Equilibrium
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 m_{1 }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 m_{2 }is required for equilibrium?
Figure 7-9
A string has one tension throughout its length, no matter what pulleys it goes around. Thus the tension T_{1 }= m_{1}g, and T_{2 }= m_{2}g. 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
Ï„_{net }= 0
Ï„_{1 }â€“ Ï„_{2 }= 0
(0.3 m)(2 kg 10 m/s^{2})(1) â€“ (0.1 m)(m_{2} 10 m/s^{2})(1) = 0
m_{2 }= 6 kg
For problems involving torque balance, the following methods often work:
1. DRAW A DIAGRAM.
2. Label all forces acting on the system.
3. Choose a pivot.
4. Calculate torques and set Ï„_{net }= 0.
5. Use (F_{net})_{x} = 0 and (F_{net})_{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 Î¸ with the pole. A lamp of mass m2 hangs from a wire at the end of the pole. (See Figure 7-10.)
- What is the tension in the wire from B to C in terms of m1, m2, g, L, and Î¸?
- What is the vertical force exerted by the wall on the pole?
Figure 7-10
First, we DRAW A DIAGRAM with all the forces on the pole (Figure 7-11).
Figure 7-11
We know m_{1}g and m_{2}g but not T, F_{x}, and F_{y}. Let us make a chart showing the torques about A and C.
force | torque about A | torque about C |
m_{1}g | â€“Lm_{1}g/2 | |
m_{2}g | ||
T | LTsinÎ¸ | |
F_{x} | ||
F_{y} |
force | torque about A | torque about C |
m_{1}g | â€“Lm_{1}g/2 | Lm_{1}g/2 |
m_{2}g | â€“Lm_{2}g | 0 |
T | LTsinÎ¸ | 0 |
F_{x } | 0 | 0 |
F_{y} | 0 | â€“LF_{y} |
- In order to find T, we can take torques about point A, since that eliminates F_{x }and F_{y}. 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
- In order to find F_{y}, we can use one of two methods. Using torques and choosing point C as a pivot yields
Lm_{1}g/2 â€“ LF_{y} = 0
F_{y} = m_{1}g/2
On the other hand, we could add up all the vertical forces and use (F_{net})_{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 Î± 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)
- What is the tension in the horizontal string?
- What is the magnitude of the force of the wall on the pole at point A?
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 F_{s}: 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.
Figure 7-13
- We choose A for the pivot, since that choice kills the force F_{s }but keeps T and mg. Setting the sum of the torques equal to zero gives
- In order to find F_{s}, we set the sum of vertical forces to zero, so that
(F_{net})_{y} = 0
â€“F_{s }cosÎ± â€“ mg = 0
F_{s} = â€“mg/cosÎ±
The negative sign tells us that we drew the F_{s }vector the wrong way. (Perhaps you knew this already.) The force is compressional.