Force F = (8.0 N) + (-8.0 N) acts on a pebble with position vector r = (2.60 m)

Question

Force F = (8.0 N) + (-8.0 N) acts on a pebble with position vector r = (2.60 m) + (-1.8 m) , relative to the origin. What is the resulting torque acting on the pebble about the origin? Your answer cannot be understood or graded. More Information N m) + ( N.m) J + ( N m) k What is the resulting torque acting on the pebble about a point with coordinates (-4.0 m, -6.0 m, 2.0 m)? Torque is the cross product of a position vector (extending from a chosen point to the particle) and a force vector. Do you remember how to find the position vector relative to a point that is not the origin? How far parallel to the x axis is the particle? How far parallel to the y axis?

Explanation / Answer

a) = F x r = ( 8i + 0j -8k) x ( 0i + 2.6 j - 1.8 k) =20.8 i + 64j -20.8k

b) = F x = ( 8i + 0j -8k) x ( -4 i - 6j + 2k) = -48i + 16j -48k

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