Aparticle is located on the x axis 4.9 m from the origin. a force of 38 n, directed 30° above the x axis in the x-y plane, acts on the particle. what is the torque about the origin on the particle? round your answer to the nearest whole number. use a minus sign to indicate a negative direction and no sign to indicate a positive direction.

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

        Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

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