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Torques are useful in problems involving rolling and spinning, although most such problems lie outside the scope of the MCAT. Torques are also useful in solving for forces in structural problems, even if there is no motion. For these problems we use the following principle:

If a system is in static equilibrium, then


no matter which point you choose as the pivot.


Equations (5a) and (5b) assure translational equilibrium, and equation (5c) assures rotational equilibrium. The following examples illustrate methods for calculating forces in static equilibrium.



Consider the pulley system shown in Figure 7-9. Mass m1 is 2 kg, the radius of pulley A is 0.15 m, the radii of pulley B are 0.3 m and 0.1 m, and the radius of pulley C is 0.05 m. What mass m2 is required for equilibrium?


..\art 7 jpg\figure 7-ti.jpg
Figure 7-9


A string has one tension throughout its length, no matter what pulleys it goes around. Thus the tension T1 m1g, and T2 m2g. The radii of pulleys A and C are irrelevant.
For equilibrium, we can take torques about the pivot of pulley B, keeping in mind that counterclockwise is positive and clockwise is negative, so that

τnet = 0
τ1 – τ2 = 0
(0.3 m)(2 kg 10 m/s2)(1) – (0.1 m)(m2 10 m/s2)(1) = 0
m2 = 6 kg

For problems involving torque balance, the following methods often work:

2. Label all forces acting on the system.
3. Choose a pivot.
4. Calculate torques and set τnet = 0.
5. Use (Fnet)x = 0 and (Fnet)y = 0, or choose another pivot.


In the following example, it will be helpful for you to work out the example as you read along.


A pole of mass m1 and length L sticks out perpendicularly from a wall at point A. A wire connects the end of the pole to a point B above point A, making an angle θ with the pole. A lamp of mass m2 hangs from a wire at the end of the pole. (See Figure 7-10.)

  1. What is the tension in the wire from B to C in terms of m1, m2, g, L, and θ?
  2. What is the vertical force exerted by the wall on the pole?
..\art 7 jpg\figure 7-tj.jpg
Figure 7-10



First, we DRAW A DIAGRAM with all the forces on the pole (Figure 7-11).


..\art 7 jpg\figure 7-tk.jpg

Figure 7-11

We know m1g and m2g but not TFx, and Fy. Let us make a chart showing the torques about A and C.

force torque about A torque about C
m1g Lm1g/2  
T LTsinθ  

Make sure you understand why these two entries are correct. Then fill in the rest of the chart on your own. The completed chart is shown on the next page.
force torque about A  torque about C
m1g  Lm1g/2  Lm1g/2
m2g Lm2g   0
T  LTsinθ 0
Fx 0 0
Fy 0 LFy

In the second column the entries for the torque due to Fx and Fy about A are zero, since r = 0 for these entries. In the third column, the entry for the torque due to Fx about C is zero because sinφ = 0. The torque due to T about C is zero because r = 0.
  1. In order to find T, we can take torques about point A, since that eliminates Fx and Fy. The net torque must be zero because the system is in equilibrium, so we have (taking the sum of the first column)

    We solve for T and divide through by L to obtain
  2. In order to find Fy, we can use one of two methods. Using torques and choosing point C as a pivot yields
    Lm1g/2 – LFy = 0
    Fy = m1g/2

    On the other hand, we could add up all the vertical forces and use (Fnet)y = 0, so that


This was a longer solution, so using torques is clearly the way to go.



A pole of length L is connected to a hinge at point A on a vertical wall, making an angle α with the wall. A horizontal string connects to the wall at point B and the end of the pole at point C. A box of candy of mass m hangs from a string at the end of the pole. (See Figure 7-12)

  1. What is the tension in the horizontal string?
  2. What is the magnitude of the force of the wall on the pole at point A?

..\art 7 jpg\figure 7-tl.jpg
Figure 7-12


First we DRAW A DIAGRAM with all the forces on the pole (Figure 7-13). We have the tension T in the horizontal string and tension mg in the vertical string. We are left with a choice for the force of the wall on the pole Fs: pushing or pulling? No matter, we can draw it either way, and physics will tell us later on if we have it right. Let's make the force tensile, that is, pulling.


..\art 7 jpg\figure 7-tm.jpg
Figure 7-13

  1. We choose A for the pivot, since that choice kills the force Fs but keeps T and mg. Setting the sum of the torques equal to zero gives

  1. In order to find Fs, we set the sum of vertical forces to zero, so that

(Fnet)y = 0
Fs cosα – mg = 0
Fs = –mg/cosα

The negative sign tells us that we drew the Fs vector the wrong way. (Perhaps you knew this already.) The force is compressional.

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