Torque/Equilibrium
From Princeton Review Book
1. In an effort to tighten a bolt, a force F is applied as shown in the figure below. If the
distance from the end of the wrench to the center of the bolt is 20 cm and F = 20
Newtons, what is the magnitude of the torque produced by F?
a. 0 Nm.
b. 1 Nm.
c. 2 Nm
d. 4 Nm
e. 10 Nm.
*D. The torque is r*Fsin = (.2m) (20 N) sin(90) = 4 Nm
2. In the figure below, what is the torque about the pendulum’s suspension point
produced by the weight of the bob, given that the length of the pendulum, L, is 80 cm and
has a mass of .5 kg?
a. 0 Nm
b. 0.49 Nm
c. 0.98 Nm
d. 2.0 Nm
e. 3.4 Nm
* D. Think carefully about which angle you want.
Torque = R* F sin, where is the angle between the vector radiating out from
the pivot and the Force vector. The force vector here is the weight.
One of the components of the weight causes torque, the other causes tension in
the rope.
See the diagram:
3. A uniform meter stick of mass 1 kg is hanging from a thread attached at the stick’s
midpoint. One block of mass m = 3 kg hangs from the left end of the stick, and another
block, of unknown mass M, hangs directly below the 80 cm mark on the meter stick. If
the stick remains at rest in the horizontal position shown above, what is M?
(Hint: Label 0 meters, 50 meters, 80 meters, and 100 m on the meter stick)
a. 4 kg.
b. 5 kg.
c. 6 kg.
d. 8 kg.
e. 9 kg.
*B See Diagram
4. If the rod in the figure below is uniform and has mass m, what is the tension in the
supporting string?
a.
1
mg sin 
2
b. mg sin 
c.
1
mg cos 
2
d.
1
mg
2
e. mg.
*D.
5. The figure below shows a uniform bar of mass M resting on two supports. A block of
mass 1/2M is placed on the bar twice as far from Support 2 as from Support 1. If F1 and
F2 denote the downward forces on Support 1 and Support 2 respectively, what is the
value of F2/F1?
a. ½
b. 2/3
c. ¾
d. 4/5
e. 5/6
*D. Method One (Logic):
Don’t fall into the trap of thinking that the mass of the bar does not count since it is
evenly distributed. It adds to the total mass and decreases the effect that the block has.
Support 1 is supporting half of the bar (M/2) and 2/3 of the block 2/3 (1/2M) for a total of
5/6M.
Support 2 is supporting half of the bar (M/2) and 1/3 of the block 1/3(1/2M) for a total of
4/6 M.
Taking the ratio yields 4/5.
Method Two: You may also set the torque about each support contact point equal to zero,
solve for the support forces, then take the ratio.
Analyzing the contact point above support 1:
There are three torques—weight of the block, weight of the bar, support force:
  0
rF sin(  )  0
(1)(( L / 3)(1 / 2Mg )  (1)( L / 2)( Mg )  ( L)( S 2 )  0
S 2  2 / 3Mg
Analyzing the contact point above support 2:
There are three torques: weight of the block, weight of the bar, support force #1:
(but the lever arm of the block is different)
  0
rF sin   0
(2 / 3L)(1 / 2 Mg )  ( L / 2)( Mg )  ( L)( S1 )  0
S1  5 / 6 Mg
9. The figure below shows a square metal plate of side length 40 cm and uniform density,
lying flat on a table. A force F of magnitude 10 N is applied at one of the corners, as
shown. Determine the torque produced by F relative to the center of rotation.
a. 0 Nm.
b. 1.0 Nm.
c. 1.4 Nm.
d. 2.0 Nm.
e. 4.0 Nm
*D. The center of rotation will be the center of the square, since it is symmetric. The
force is acting through a line which is .2 meters away from the center, so:
  rF sin   (.2m)(10 N ) sin( 90)  2 Nm
From old AP’s
10. (1984) Torque is the rotational analogue of:
a. kinetic energy.
b. linear momentum.
c. acceleration.
d. force.
e. mass.
* D. Rotational analogue means that the two are similar. Just as force causes changes in
linear motion (speeding up, slowing down), torque causes changes in rotational motion
(rotating faster or slower).
11. (1984) A square piece of plywood on a horizontal tabletop is subjected to the two
horizontal forces shown above. Where should a third force of magnitude 5 Newtons be
applied to put the piece of plywood in equilibrium? (All are bird’s eye views)
* A. To be in vertical equilibrium, the additional 5 N force must be down, which throws
out options B, C, and E.
The 10 N force passes through the pivot, and thus causes no net torque.
So, the additional 5 N force must only cancel the 5 N force acting on the upper right
corner. The 5 N force in option A has the same magnitude force, same lever arm,
opposite rotational tendency, so this what you want.
12. To weigh a fish, a person hangs a tackle box of mass 3.5 kg and a cooler of mass 5 kg
from the ends of a uniform rigid pole that is suspended by a rope attached to its center.
The system balances when the fish hangs at a point ¼ of the rod’s length from the tackle
box. What is the mass of the fish?
a. 1.5 kg.
b. 2 kg.
c. 3 kg.
d. 6 kg.
e. Cannot be determined without knowing the mass of the pole.
*C.
13. (1993) A rod of negligible mass is pivoted at a point that is off-center, so that length
l1 is different from length l2. The figures above show two cases in which masses are
suspended from the ends of the rod. In each case the unknown mass m is balanced by a
known mass, M1 or M2, so that the rod remains horizontal. What is the value of m in
terms of the known masses?
a. M1 + M2
b.
M1  M 2
2
c. M1M2
d.
e.
M 1M 2
2
M 1M 2
*E. The quickest way to do this problem is to recognize that since the lever arms are the
same on both diagrams, the ratio of each pair of masses is the same:
M
m
 2
M1
m
m 2  M 1M 2
So : m  M 1 M 2
You could also set up sum of torque is zero for each system and solve for m, but it involves
quite a bit of algebra.
14. A massless rigid rod of length 3d is pivoted at a fixed point W, and two forces each
of magnitude F are applied vertically upward as shown above. A third vertical force of
magnitude F maybe applied, either upward or downward, at one of the labeled
points. With the proper choice of direction at each point, the rod can be in equilibrium if
the third force of magnitude F is applied at point:
a. W only.
b. Y only.
c. V or X only.
d. V or Y only.
e. V, W, or X.
* C.
To solve this problem, calculate the current value of torque, so you know how much you
need to cancel:
  rF sin   (2d)(F)sin 90 - (d)Fsin90  dF
(countercl ockwise)
Thus, we need to create a torque of dF CLOCKWISE.
This could be achieved by applying a force of F upward at point V OR downward at point X.
15. In which of the following diagrams is the torque about point O equal in magnitude to
the torque about point X in the diagram above? (All forces lie in the plane of the paper,
all views are bird’s eye views.)
e. None of the above.
* C. The torque on the initial diagram is lFsin(90) = lF
Be careful that you don’t plug and chug into the torque formula without thinking
carefully about which angle you need.
When calculating torque, you want to pull out the component of the vector that is
perpendicular to the lever arm, which is the COSINE term in diagrams C and D.
So—Diagram C will yield a torque of: (l)(2F)cos(60) = lF.
16.
(2004B, 52%)
A horizontal, uniform board of weight 125 N and length 4 meters is supported by vertical
chains at each end. A person weighing 500 N is sitting on the board. The tension in the
right chain is 250 N. How far from the left end of the board is the person sitting?
a. 0.4 meters.
b. 1.5 meters.
c. 2 meters.
d. 2.5 meters.
e. 3 meters.
*B
Draw a force diagram: