Question

Moment of a force of magnitude 20 N acting along positive X direction at (3m,0,0) about the points (0,2,0) (in Nm) is?

the answer is *40*

Answer 1

By the definition of torque,

г = r x F

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

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

r = -3m i + 2 j

Now,

г = (-3m i + 2 j) x (20 i)

г = (0-0) i - (0-0) j + (0-40) k

г = - 40 k

Ι г Ι = 40 N

which is the required answer