Download Physics 2A

Physics 11
HW #4 Solutions
Chapter 4:
Focus On Concepts: 5, 7, 12, 23
Problems: 4, 49, 54, 64, 74, 98, 110, 115
Focus On Concepts 4-5
(c) Newton’s second law gives the answer directly, provided the net force is calculated
by vector addition of the two given forces. The direction of the net force gives the
direction of the acceleration.
Focus On Concepts 4-7
(e) Answers a and b are false, according to the third law, which states that whenever one
body exerts a force on a second body, the second body exerts an oppositely directed force
of equal magnitude on the first body. It does not matter whether one of the bodies is
stationary or whether it collapses. Answer c is false, because according to the third law,
Sam and his sister experience forces of equal magnitudes during the push-off. Since Sam
has the greater mass, he flies off with the smaller acceleration, according to the second
law. Answer d is false, because in catching and throwing the ball each astronaut applies a
force to it, and, according to the third law, the ball applies an oppositely directed force of
equal magnitude to each astronaut. These reaction forces accelerate the astronauts away
from each other, so that the distance between them increases.
Focus On Concepts 4-12
(d) What matters is the direction of the elevator’s acceleration. When the acceleration is
upward, the apparent weight is greater than the true weight. When the acceleration is
downward, the apparent weight is less than the true weight. In both possibilities the
acceleration points upward.
Focus On Concepts 4-23
(b) Since the boxes move at a constant velocity, they have no acceleration and are,
therefore, in equilibrium. According to Newton’s second law, the net force acting on
each box must be zero. Thus, Newton’s second law applied to each box gives two
equations in two unknowns, the magnitude of the tension in the rope between the boxes
and the kinetic frictional force that acts on each box. Note that the frictional forces acting
on the boxes are identical, because the boxes are identical. Solving these two equations
shows that the tension is one-half of the applied force.
Problem 4-4
REASONING According to Newton’s second law, Equation 4.1, the average net force ΣF
is equal to the product of the object’s mass m and the average acceleration a . The average
acceleration is equal to the change in velocity divided by the elapsed time (Equation 2.4),
where the change in velocity is the final velocity v minus the initial velocity v0.
SOLUTION The average net force exerted on the car and riders is
)
(
v − v0
45 m/s − 0 m/s
ma =
m
=
=
5.5 × 103 kg
3.5 × 104 N
∑F =
t − t0
7.0 s
Problem 4-49
SSM REASONING AND SOLUTION The free-body diagram is shown
at the right. The forces that act on the picture are the pressing force P, the
normal force FN exerted on the picture by the wall, the weight mg of the
picture, and the force of static friction fsMAX . The maximum magnitude
for the frictional force is given by Equation 4.7: fsMAX = µs FN . The
picture is in equilibrium, and, if we take the directions to the right and up
as positive, we have in the x direction
∑ Fx =P − FN =0
or
P =FN
fsMAX − mg= 0
∑ F=
y
or
fsMAX= mg
and in the y direction
Therefore,
MAX
fs=
m=
mg
s FN
But since FN = P , we have
ms P = mg
Solving for P, we have
mg (1.10 kg)(9.80 m/s 2 )
P =
=
=
0.660
ms
16.3 N
Problem 4-54
REASONING The drawing assumes that upward is the +y direction
and shows the I-beam together with the three forces that act on it: its
weight W and the tension T in each of the cables. Since the I-beam is
moving upward at a constant velocity, its acceleration is zero and it is in
equilibrium. Therefore, according to Equation 4.9b, the net force in the
vertical (or y) direction must be zero, so that ∑ Fy = 0 . This relation
T
T
70.0°
70.0°
will allow us to find the magnitude of the tension.
W = –8.00 × 103 N
SOLUTION With up as the +y direction, the vertical component of the tension in each
cable is T sin 70.0° , and Equation 4.9b becomes
Solving this equation for the tension gives T = 4260 N .
