Question

The moment of the force, vector $$F=4\hat{i}+5\hat{j}-6\hat{k}$$at (2,0,-3) about the point (2,-2,-2) is given by
A-$$-8\hat{i}-4\hat{j}-7\hat{k}$$
B-$$-4\hat{i}-\hat{j}-8\hat{k}$$
C-$$-7\hat{i}-8\hat{j}-4\hat{k}$$
D-$$-7\hat{i}-4\hat{j}-8\hat{k}$$

Answer 1

Hint: Here we need to find the moment of force at a point. First of all, let us define what is a moment of force. The moment of force is a measure of the ability of the force acting on a particular body to rotate it around a given axis.

Complete step by step answer:
We are given the coordinates of two points, say A (2,0,-3) and point B (2,-2,-2). In order to find the moment of force about point B, we need to have the distance between the point of application of force and the perpendicular distance between this and the axis of rotation. So we need to find first the displacement vector, $$\overrightarrow{r}$$. After this we know the moment of force commonly called torque is given by the cross product of the force and the displacement vector. So, we make use of matrix algebra to find out the value of the cross product.
The force is $$F=4\hat{i}+5\hat{j}-6\hat{k}$$and the distance vector is given by,
$$\begin{array}{rl}& \overrightarrow{r}=(2-2)\hat{i}+(0+2)\hat{j}+(-3+2)\hat{k}\\ & =2\hat{j}-\overrightarrow{k}\end{array}$$
So, torque is given by the formula, Torque = $$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$$
$$\begin{array}{rl}& \overrightarrow{\tau }=(2\hat{j}-\hat{k})\times (4\hat{i}+5\hat{j}-6\hat{k})\\ & =\left(\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& -1\\ 4& 5& -6\end{array}\right)\\ & =-7\hat{i}-4\hat{j}-8\hat{k}\end{array}$$

So, the correct option is (D)

Note:While calculating the torque we have to keep in our mind that it is a vector quantity and is given by the cross product of force acting on the body and the perpendicular distance from the axis of rotation. So, we have to consider the angle between the two vectors