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 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. Thanks.

Answer 2

Answer:

r =40 N

Explanation:

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