Problem 4-64
REASONING The block is in equilibrium in each case, because the block has a constant
velocity and is not accelerating. A zero acceleration is the hallmark of equilibrium. At
equilibrium, the net force is zero (i.e., the forces balance to zero), and we will obtain the
magnitude of the pushing force by utilizing this fact as it pertains to the vertical or y
direction.
We will use Equation 4.9b
0 ) for this purpose.
( ΣFy =
+y
+y
+x
+x
Note that the direction of the kinetic frictional
force is not the same in each case. The
fk
frictional force always opposes the relative
θ
θ
fk
P
motion between the surface of the block and P
W
W
the wall. Therefore, when the block slides
upward, the frictional force points downward. Free-body diagram for
Free-body diagram for
upward motion
downward motion
When the block slides downward, the
frictional force points upward.
These
directions are shown in the free-body diagrams (not to scale) for the two cases. In these
drawings W is the weight of the block and fk is the kinetic frictional force.
In each case the magnitude of the frictional force is the same. It is given by Equation 4.8 as
fk = μkFN, where μk is the coefficient of kinetic friction and FN is the magnitude of the
normal force. The coefficient of kinetic friction does not depend on the direction of the
motion. Furthermore, the magnitude of the normal force in each case is the component of
the pushing force that is perpendicular to the wall, or FN = P sin θ.
SOLUTION Using Equations 4.9b to describe the balance of forces that act on the block in
the y direction and referring to the free-body diagrams, we have
Upward motion
Σ=
Fy P cos θ − W −=
fk 0
Downward motion
Σ=
Fy P cos θ − W +=
fk 0
According to Equation 4.8, the magnitude of the kinetic frictional force is fk = μkFN, where
we have pointed out in the REASONING that the magnitude of the normal force is
FN = P sin θ. Substituting into the equations for ΣFy in the two cases, we obtain
Upward motion
Σ=
Fy P cos θ − W − µk P sin=
θ 0
Downward motion
Σ=
Fy P cos θ − W + µk P sin=
θ 0
Solving each case for P, we find that
a. Upward motion
=
P
W
39.0 N
=
= 52.6 N
cos θ − µk sin θ cos 30.0° − ( 0.250 ) sin 30.0°
b. Downward motion
=
P
W
39.0 N
=
= 39.4 N
cos θ + µk sin θ cos 30.0° + ( 0.250 ) sin 30.0°
Problem 4-74
REASONING In the absence of air resistance, the two forces acting on the sensor are its
weight W and the tension T in the towing cable (see the free-body diagram). We see that Tx
is the only horizontal force acting on the sensor, and therefore Newton’s second law
ΣFx = max (Equation 4.2a) gives Tx = max. Because the vertical component of the sensor’s
acceleration is zero, the vertical component of the cable’s tension T must balance the
sensor’s weight: Ty = W = mg. We thus have sufficient information to calculate the
horizontal and vertical components of the tension force T, and therefore to calculate its
2
magnitude T from the Pythagorean theorem: T=
Tx2 + Ty2 .
SOLUTION Given that Tx = max and that Ty = mg, the Pythagorean theorem yields the
magnitude T of the tension in the cable:
T=
Tx2 + Ty2 =
= (129 kg)
( ma )2 + ( mg )2 =
m2 a 2 + m2 g 2 = m a 2 + g 2
m/s 2 )
( 2.84 m/s2 ) + (9.80=
2
2
1320 N
Problem 4-98
REASONING Newton’s second law gives the acceleration as a = (ΣF)/m. Since we seek
only the horizontal acceleration, it is the x component of this equation that we will use;
ax = (ΣFx)/m. For completeness, however, the free-body diagram will include the vertical
forces also (the normal force FN and the weight W).
SOLUTION
The free-body diagram is
shown at the right, where
+y
F1 = 59.0 N
F2 = 33.0 N
FN
θ = 70.0°
When F1 is replaced by its x and y
components, we obtain the free body
diagram in the following drawing.
