HW #2 Chapter 9.1-9.4
Q 2,3,6,8,9,12,14
P 1-3,6,8,15,16,19,20,21,57,59 + 26 recommended for AP Test
Questions
2.
The bungee jumper is not in equilibrium, because the net force on the jumper is not zero. If the jumper were at rest
and the net force were zero, then the jumper would stay at rest by Newton’s 1st law. The jumper has a net upward
force when at the bottom of the dive, and that is why the jumper is then pulled back upwards.
3.
If the fingers are not the same distance from the CG, the finger closer to the CG will support a larger fraction of the
weight of the meter stick so that the net torque on the stick is zero. That larger vertical force means there will be
more friction between the stick and that closer finger, and thus the finger further from the CG will be easier to move.
The more distant finger will slide easier, and therefore move in closer to the CG. That finger, when it becomes the
one closest to the CG, will then have more friction and will “stick”. The other finger will then slide. You then repeat
the process. Whichever finger is farther from the CG will slide closer to it, until the two fingers eventually meet at
the CG.
For rotating the upper half body, the pivot point is near the waist and hips. In that
position, the
arms have a relatively small torque, even when extended, due to their smaller mass,
and the more
massive trunk–head combination has a very short lever arm, and so also has a
relatively
small torque. Thus the force of gravity on the upper body causes relatively little
torque about
the hips tending to rotate you forward, and so the back muscles need to produce
little torque to
keep you from rotating forward. The force on the upper half body due to the back
muscles is
small, and so the (partially rightward) force at the base of the spinal column, to keep the spine in equilibrium, will be
small.
6.
8.
When standing and bending over, the lever arm for the upper body is much
larger than while
sitting, and so causes a much larger torque. The CM of the arms is also
further from the
support point, and so causes more torque. The back muscles, assumed to act
at the center of the
back, do not have a very long lever arm. Thus the back muscles will have to
exert a large force
to cause a counter-torque that keeps you from falling over. And accordingly,
there will have to
be a large force (mostly to the right in the picture) at the base of the spine to
keep the spine in
equilibrium.
The mass of the meter stick is equal to that of the rock. For purposes of calculating torques, the meter stick can be
treated as if all of its mass were at the 50 cm mark. Thus the CM of the meter stick is the same distance from the
pivot point as the rock, and so their masses must be the same in order to exert the same torque.
9.
If the sum of the forces on an object are not zero, then the CM of the object will accelerate in the direction of the net
force. If the sum of the torques on the object are zero, then the object has no angular acceleration. Some examples
are:
a) A satellite in a circular orbit around the Earth.
b) A block sliding down an inclined plane.
c) An object that is in projectile motion but not rotating
d) The startup motion of an elevator, changing from rest to having a non-zero velocity.
12. When walking, you must keep your CG over your feet. If you have a heavy load in your arms, your CG is shifted
forward, and so you must lean backwards to realign your CG over your feet.
14. When you start to stand up from a normal sitting position, your CM is not over your point of support (your feet), and
so gravity will exert a torque about your feet that rotates you back down into the chair. You must lean forward in
order that your CM be over your feet so that you can stand up.
Problems
1.
If the tree is not accelerating, then the net force in all directions is 0.
 Fx  FA  FB cos110  FC x  0 
110o
FB
FC x   FA  FB cos110  310 N   425 N  cos110  164.6 N
F
y
FA

 FB sin  FC y  0 
FC y   FB sin110    425 N  sin110  399.4 N
FC 
FC2 x  FC2 y 
  tan 1
FC y
FC x
 tan 1
FC
 164.6 N    399.4 N 
2
399.4 N
164.6 N

2
 432.0 N  4.3  102 N
 67.6o ,   180o  67.6o  112.4o  112o
And so FC is 430 N, at an angle of 112o clockwise from FA .
2.
3.
The torque is the force times the lever arm.
  Fr   58 kg   9.8 m s 2   3.0 m   1.7 103 m N , clockwise
Because the mass m is stationary, the tension in the rope
pulling up on the sling must be mg, and so the force of the
leg must be mg, upward. Calculate torques about the hip joint,
counterclockwise torque taken as positive. See the free-body
the leg. Note that the forces on the leg exerted by the hip joint
because they do not exert a torque about the hip joint.
  mgx
2
 Mgx1  0  m  M
x1
x2
 15.0 kg 
 35.0 cm 

