A solid cylinder is pivoted at its center about

a frictionless axle. A force is applied to the
outer radius of 1.27 m at an angle of 30 ◦
above the tangential and exerts a force of 5 N.
A second force is applied by wrapping rope
around the inner radius of 0.576 m, which
exerts a force of 8.36 N tangent to the cylinder.
The moment of inertia of the cylinder is
325 kg · m2
.What is the net Torque

To determine the net torque acting on the cylinder, you need to calculate the torque produced by each force and then find their sum.

The torque produced by a force acting on a rotating object is given by the equation:

τ = r * F * sinθ,

where τ is the torque, r is the perpendicular distance from the force to the axis of rotation, F is the magnitude of the force, and θ is the angle between the force and the line connecting the point of application to the axis of rotation.

Let's calculate the torque produced by each force:

1. Torque due to the force applied to the outer radius:
The perpendicular distance from the center of the cylinder to the point of application of the force is half of the outer radius, which is 1.27 m / 2 = 0.635 m.
Using the given force of 5 N and an angle of 30 degrees, we have:
τ1 = 0.635 m * 5 N * sin(30 degrees).

2. Torque due to the force applied to the inner radius:
The perpendicular distance from the center of the cylinder to the point of application of the force is half of the inner radius, which is 0.576 m / 2 = 0.288 m.
Using the given force of 8.36 N, we have:
τ2 = 0.288 m * 8.36 N.

Now, we can calculate the net torque by summing up the individual torques:
Net Torque = τ1 + τ2.

Finally, substitute the values into the equation and calculate the net torque.