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Motion In A Plane

Question
CBSEENPH11020855

The moment of the force, F = 4i^ + 5j^ -6k^  at (2, 0, 3) about the point (2,-2,2), is given by

  • -8i^ -4j^ -7k^

  • -4i^ -j^ -8k^

  • -7i^ -4j^ -8k^

  • -7i^ -8j^ -4k^

Solution

C.

-7i^ -4j^ -8k^

Moment of force,
τ = r x F

τ = (r -r0) x Fr -r0 = (2i^ + 0j^ - 3k^) - (2i^ -2j^-2k^)=0i^ + 2j^ -k^τ = (0i^ + 2j^ -k^) (4i^ + 5j^-6k^)τ = i^j^k^02-145-6 = - 7i^ - 4j^ -8k^