80.5 cm 
mg
x2
x1
6.52 kg
Mg
sling on the
with
diagram for
are not drawn,
6.
(a) Let m = 0. Calculate the net torque about the left end of the
diving board, with counterclockwise torques positive. Since the
equilibrium, the net torque is zero.
FB
FA
  F 1.0 m   Mg  4.0 m   0 
F  4Mg  4  58 kg   9.80 m s   2274 N  2.3 10 N
1.0 m
B
2
board is in
mg
Mg
2.0 m
4.0 m
3
B
Use Newton’s 2nd law in the vertical direction to find FA .
F
 FB  Mg  FA 
y


FA  FB  Mg  4Mg  Mg  3Mg  3  58 kg  9.80 m s 2  1705 N  1.7 103 N
(b) Repeat the basic process, but with m = 35 kg. The weight of the board will add more clockwise
torque.
  FB 1.0 m   mg  2.0 m   Mg  4.0 m   0 


FB  4 Mg  2mg   4  58 kg   2  35 kg  9.80 m s 2  2960 N  3.0  10 3 N
F
y
 FB  Mg  mg  FA 
FA  FB  Mg  mg  4 Mg  2mg  Mg  mg  3Mg  mg


 3  58 kg   35 kg  9.80 m s 2  2048 N  2.0  103 N
8.
Let m be the mass of the beam, and M be the mass of the piano. Calculate torques about the left end of the beam,
with counterclockwise torques positive. The conditions of equilibrium for the beam are used to find the forces that
the support exerts on the beam.
  FR L  mg  12 L   Mg  14 L   0


FR   12 m  14 M  g   12 140 kg   14  320 kg  9.80 m s 2  1.47  103 N
F
y
 FL  FR  mg  Mg  0


FL   m  M  g  FR   460 kg  9.80 m s 2  1.47  103 N  3.04  103 N
The forces on the supports are equal in magnitude and opposite in direction to the above two results.
FR  1.5  103 N down
FL  3.0  103 N down
15. The beam is in equilibrium, and so both the net torque and net force on it must be zero. From the free-body diagram,
calculate the net torque about the center of the left support, with counterclockwise torques as positive. Calculate the
net force, with upward as positive. Use those two equations to find FA and FB .
  F  x  x
B
FB 

1
2
 x3  x4   F1 x1  F2  x1  x2   F3  x1  x2  x3   mgx5
F1 x1  F2  x1  x2   F3  x1  x2  x3   mgx5
 x1  x2  x3  x4 
 4300 N  2.0 m    3100 N  6.0 m    2200 N  9.0 m    250 kg   9.8 m s2   5.0 m 
10.0 m
 5925 N  5.9 10 N
3
F  F
A
 FB  F1  F2  F3  mg  0


FA  F1  F2  F3  mg  FB  9600 N   250 kg  9.8 m s 2  5925 N  6125 N  6.1 103 N
16. From the free-body diagram, the conditions of equilibrium
find the location of the girl (mass mC ). The 50-kg boy is
mA , and the 35-kg girl by mB . Calculate torques about the
see-saw, and take counterclockwise torques to be positive.
force of the fulcrum on the see-saw  F  causes no torque
center.
are used to
represented by
L
x
mA g
mC g
F
center of the
mB g The upward
about the
  m g  L   m gx  m g  L   0
A
x
1
2
 mA  mB 
mC
C
 L 
1
2
B
1
2
 50 kg  35 kg  1
25 kg
2
 3.6 m   1.1 m
19. The person is in equilibrium, and so both the net torque and net force
From the free-body diagram, calculate the net torque about the center
with counterclockwise torques as positive. Use that calculation to
location of the center of gravity, a distance x from the feet.
x
L-x
FA
FB
must be zero.
of gravity,
find the
mg
  F x  F  L  x   0
B
FA
x
A
L
mA g
L
mA
L
35.1 kg
FA  FB
mA g  mB g
mA  mB
31.6 kg  35.1 kg
The center of gravity is about 90.5 cm from the feet.
1.72 m  
20. The beam is in equilibrium. Use the conditions of equilibrium to
tension in the wire and the forces at the hinge. Calculate torques about
take counterclockwise torques to be positive.
   FT sin   l2  m1 g l1 2  m2 gl1  0 
1
2
FT 
m1 gl1  m2 gl1
l2 sin 

