Question

In FIGURE EX12.19, what magnitude force provides $5.0\text{Nm}$net torque about the axle?

Short answer

The magnitude of the force is$50\text{N}$

Step-by-step solution

Step 1. Given information

The magnitude of the force is 5.0 Nm. The distance between couple forces is$r=10\text{\hspace{0.17em}cm}$

Step 2. Explanation

The expression for the torque due the couple force

$\tau =rF$

After rearranging

$\begin{array}{l}F=\frac{\tau }{r}\\ F=\frac{5.0\text{N}\cdot \text{m}}{10\text{cm}}\\ =\frac{5.0\text{N}\cdot \text{m}}{10\text{cm}\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)}=50\text{N}\end{array}$

Hence, the magnitude of the force is $50\text{N}$.