Question

A rod is pivoted about its center. A $5\text{N}$ force is applied $4\text{m}$ from the pivot and another $5\text{N}$ force is applied $2$ $\text{m}$ from the pivot, as shown. The magnitude of the total torque about the pivot (in $\text{N-m}$) is :

• A. $0$
• B. $5$
• C. $15$
• D. $8.7$

Answer 1

The correct option is C $15$

For the given condition in question,
Total torque is given as :

${\overrightarrow{\tau }}_{net}={\overrightarrow{F}}_1\times {\overrightarrow{r}}_1+{\overrightarrow{F}}_2\times {\overrightarrow{r}}_2$

$\Rightarrow |{\tau }_{net}|=5\sin {30}^{\circ }\times 4+5\sin {30}^{\circ }\times 2$

$\Rightarrow |{\tau }_{net}|=15\text{N -m}$

Hence , option $\left(D\right)$ is correct.