Question

The 20-cm-diameter disk in FIGURE EX12.20 can rotate on an axle through its center. What is the net torque about the axle?

Short answer

The net torque on the axle is$-0.94\text{N}\text{m}$ , the direction is clockwise.

Step-by-step solution

Step 1. Given information

The radius of the disk is 10 cm and the disk rotates on an axle through its center.

Step 2. Explanation

The net torque on the axle is

$\begin{array}{l}\tau =F_{\text{A}}{r}_{\text{A}}\sin {\varphi }_{\text{A}}+F_{\text{B}}{r}_{\text{B}}\sin {\varphi }_{\text{B}}+F_{\text{C}}{r}_{\text{C}}\sin {\varphi }_{\text{C}}+F_{\text{D}}{r}_{\text{D}}\sin {\varphi }_{\text{D}}\\ =(30\text{N})(0.10\text{m})\sin \left(-{90}^{°}\right)+(20\text{N})(0.050\text{m})\sin {90}^{°}+(30\text{N})(0.050\text{m})\sin {135}^{°}+(20\text{N})(0.10\text{m})\sin 0^{°}\\ =-3\text{N}\text{m}+1\text{N}\text{m}+1.0607\text{N}\text{m}\\ =-0.94\text{N}\text{m}\end{array}$

The net torque on the axle is $-0.94\text{N}\text{m}$, the direction is clockwise.