Rotational Motion Question 91
Question: The moment of the force, $ \overrightarrow{{}F}\text{=4}\widehat{i}\text{+5}\widehat{j}\text{-6}\widehat{k} $ at (2, 0, -3), about the point (2, -2, -2), is given by
Options:
A) $ \text{-7}\widehat{i}\text{-8}\widehat{j}\text{-4}\widehat{k} $
B) $ \text{-4}\widehat{i}\text{-}\widehat{j}\text{-8}\widehat{k} $
C) $ \text{-8}\widehat{i}\text{-4}\widehat{j}\text{-7}\widehat{k} $
D) $ \text{-7}\widehat{i}\text{-4}\widehat{j}\text{-8}\widehat{k} $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d]
$ \overrightarrow{{}\tau }=(\overrightarrow{{}r}-{{\overrightarrow{{}r}}_0})\times \overrightarrow{{}F} $ (i)
$ \overrightarrow{{}r}-{{\overrightarrow{{}r}} _{_0}}=(2\widehat{i}-0\widehat{j}-3\widehat{k})-(2\widehat{i}-2\widehat{j}-2\widehat{k}) $
$ =0\widehat{i}+2\widehat{j}-\widehat{k} $
$ \overrightarrow{{}\tau }= \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \\ \end{vmatrix} =-7\widehat{i}-4\widehat{j}-8\widehat{k} $
