Torque is equal to the cross product as stated above. In general, one can simplify by saying that the torque is equal to the force acting on the object multiplied by the perpendicular distance from the application of the force to the rotational axis. Say you had a seesaw. It is easier to exert torque, get the seesaw to move, if you pushed on the board near the end rather than near the middle. It is the rotational version of Force.

Individual torques are determined by multiplying the force applied by the *perpendicular* component of the moment arm

#### Example

A 1 m long 2 kg rod is attached to the side of a building with a hinge at one end. The rod is held level by a cable that makes an angle of 45 degrees with the rod and is also attached to the building above the rod. What is the tension in the cable? The situation is illustrated in a diagram below.

There will be multiple steps to this problem. First we'll make a free body diagram for the rod; making free body diagrams in torque problems is just as important as in Newton's Laws problems. In the diagram, the force of tension from the cable, the rod's weight, and the forces from the hinge are shown. For the purposes of calculating torque, an object's weight acts from its center of mass (half way along the rod in this case). In the diagram below, the force on the rod from the hinge is already broken into it's components to make it easier to visualize.

Now, in order to find the tension in the cable, we'll sum the torques on the rod with the hinge as the axis of rotation. Notice that the sign of the torque due to tension in the cable and the torque due to the rod's weight will be opposite signs because they would cause the rod to rotate in opposite directions.

\begin{align*} \Sigma\tau&=0\\ mg\frac{r}{2} - T\sin(\theta)r&=0\\ T&=\frac{mgr}{2\sin(\theta)r}\\ T&=\frac{mg}{2\sin(\theta)}\\ T&=\frac{2\;\text{kg} * 9.8\;\text{m/s}^2}{2\sin(45)}\\ T&=13.85\;\text{N}\\ \end{align*}

### Simulation

### Review

- Consider hitting someone with a Wiffle ball bat. Will it hurt them more if you grab the end or the middle of the bat when you swing it? Explain your thinking, but do so using the vocabulary of
*moment of inertia*(treat the bat as a rod) and*torque*(in this case, torques caused by the contact forces the other person’s head and the bat are exerting on each other). - A wooden plank is balanced on a pivot, as shown below. Weights are placed at various places on the plank. Consider the torque on the plank caused by weight \begin{align*}A\end{align*}.
- What force, precisely, is responsible for this torque?
- What is the magnitude (value) of this force, in Newtons?
- What is the moment arm of the torque produced by weight \begin{align*}A\end{align*}?
- What is the magnitude of this torque, in \begin{align*}N \cdot m\end{align*}?
- Repeat parts (a – d) for weights \begin{align*}B\end{align*} and \begin{align*}C\end{align*}.
- Calculate the net torque. Is the plank balanced? Explain.

- There is a uniform rod of mass \begin{align*}2.0 \;\mathrm{kg}\end{align*} of length \begin{align*}2.0 \;\mathrm{m}\end{align*}. It has a mass of \begin{align*}2.6 \;\mathrm{kg}\end{align*} at one end. It is attached to the ceiling \begin{align*}.40 \;\mathrm{m}\end{align*} from the end with the mass. The string comes in at a \begin{align*}53\end{align*} degree angle to the rod.
- Calculate the total torque on the rod.
- Determine its direction of rotation.
- Explain, but don’t calculate, what happens to the angular acceleration as it rotates toward a vertical position.

**Review (Answers)**

- At End. Discuss in class. Moment of inertia at the end \begin{align*}1/3 \;\mathrm{ML}^2\end{align*} at the center \begin{align*}1/12 \;\mathrm{ML}^2\end{align*} and torque, \begin{align*}\tau = I\alpha\end{align*} change the in the same way
- a. weight b. \begin{align*}19.6 \;\mathrm{N}\end{align*} c. plank’s length \begin{align*}(0.8\mathrm{m})\end{align*} left of the pivot d. \begin{align*}15.7 \;\mathrm{N \ m}\end{align*}, e. Ba. weight, Bb. \begin{align*}14.7 \;\mathrm{N}\end{align*}, Bc. plank’s length \begin{align*}(0.3\mathrm{m})\end{align*} left of the pivot, Bd. \begin{align*}4.4 \;\mathrm{N \ m}\end{align*}, Ca. weight, Cb. \begin{align*}13.6\;\mathrm{N}\end{align*}, Cc. plank’s length \begin{align*}(1.00 \;\mathrm{m})\end{align*} right of the pivot, Cd. \begin{align*}13.6 \;\mathrm{N \ m}\end{align*}, f) \begin{align*}6.5 \;\mathrm{N \ m \ CC}\end{align*}, g) no, net torque doesn’t equal zero
- \begin{align*}\;\mathrm{CCW}\end{align*}