The moment of the force, F = 4 i + 5 j - 6 k at (2, 0, −3), about the point (2, −2, −2), is given by
- -8 i - 4 j - 7 k
- -4 i - j - 8 k
- -7 i - 8 j - 4 k
- -7 i - 4 j - 8 k
Answer
(4) -7 i - 4 j - 8 k
Solution
Position vector of the point of application relative to (2, −2, −2):
r = (2-2) i + (0+2) j + (-3+2) k = 2 j - k
Moment of force: = r F
= vmatrix i & j & k \ 0 & 2 & -1 \ 4 & 5 & -6 vmatrix
= i(-12 + 5) - j(0 + 4) + k(0 - 8)
= -7 i - 4 j - 8 k
Answer: (4)