  n2 on 1.
Solve:
The y-component of Newton’s second law for block m2 is
 F 
on 2 y
 n1 on 2 cos   FG 2  0 N  n1 on 2 
m2 g
cos
Combining this equation with the x-component of Newton’s second law yields:
 F 
on 2 x
 n1 on 2 sin  m2a  a 
n1 on 2 sin
 g tan
m2
Now, Newton’s second law for the wedge is
 F 
on 1 x
 F  n2 on 1 sin  m1a
 F  m1a  n2 on 1 sin  m1a  m2a  (m1  m2 )a  (m1  m2 ) g tan