Physics II
Homework III
CJ
Chapter 10; 76
Chapter 11; 11, 12, 18, 52, 62, 82
Chapter 13; 17, 66
10.76. IDENTIFY: Apply conservation of energy to the motion of the ball as it rolls up the hill.
After the ball leaves the edge of the cliff it moves in projectile motion and constant
acceleration equations can be used.
(a) SET UP: Use conservation of energy to find the speed v2 of the ball just before it leaves
the top of the cliff. Let point 1 be at the bottom of the hill and point 2 be at the top of the
hill. Take y  0 at the bottom of the hill, so y1  0 and y2  28.0 m.
EXECUTE:
1
2
K1  U1  K 2  U 2
mv  I   mgy2  12 mv22  12 I 22
2
1
2
1
1
2
Rolling without slipping means   v / r and
7
10
1
2
I 2 
1
2

2
5
mr 2  (v / r ) 2  15 mv 2
mv12  mgy2  107 mv22
v2  v12  107 gy2  15.26 m/s
SET UP: Consider the projectile motion of the ball, from just after it leaves the top of the
cliff until just before it lands. Take  y to be downward. Use the vertical motion to find the
time in the air:
v0 y  0, ay  9.80 m/s2 , y  y0  28.0 m, t  ?
EXECUTE: y  y0  v0 yt  12 ayt 2 gives t  2.39 s
During this time the ball travels horizontally
x  x0  v0 xt  (15.26 m/s)(2.39 s)  36.5 m.
Just before it lands,
v y  v0 y  a yt  23.4 m/s
and
vx  v0 x  15.3 m/s
v  v  v  28.0 m/s
2
x
2
y
(b) EVALUATE: At the bottom of the hill,   v / r  (25.0 m/s)/ r. The rotation rate doesn’t
change while the ball is in the air, after it leaves the top of the cliff, so just before it lands
  (15.3 m/s)/ r. The total kinetic energy is the same at the bottom of the hill and just before it
lands, but just before it lands less of this energy is rotational kinetic energy, so the
translational kinetic energy is greater.
11.11. IDENTIFY: The system of the person and diving board is at rest so the two conditions of
equilibrium apply.
(a) SET UP: The free-body diagram for the diving board is given in Figure 11.11. Take the
origin of coordinates at the left-hand end of the board (point A).
is the force applied
at the support point
and F2 is the force at
the end that is held
down.
F1
Figure 11.11
EXECUTE:

A
0
gives
 F1 (1.0 m)  (500 N)(3.00 m)  (280 N)(1.50 m)  0
F1 
(500 N)(3.00 m)  (280 N)(1.50 m)
 1920 N
1.00 m
(b)
F
y
 may
F1  F2  280 N  500 N  0
F2  F1  280 N  500 N  1920 N  280 N  500 N  1140 N
EVALUATE: We can check our answers by calculating the net torque about some point and
checking that  z  0 for that point also. Net torque about the right-hand of the board:
(1140 N)(3.00 m)+(280 N)(1.50 m)  (1920 N)(2.00 m)  3420 N  m  420 N  m  3840 N  m  0, which checks.
11.12. IDENTIFY: Apply the first and second conditions of equilibrium to the beam.
SET UP: The boy exerts a downward force on the beam that is equal to his weight.
EXECUTE:
(a) The graphs are given in Figure 11.12.
(b)
x  6.25 m when FA  0, which
is 1.25 m beyond point B.
FA  0, so FB  900 N. The
(300 N)(4.50 m)
 1.50 m.
(900 N)
(c) Take torques about the right end. When the beam is just balanced,
distance that point B must be from the right end is then
EVALUATE: When the beam is on the verge of tipping it starts to lift off the support A and
the normal force FA exerted by the support goes to zero.
Figure 11.12
11.18. IDENTIFY: Apply the conditions for equilibrium to the crane.
SET UP: The free-body diagram for the crane is sketched in Figure 11.18. Fh and Fv are the
components of the force exerted by the axle. T pulls to the left so Fh is to the right. T also
pulls downward and the two weights are downward, so Fv is upward.
EXECUTE:
(a)  z  0 gives
T ([13 m]sin 25°  wc ([7.0 m]cos55°)  wb ([16.0 m]cos55°  0 .
T
(11,000 N)([16.0 m]cos55°)  (15,000 N)([7.0 m]cos55°)
 2.93  104 N .
(13.0 m)sin 25°
(b)
F
x
F
y
 0 gives Fh  T cos30°  0 and Fh  2.54  104 N .
 0 gives Fv  T sin30°  wc  wb  0 and Fv  4.06  104 N .
EVALUATE:
tan 
along the crane.
Fv 4.06 104 N

