Question

Moment of force of magnitude 20n acting along positive x direction at a point (3m,0,0)

Answer 1

your question is incomplete. A complete question is ---> 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?

solution : force acting along positive x direction at point (3, 0, 0).

so, vector form of force, $\vec{F}=|F|\hat{n}$

= 20 × (3i)/√{(3)² + 0² + 0²}

= 20i N

now, position vector , $\vec{r}$ = (0 - 3)i + (2 - 0)j + (0 - 0)k

= -3i + 2j

now, moment of force, $\tau=\vec{r}\times\vec{F}$

= (-3i + 2j) × (20i)

= -3i × 20i + 2j × 20i

[ as you know, i × i = 0, j × i = -k ]

so, $\tau$ = -40k

and magnitude of moment of force, $|\tau|$ = 40 Nm

Answer 2

Explanation:

force acting in the direction given..

torque is 40Nm