Question

Two forces $F_1=(2\hat{i}–5\hat{j}-6\hat{k})\text{N}$ and $F_2=(-\hat{i}+2\hat{j}-\hat{k})\text{N}$ are acting on a body at the points $(1,1,0)$ and $(0,1,2)$ respectively. Find resultant torque acting on the body about point $(-1,0,1)$ in $\left(\text{N-m}\right)$.

• A. $-10\hat{i}+14\hat{j}-9\hat{k}$
• B. $-14\hat{i}+5\hat{j}-9\hat{k}$
• C. $14\hat{i}-9\hat{j}+10\hat{k}$
• D. $-14\hat{i}+10\hat{j}-9\hat{k}$

Answer 1

The correct option is D $-14\hat{i}+10\hat{j}-9\hat{k}$

Let the position vector of $(1,1,0)$ and $(0,1,2)$ w.r.t point $(-1,0,1)$ be $r_1$ and $r_2$ .
Therefore, $r_1=(\hat{i}+\hat{j})-(-\hat{i}+\hat{k})=2\hat{i}+\hat{j}-\hat{k}$
$r_2=(\hat{j}+2\hat{k})-(-\hat{i}+\hat{k})=\hat{i}+\hat{j}+\hat{k}$
Then, net torque is given by
$\tau =r_1\times F_1+r_2\times F_2$

Now, $r_1\times F_1=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 2& -5& -6\end{array}\right|=-11\hat{i}+10\hat{j}-12\hat{k}$
and
$r_2\times F_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ -1& 2& -1\end{array}\right|=-3\hat{i}+3\hat{k}$
So, resultant torque $=(-11\hat{i}+10\hat{j}-12\hat{k})+(-3\hat{i}+3\hat{k})$
$=-14\hat{i}+10\hat{j}-9\hat{k}\text{N-m}$