Question

A circular-shaped object of mass 10 kg has an inner radius of 13 cm and an outer radius of 26 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 12 N, 26 N, and 16 N act on the object, as shown. The force of magnitude 26 N acts 30❦ below the horizontal. 12 N 16 N 26 N 30❦ ϝ Find the magnitude of the net torque on the wheel about the axle through the center of the object. Answer in units of N · m

Answer 1

Explanation:

It is known that it does not matter at what angle the force has been applied on an object because they all are acting perpendicular to the axis.

And, expression for torque from each force is as follows.

                  $\tau = F \times r$

where,   F = magnitude of force

             r = distance from the center

Total torque will be calculated as follows.

    $\tau = [(12 + 16) \times 0.26 - (0.26 \times 0.13)]$ Nm

                   = 3.9 Nm

Therefore, we can conclude that the magnitude of the net torque on the wheel about the axle through the center of the object is 3.9 Nm.