First slide

Torque

Question

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

Moderate

A

$- 7 \hat{i} - 8 \hat{j} - 4 \hat{k}$
B

$- 4 \hat{i} - \hat{j} - 8 \hat{k}$
C

$- 8 \hat{i} - 4 \hat{j} - 7 \hat{k}$
D

$- 7 \hat{i} - 4 \hat{j} - 8 \hat{k}$

Solution

Moment of force is defined as the cross product of the force and the force arm.
Given, $F = 4 \hat{i} + 5 \hat{j} - 6 \hat{j}, r_{1} = 2 \hat{i} - 2 \hat{j} - 2 \hat{k}$
and $r_{2} = 2 \hat{i} + 0 \hat{j} - 3 \hat{k}$
Moment of force $= r \times F = (r_{2} - r_{1}) \times F$
$= [(2 \hat{i} + 0 \hat{j} - 3 \hat{k}) - (2 \hat{i} - 2 \hat{j} - 2 \hat{k})] \times (4 \hat{i} + 5 \hat{j} - 6 \hat{k})$
$= (0 \hat{i} + 2 \hat{j} - \hat{k}) \times (4 \hat{i} + 5 \hat{j} - 6 \hat{k}) = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & - 1 \\ 4 & 5 & - 6 \end{array}|$
$= \hat{i} (- 6 \times 2) - (- 1 \times 5)] - \hat{j} [(- 6 \times 0) - (- 1 \times 4)] + \hat{k} [(0 \times 5) - 2 \times 4]$
$= - 7 \hat{i} - 4 \hat{j} - 8 \hat{k}$

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