Question

Moment of a force of magnitude 20 N acting along positive x-direction at point (3m,0,0) about the point (0,2,0) (in N-m) is

Answer 1

CORRECT QUESTION :

• Moment of a force of magnitude 20N acting along positive x direction at point (3m,0,0) about the point (0,2,0) .What is the torque?

ANSWER :

By the definition of torque

$\large \boxed{ \sf \green{ \tau = r \times F}}$

Given points are  (3m,0,0) and (0,2,0)

$\sf r = (0 - 3m) i + (2-0) j + (0-0) k$

$\sf\tau = -3m \: i + 2 \: j$

Now

$\sf \tau = (-3m \: i + 2 j) \times (20 i)$

$\sf \tau = (0-0) i - (0-0) j + (0-40) k$

$\large \boxed{ \sf \orange{\tau = - 40 \:n}}$