Question

A system of two cylinders fixed to each other is free to rotate about a frictionless axis through the common center of the cylinders and perpendicular to the page. A rope wrapped around the cylinder of radius 2.50 m exerts a force of 4.49 N to the right on the cylinder. A rope wrapped around the cylinder of radius 1.14 m exerts a force of 9.13 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinders about the rotation axis? Answer in three decimal places.

Answer 1

The net torque is  331.402 N-m

Explanation:

Because torque is a vector quantity , the direction of torque is considered along the axis of rotation .

In first case force is applied in the right direction , thus the torque will be in upwards direction .

The magnitude of torque is = force x perpendicular distance between point of application of force and axis of rotation .

Thus in this case τ₁ = 4.49 x 2.50 = 11.25 N-m

Similarly , torque on second cylinder τ₂ = 91.3 x 1.14 = 104.08 N-m

Here force is applied downwards direction , thus the direction of torque is in the plane of page .

The torque is a vector quantity . The net torque can be found

τ = $\sqrt{(11.25)^2 + (104.08)^2}$ =  331.402 N-m