Question

In FIGURE EX12.18, what is the net torque about the axle?

Short answer

The net torque on the pulley about the axle is $-0.20\text{N}\text{m}$.

Direction of torque is clockwise.

Step-by-step solution

Step 1. Given information

Torque by a force is defined as $\tau =Fr\sin \varphi$,

Here $\varphi$is measured counterclockwise from the $\overrightarrow{r}$vector to the $\overrightarrow{F}$vector.

Step 2. Explanation

The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:

$\begin{array}{l}\tau =(30\text{N})r_1\sin {\varphi }_1+(20\text{N})r_2\sin {\varphi }_2\\ =(30\text{N})(0.02\text{m})\sin \left(-{90}^{°}\right)+(20\text{N})(0.02\text{m})\sin \left({90}^{°}\right)\\ =(-0.60\text{Nm})+(0.40\text{N}\text{m})\\ =-0.20\text{N}\text{m}\end{array}$

The net torque on the pulley about the axle is$-0.20\text{N}\text{m}$

Direction of torque is clockwise.