Question

The moment of the force f=3i+j-2k acting at the point (1,2,3) about the point (1,0,-1) is​

Answer 1

We have to find the moment of the force f = 3i + j - 2k acting at the point (1, 2, 3) about the point (1, 0, -1) is ...

solution : first find position vector.

we have to find moment of force at (1, 2, 3) about (1, 0, -1)

so position vector, r = (1, 2, 3) - (1, 0, -1)

= (0, 2, 4) = 2j + 4k

force vector, F = 3i + j - 2k

we know, the moment of force ( torque) is the cross product of position vector and force.

so, moment of force = r × F

= (2j + 4k) × (3i + j - 2k)

= 2j × (3i + j - 2k) + 4k × (3i + j - 2k)

= ( 2j × 3i + 2j × j - 2j × 2k ) + (4k × 3i + 4k × j - 4k × 2k)

= (-6k + 0 - 4i) + (12j - 4i - 0)

= - 8i + 12j - 6k

Therefore the moment of force is -8i + 12j - 6k

Answer 2

$\huge \tt - 8i + 12j - 6k$