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

Options

A. 20
B. 60
C. 40
D. 30

Answer

C

Explanation

Torque, `tau=rxxF`
`=[(0-3)hat(i)+(2-0)hat(j)+(0-0)hat(k)]xx[20 hat(i)]`
`=[3 hat(i)+2hat(j)]xx[20 hat(i)]=-40 hat(k)`.