Question

A uniform disk of mass m = 2.4 kg and radius R = 25 cm can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 8.5 N, F2 = 1.5 N, F3 = 6.5 N and F4 = 6.5 N. F2 and F4 act a distance d = 3.5 cm from the center of mass. These forces are all in the plane of the disk. Write an expression of the magnitude of the torque due to the force F3

Answer 1

Answer:

0 Nm

Explanation:

Torque is the cross product of the radius vector and the force vector.

τ = r × F

In other words, the magnitude of the torque is equal to the magnitude of the radius times the magnitude of the force times the sine of the angle between them.

τ = rF sin θ

Since F₃ is parallel to the radius vector, θ = 0, so τ = 0.