F2
+x
θ
W
F1
Choosing right to be the positive direction, we have
ΣFx F1 cos θ − F2
ax =
=
m
m
ax =
( 59.0 N ) cos 70.0° − ( 33.0 N ) =
7.00 kg
−1.83 m/s
+y
FN
2
F1cos θ
F2
Thus, the horizontal acceleration has a magnitude of
1.83 m / s2 , and the minus sign indicates that it points to
+x
F1sin θ
the left .
W
Problem 4-110
REASONING Since the mountain climber is at rest, she is in equilibrium and the net force
acting on her must be zero. Three forces comprise the net force, her weight, and the tension
forces from the left and right sides of the rope. We will resolve the forces into components
and set the sum of the x components and the sum of the y components separately equal to
zero. In so doing we will obtain two equations containing the unknown quantities, the
tension TL in the left side of the rope and the tension TR in the right side. These two
equations will be solved simultaneously to give values for the two unknowns.
SOLUTION Using W to denote the weight of the mountain climber and choosing right and
upward to be the positive directions, we have the following free-body diagram for the
climber:
For the x components of the forces we have
=
ΣFx TR sin 80.0° − TL =
sin 65.0° 0
+y
TL
65.0º 80.0º
For the y components of the forces we have
=
ΣFy TR cos80.0° + TL cos 65.0
=
° −W 0
TR
+x
W
Solving the first of these equations for TR, we find that
TR = TL
sin 65.0°
sin 80.0°
Substituting this result into the second equation gives
TL
sin 65.0°
cos80.0° + TL cos=
65.0° − W 0=
or TL 1.717 W
sin 80.0°
Using this result in the expression for TR reveals that
sin 65.0°
=
TR T=
L
sin 80.0°
sin 65.0°
1.580 W
sin 80.0°
(1.717W=
)
Since the weight of the climber is W = 535 N, we find that
=
=
TL 1.717
W 1.717 (=
535 N ) 919 N
=
TR 1.580
=
W 1.580 (=
535 N ) 845 N
Problem 4-115
SSM REASONING Let us assume that the skater is moving horizontally along the +x
axis. The time t it takes for the skater to reduce her velocity to vx = +2.8 m/s from
v0x = +6.3 m/s can be obtained from one of the equations of kinematics:
v=
v0 x + ax t
x
(3.3a)
The initial and final velocities are known, but the acceleration is not. We can obtain the
acceleration from Newton’s second law ( ΣFx =
max , Equation 4.2a ) in the following
manner. The kinetic frictional force is the only horizontal force that acts on the skater, and,
since it is a resistive force, it acts opposite to the direction of the motion. Thus, the net force
in the x direction is ΣFx =
− f k , where fk is the magnitude of the kinetic frictional force.
Therefore, the acceleration of the skater is ax =
ΣFx /m =
− fk / m .
The magnitude of the frictional force is f k = µk FN (Equation 4.8), where µk is the
coefficient of kinetic friction between the ice and the skate blades and FN is the magnitude
of the normal force. There are two vertical forces acting on the skater: the upward-acting
normal force FN and the downward pull of gravity (her weight) mg. Since the skater has no
vertical acceleration, Newton's second law in the vertical direction gives (taking upward as
the positive direction) ΣFy = FN − mg = 0 . Therefore, the magnitude of the normal force is
FN = mg and the magnitude of the acceleration is
=
ax
− f k − mk FN − mk m g
=
=
= − mk g
m
m
m
SOLUTION
Solving the equation v=
v0 x + ax t for the time and substituting the expression above for
x
the acceleration yields
=
t
vx − v0 x vx − v0 x
2.8 m/s − 6.3 m/s
= =
= 4.4 s
ax
− mk g
− ( 0.081) ( 9.80 m/s 2 )