A particle 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.

Question

Isabella Reed

Physics

19 March, 20:56

A particle 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.

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Answers (1)

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Jayvion · Answer 1

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.