Question

A force $\overrightarrow{F}=4\hat{i}+3\hat{j}+4\hat{k}$ is applied on an intersection point of $x=2$ plane and $x$-axis. The magnitude of torque of this force about a point $(2,3,4)$ is _______. (Round off to the Nearest Integer)

Answer 1

Step 1: Given Data and draw diagram

Force, $\overrightarrow{F}=4\hat{i}+3\hat{j}+4\hat{k}$

Step 2: Find radius

Radius is,

$$\begin{gathered} \overrightarrow{r_A}+\overrightarrow{r}=\overrightarrow{r_B} \\ \overrightarrow{r}=\overrightarrow{r_B}-\overrightarrow{r_A} \\ \overrightarrow{r}=(-3\hat{j}-4\hat{k}) \end{gathered}$$

Step 3: Find torque

Torque is the rotational equivalent of linear force.

$$\begin{gathered} \left|\overrightarrow{\tau }\right|=\left|\overrightarrow{r}\times \overrightarrow{F}\right| \\ \left|\overrightarrow{\tau }\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& -3& -4\\ 4& 3& 4\end{array}\right| \\ \left|\overrightarrow{\tau }\right|=-16\hat{j}-12\hat{k} \\ \left|\overrightarrow{\tau }\right|=\sqrt{{\left(-16\right)}^2+{(-12)}^2} \\ \left|\overrightarrow{\tau }\right|=20 \end{gathered}$$

Hence, the magnitude of torque of this force about a point $(2,3,4)$ is $20$.