Download 13 torque problems w/Key File

Student ________________________________
AP Physics 1
Date ______________
Static Equilibrium
1.
The wheel of a car, that is shown above, has a radius of 0.350 m. The engine of the car applies
a torque of 295 𝑁 ∙ 𝑚 to this wheel, which does not slip against the road surface. Since the
wheel does not slip, the road must be applying a force of static friction to the wheel that
produces a counter-torque. Moreover, the car has a constant velocity, so this counter-torque
balances the applied torque. What is the magnitude of the static frictional force?
1
#2
Two children hang by their hands from the same tree branch. The branch is straight, and
grows out from the tree trunk at an angle of 27.0𝑜 above the horizontal. One child, with a
mass of 44.0 kg, is hanging 1.30 m along the branch from the tree trunk. The other child, with
a mass of 35.0 kg, is hanging 2.10 m from the tree trunk. What is the magnitude of the net
torque exerted on the branch by the children? Assume that the axis is located where the
branch joins the tree trunk and is perpendicular to the plane formed by the branch and the
trunk.
2
F2 = 55.0 N
x
Rod


F1 = 38.0 N
Hinge (axis of
rotation)
2
90º
3.
A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod
can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at
the same place. One force is directed perpendicular to the rod and has a magnitude of 38.0
N. The second force has a magnitude of 55.0 N and is directed at an angle  with respect to
the rod. If the sum of the torques due to the two forces is zero, what must be the angle  ?
3
+y
+
+x
4.
A hiker, who weighs 985 N, is strolling through the woods and crosses a small horizontal
bridge. The bridge is uniform, weighs 3610 N, and rests on two concrete supports, one at
each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force
that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?
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Answer Key
1.
The drawing shows the wheel as it rolls to the right, so the torque applied by the engine is
assumed to be clockwise about the axis of rotation. The force of static friction that the ground
applies to the wheel is labeled as fs. This force produces a counterclockwise torque  about the
axis of rotation, which is given by  = fs , where is the lever arm. Using this relation we can
find the magnitude fs of the static frictional force.
The counter-torque is given as  = fs , where fs is the magnitude of the static frictional force
and is the lever arm. The lever arm is the distance between the line of action of the force
and the axis of rotation; in this case the lever arm is just the radius r of the tire.
Solving for fs gives
fs 


295 N  m
 843 N
0.350 m
2.
The net torque on the branch is the sum of the torques exerted by the children. Each
individual torque  is given by   F , where F is the magnitude of the force exerted on the
branch by a child, and is the lever arm (see the diagram). The branch supports each
child’s weight, so, by Newton’s third law, the magnitude F of the force exerted on the
branch by each child has the same magnitude as the child’s weight: F = mg. Both forces are
directed downwards. The lever arm for each force is the perpendicular distance between
the axis and the force’s line of action, so we have  d cos (see the diagram).
The mass of the first child is m1 = 44.0 kg. This child is a distance d1 = 1.30 m from the tree
trunk. The mass of the second child, hanging d2 = 2.10 m from the tree trunk, is
m2 = 35.0 kg. Both children produce positive (counterclockwise) torques. The net torque
exerted on the branch by the two children is then
  1   2  F1 1  F2
2
 m1g d1 cos   m2 g d 2 cos   g cos   m1d1  m2d1 
F1
1
F2
2
Substituting the given values, we obtain



  9.80 m/s 2 cos 27.0  44.0 kg 1.30 m    35.0 kg  2.10 m   1140 N  m
5
3.
Each of the two forces produces a torque about the axis of rotation, one clockwise and the
other counterclockwise. By setting the sum of the torques equal to zero, we will be able to
determine the angle  in the drawing.
The two forces act on the rod at a distance x from the hinge. The torque 1 produced by the
Force F1 is given by 1 = +F1
1
,where F1 is the magnitude of the force and
1
is
the lever arm. It is a positive torque, since it tends to produce a counterclockwise rotation.
Since F1 is applied perpendicular to the rod, 1  x. The torque 2 produced by F2 is
2 = F2
2,
where
2
 x sin  (see the drawing). It is a negative torque, since it tends to
produce a clockwise rotation.
Setting the sum of the torques equal to zero, we have
+ F1
1
+   F2
2
 =0
+F1x  F2  x sin   = 0
or
2
The distance x in this relation can be eliminated algebraically.
Solving for the angle  gives
sin  
F1
F2
 F1 
  38.0 N 
  sin 
  43.7
F
55.0
N


 2
  sin  
or
4.
a. We will begin by taking the axis of rotation about the right end of the bridge. The torque
produced by F2 is zero, since its lever arm is zero. When we set the sum of the torques equal
to zero, the resulting equation will have only one unknown, F1, in it. Setting the sum of the
torques produced by the three forces equal to zero gives
   F1L  Wh
 54 L   Wb  12 L   0
Algebraically eliminating the length L of the bridge from this equation and solving for F1
gives
F1  54 Wh  12 Wb 
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5
 985 N   12  3610 N  
2590 N
b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be
zero:
 Fy  F1  Wh  Wb  F2  0
Solving for F2 gives
F2   F1  Wh  Wb  2590 N + 985 N + 3610 N = 2010 N
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