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