A force given by F rightarrow = (-2i + 6j + 3k)kN acts at a point...

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Question

science physics

Answer 1

Answer #1

in the first case,

position vector of point where force is acting, r1 = (4-0)i + (2-0)i + (2-3)k

= 4i + 2j - 1k

Moment of the force = r1 cross F

= (4i + 2j - 1k) cross (-2i + 6j + 3k)*10^3

= (24*k + 12*(-j) - 4*(-k) + 6*i + 2j - 6*(-i))*10^3

= (12i - 10j + 20k)*10^3 N.m

in the second case,

position vector of point where force is acting, r1 = (4-0)i + (2-4)i + (2-0)k

= 4i - 2j + 2k

Moment of the force = r1 cross F

= (4i - 2j - 2k) cross (-2i + 6j + 3k)*10^3

= (24*k + 12*(-j) - 12*(-k) - 6*i + 4j - 12*(-i))*10^3

= (6i - 8j + 36k)*10^3 N.m

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Answer 2

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