Question

Suppose that a rigid body is free to rotate about a fixed point $O$. If a force $\overrightarrow{F}$ acts on the body at a point $P$, then the body will tend to rotate about an axis through $O$. This effect is measured by the torque vector $t$ which is given by $\overrightarrow{t}=\overrightarrow{r}\times \overrightarrow{F}$. Where $\overrightarrow{r}$ is the position vector$\overrightarrow{OP}=25\hat{i}$and$\overrightarrow{F}=10\hat{i}+10\sqrt{3}\hat{j}$. The straight line through $O$ determined by $\overrightarrow{t}$ is the axis of rotation. The vector $\overrightarrow{r},\overrightarrow{F}$ and $\overrightarrow{t}$ form a right handed system.
$\overrightarrow{t}$ is the torque, $\overrightarrow{F}$ is the force and $\overrightarrow{r}$ is the position vector, then $\left[\overrightarrow{t}\overrightarrow{r}\overrightarrow{F}\right]$ is equal to

• A. $62500$
• B. $187500$
• C. $18750$
• D. $625000$

Answer 1

The correct option is B $187500$

$\overrightarrow{t}=\overrightarrow{r}\times \overrightarrow{F}$
$\overrightarrow{t}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 25& 0& 0\\ 10& 10\sqrt{3}& 0\end{array}\right|=(0-0)\hat{i}-(0-0)\hat{j}+(250\sqrt{3}-0)\hat{k}$
$\overrightarrow{t}=250\sqrt{3}\hat{k}$
$\left[\overrightarrow{t}\overrightarrow{r}\overrightarrow{F}\right]=\left|\begin{array}{ccc}0& 0& 250\sqrt{3}\\ 25& 0& 0\\ 10& 10\sqrt{3}& 0\end{array}\right|=250\sqrt{3}(250\sqrt{3}-0)=187500$
$\left[\overrightarrow{t}\overrightarrow{r}\overrightarrow{F}\right]=187500$