Fh 2.54 104 N
and   58° . The force exerted by the axle is not directed
Figure 11.18
11.52. IDENTIFY: Apply the first and second conditions for equilibrium to the bridge.
SET UP: Find torques about the hinge. Use L as the length of the bridge and wT and wB for
the weights of the truck and the raised section of the bridge. Take  y to be upward and  x
to be to the right.
EXECUTE:
(a) TL sin70  wT  34 L cos30  wB  12 L cos30 , so
T
 34 mT  12 mB  (9.80 m s 2 )cos30  2.57  105 N.
sin 70
(b) Horizontal: T cos  70  30  1.97 105 N (to the right). Vertical: wT  wB  T sin 40  2.46  105 N
(upward).
EVALUATE: If  is the angle of the hinge force above the horizontal,
tan  
2.46 105 N
1.97 105 N
and
  51.3° . The hinge force is not directed along the bridge.
11.62. IDENTIFY: Apply  F  0 to each object, including the point where D, C and B are joined.
Apply  z  0 to the rod.
SET UP: To find
TC and TD ,
use a coordinate system with axes parallel to the cords.
EXECUTE: A and B are straightforward, the tensions being the weights suspended:
Τ A  (0.0360 kg)(9.80 m/s 2 )  0.353 N and TB  (0.0240 kg  0.0360 kg)(9.80 m s 2 )  0.588 N . Applying
 Fx  0 and  Fy  0 to the point where the cords are joined, TC  TB cos36.9  0.470 N and
TD  TB cos53.1  0.353 N . To find TE , take torques about the point where string F is attached.
TE (1.00 m)  TD sin 36.9 (0.800 m)  TC sin 53.1 (0.200 m)  (0.120 kg)(9.80 m s 2 )(0.500 m) and TE  0.833 N. TF
may be found similarly, or from the fact that TE  TF must be the total weight of the ornament.
(0.180kg)(9.80m s 2 )  1.76 N, from which TF  0.931 N.
EVALUATE: The vertical line through the spheres is closer to F than to E, so we expect
TF  TE , and this is indeed the case.
11.82. IDENTIFY: Apply the equilibrium conditions to the pole. The horizontal component of the
tension in the wire is 22.0 N.
SET UP: The free-body diagram for the pole is given in Figure 11.82. The tension in the
cord equals the weight W. Fv and Fh are the components of the force exerted by the hinge. If
either of these forces is actually in the opposite direction to what we have assumed, we will
get a negative value when we solve for it.
EXECUTE:
(a) T sin 37.0°  22.0 N so
T  36.6 N .
W
(22.0 N)(1.75 m)
 28.5 N .
1.35 m
(b)
F
y

 0 gives Fv  T cos37.0°  W  0
z
 0 gives (T sin37.0°)(1.75 m)  W (1.35 m)  0 .
and
Fv  (36.6 N)cos37.0°  55.0 N  84.2 N .
W  T sin37.0°  Fh  0 and Fh  28.5 N  22.0 N  6.5 N .
F  F  F  84.5 N
2
h
2
v
F
x
 0 gives
The magnitude of the hinge force is
.
EVALUATE: If we consider torques about an axis at the top of the plate, we see that
be to the left in order for its torque to oppose the torque produced by the force W.
Fh must
Figure 11.82
T  2
13.17. IDENTIFY:
m
k
k
m
. ax   x so amax 
k
A . F  kx .
m
SET UP: ax is proportional to x so ax goes through one cycle when the displacement goes
through one cycle. From the graph, one cycle of ax extends from t  0.10 s to t  0.30 s, so the
period is T  0.20 s. k  2.50 N/cm  250 N/m. From the graph the maximum acceleration is
12.0 m/s2 .
EXECUTE:
(a) T  2 m gives
k
2
2
 T 
 0.20 s 
m  k
  (250 N/m) 
  0.253 kg
 2 
 2 
mamax (0.253 kg)(12.0 m/s 2 )

 0.0121 m  1.21 cm
k
250 N/m
(b)
A
(c)
Fmax  kA  (250 N/m)(0.0121 m)  3.03 N
EVALUATE: We can also calculate the maximum force from the maximum acceleration:
Fmax  mamax  (0.253 kg)(12.0 m/s 2 )  3.04 N, which agrees with our previous results.
13.66. (a) IDENTIFY and SET UP:
Combine Eqs. (13.12) and (13.21) to relate
vx
and x to T.
EXECUTE: T  2 m/k
We are given information about v x at a particular x. The expression relating these two
quantities comes from conservation of energy: 12 mvx2  12 kx 2  12 kA2
We can solve this equation for m/k , and then use that result to calculate T. mvx2  k ( A2  x 2 )
m

k
Then
(0.100 m) 2  (0.060 m) 2
A2  x 2

 0.267 s
vx
0.300 m/s
T  2 m/k  2  0.267 s   1.68 s.
(b) IDENTIFY and SET UP:
equation: 12 mvx2  12 kx 2  12 kA2
We are asked to relate x and
vx ,
so use conservation of energy
kx 2  kA2  mvx2
x
A2  (m/k )vx2  (0.100 m) 2  (0.267 s) 2 (0.160 m/s) 2  0.090 m
EVALUATE: Smaller
vx
means larger x.
(c) IDENTIFY: If the slice doesn't slip the maximum acceleration of the plate (Eq.13.4)
equals the maximum acceleration of the slice, which is determined by applying Newton's
2nd law to the slice.
SET UP: For the plate, kx  max and ax  (k/m) x. The maximum x is A, so amax  (k/m) A. If
the carrot slice doesn't slip then the static friction force must be able to give it this much
acceleration. The free-body diagram for the carrot slice (mass m ) is given in Figure 13.66.
EXECUTE:
F
y
 may
n  mg  0
n  mg 
Figure 13.66
F
x
 max
s n  ma
s mg  ma and a  s g
2
But we require that
a  amax  (k/m) A  s g
and s 
k A  1   0.100 m 

 
  0.143
m g  0.267 s   9.80 m/s 2 
EVALUATE: We can write this as s   2 A/g. More friction is required if the frequency or
the amplitude is increased.