Question

If a rigid body is subjected to two forces $\overrightarrow{F_1}=2\hat{i}+3\hat{j}+4\hat{k}$ acting at $(3,3,4)$ and $\overrightarrow{F_2}=-2\hat{i}-3\hat{j}-4\hat{k}$ acting at $(1,0,0)$, then which of the following is true ?

• A. The body is in equilibrium.
• B. The body is under the influence of a torque only.
• C. The body is under the influence of a single force.
• D. The body is under the influence of a force as well as a torque.

Answer 1

The correct option is A The body is in equilibrium.

Body will be in equilibrium if $\overrightarrow{F_{net}}=0,\overrightarrow{{\tau }_{net}}=0$
Net force on body, $\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}$
$\overrightarrow{F}=2\hat{i}+3\hat{j}+4\hat{k}+(-2\hat{i}-3\hat{j}-4\hat{k})$
$\overrightarrow{F}=0$

Torque due to $\overrightarrow{F_1}$ force,
$\overrightarrow{{\tau }_1}=\overrightarrow{r_1}\times \overrightarrow{F_1}$
where $\overrightarrow{r_1}=3\hat{i}+3\hat{j}+4\hat{k}$
$\Rightarrow \overrightarrow{{\tau }_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 3& 4\\ 2& 3& 4\end{array}\right|$
$=\hat{i}(12-12)-\hat{j}(12-8)+\hat{k}(9-6)$
$\overrightarrow{{\tau }_1}=-4\hat{j}+3\hat{k}$

Torque due to $\overrightarrow{F_2}$ force
$\overrightarrow{{\tau }_2}=\overrightarrow{r_2}\times \overrightarrow{F_2}$
where $\overrightarrow{r_2}=\hat{i}$
$\Rightarrow \overrightarrow{{\tau }_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 0\\ -2& -3& -4\end{array}\right|$
$=\hat{i}\left(0\right)-\hat{j}(-4)+\hat{k}(-3)$
$\overrightarrow{{\tau }_2}=4\hat{j}-3\hat{k}$

Net torque on body $\overrightarrow{\tau }=\overrightarrow{{\tau }_1}+\overrightarrow{{\tau }_2}$
$\overrightarrow{\tau }=-4\hat{j}+3\hat{k}+4\hat{j}-3\hat{k}=0$

Net force = 0, Net torque = 0. Hence body is in equilibrium.