Question

What is the torque of a force $F=(2\hat{i}-3\hat{j}+4k)$N acting at a point $r=(3\hat{i}+2\hat{j}+3\hat{k})$m about the origin in N-m ? (Given $\tau =r\times F$ )

• A. $6\hat{i}-6\hat{j}+12\hat{k}$
• B. $17\hat{i}-6\hat{j}-13\hat{k}$
• C. $-6\hat{i}+6\hat{j}-2\hat{k}$
• D. $-17\hat{i}+6\hat{j}+13\hat{k}$

Answer 1

The correct option is B $17\hat{i}-6\hat{j}-13\hat{k}$

Torque of a force $\tau =r\times F$ .
Given: Force $F=(2\hat{i}-3\hat{j}+4k)$ N and position vector $r=(3\hat{i}+2\hat{j}+3\hat{k})$ m
$\tau =(3\hat{i}+2\hat{j}+3\hat{k})\times (2\hat{i}-3\hat{j}+4\hat{k})$
Thus, taking the cross product, the torque is $17\hat{i}-6\hat{j}-13\hat{k}$.