Question

A uniform rod of length 1.01m and mass 1.39kg is supported horizontally by a collar that is attached to a vertical wall, as shown. A second uniform rod of length 0.22m and mass 0.23kg is attached to the first, with the far ends of the rods aligned as shown. What is the magnitude of the total torque, in newton meters, exerted on the longer rod due to the force of gravity? Assume that the axis extends perpendicularly out of the page and is located at the collar, which has negligible length.

A) 1.36 Nm

B) 1.57 Nm

C) 2.13 Nm

D) 2.44 Nm

Answer 1

The magnitude of the total torque exerted on the longer rod due to the force of gravity is 14.36 Nm.

Step-by-step explanation:

To calculate the magnitude of the total torque exerted on the longer rod due to the force of gravity, we need to consider the torque produced by each rod separately. The torque is given by the formula:

Torque = force x distance

The torque produced by the longer rod is:

Torque = (mass of longer rod) x (acceleration due to gravity) x (distance from pivot)

Substituting the given values, we get:

Torque = 1.39 kg x 9.8 m/s^2 x 1.01 m = 13.86 Nm

The torque produced by the shorter rod is:

Torque = (mass of shorter rod) x (acceleration due to gravity) x (distance from pivot)

Substituting the given values, we get:

Torque = 0.23 kg x 9.8 m/s^2 x 0.22 m = 0.50 Nm

Therefore, the magnitude of the total torque exerted on the longer rod due to the force of gravity is 13.86 Nm + 0.50 Nm = 14.36 Nm.