1
2
155 N 1.70 m    245 N 1.70 m 
1.35 m   sin 35.0o 
9.05  101 m

m1g
l1 2
l2
l1
 708.0 N  7.08  10 N
2
F
F
calculate the
the hinge, and
FT
FH
x
 FH x  FT cos   0  FH x  FT cos    708 N  cos 35.0o  579.99 N  5.80  102 N
y
 FH y  FT sin   m1 g  m2 g  0 
FH y  m1 g  m2 g  FT sin   155 N  245 N   708 N  sin 35.0o  6.092 N  6 N  down 
m2g
21. (a) The pole is in equilibrium, and so the net torque on it must
be zero. From the free-body diagram, calculate the net
the lower end of the pole, with counterclockwise torques as
that calculation to find the tension in the cable. The length
L.
  FT h  mg  L 2  cos   MgL cos   0
m
FT 

FT

h FP y

  7.50 m  cos 37
mg
 F
Px
2  M  gL cos 
h
 6.0 kg  21.5 kg  9.80 m s 2
Mg
y
torque about
positive. Use
of the pole is
x
L cos
o
 424.8 N  4.25 102 N
3.80 m
(b) The net force on the pole is also zero since it is in equilibrium. Write Newton’s 2nd law in both
the x and y directions to solve for the forces at the pivot.
F
F
x
 FP x  FT  0  FP x  FT  4.25 102 N
y
 FP y  mg  Mg  0  FP y   m  M  g   33.5 kg  9.80 m s 2  3.28  102 N

26. Write the conditions of equilibrium for the ladder, with torques taken
bottom of the ladder, and counterclockwise torques as positive.
l
mg
  FW l sin   mg 2 cos   0  FW  12 tan 
mg
 Fx  FG x  FW  0  FG x  FW  12 tan 
 Fy  FG y  mg  0  FG y  mg
For the ladder to not slip, the force at the ground FG x must be less than

l sin 
FG y
y
mg

maximum force of static friction.
FG x   FN   FG y 
1
2
mg
tan 
FG x
l cos
  mg 
Thus the minimum angle is  min  tan 1 1 2   .
1
2
about the
FW
 tan     tan 1 1 2  
x
or equal to the
57. Each crossbar in the mobile is in equilibrium, and so the net torque about the suspension point for each crossbar must
be 0. Counterclockwise torques will be taken as positive. The suspension point is used so that the tension in the
suspension string need not be known initially. The net vertical force must also be 0.
The bottom bar:
FCD
  mD gxD  mC gxC  0 
xD
mC  mD
F
y
xC
 mD
17.50 cm
5.00 cm
 3.50mD
xD
 FCD  mC g  mD g  0  FCD   mC  mD  g  4.50mD g
xC
mC g
mD g
The middle bar:
  F
x
CD CD
mD 
mB xB
4.50 xCD
 mB gxB  0  FCD  mB g
xB
 4.50mD g  mB g
xCD
 0.885 kg  5.00 cm 

 0.06555 
 4.50 15.00 cm 
xB
FBCD
xCD
xCD
xB
2
6.56  10 kg
mB g
FCD
mC  3.50mD   3.50  0.06555 kg   2.29  101 kg
F
y
 FBCD  FCD  mB g  0  FBCD  FCD  mB g   4.50mD  mB  g
The top bar:
  mA gxA  FBCD xBCD  0 
mA 
 4.50mD  mB  gxBCD
FABCD
  4.50mD  mB 
gxA
  4.50  0.06555 kg   0.885 kg 
59.
xBCD
xA
xA
7.50 cm
FBCD
mA g
 2.94  101 kg
30.00 cm
(a)
If the wheel is just lifted off the lowest level, then the
forces on the wheel are the horizontal pull, its weight, and the
FN at the corner. Take torques about the corner point, for the
barely off the ground, being held in equilibrium. The contact
corner exerts no torque and so does not enter the calculation.
force has a lever arm of R  R  h  2 R  h , and gravity has a
x , found from the triangle shown.
F
2R  h
Mg
2
F
.
h  2R  h 
Mgx
Rh

Mgx
Rh
 Mg
R
x
h
 Mg
 Mg
2R  h
2R  h
2R  h
(b)
The only difference is that now the pulling force has a lever arm
of R  h .
Rh
  Mgx  F  R  h   0 
F
FN
x
x  R 2   R  h   h  2R  h 
  Mgx  F  2 R  h   0
xBCD
F
Mg
h  2R  h 
Rh
x
30o
Mg
 down 
mg  down 
only
contact force
wheel just
force at the
The pulling
lever arm of