Question

ate around its central axis. A rope wrapped around the drum of radius 1.24 m exerts a force of 4.56 N to the right on the cylinder. A rope wrapped around the core of radius 0.57 m exerts a force of 7.58 N downward on the cylinder. 4.56 N 7.58 N What is the magnitude of the net torque acting on the cylinder about the rotation axis?

Answer 1

Magnitude the net torque about its axis of rotation is 1.3338 Nm

Step-by-step explanation:

The radius of the wrapped rope around the drum, r = 1.24 m

Force applied to the right side of the drum, F = 4.56 N

The radius of the rope wrapped around the core, r' = 0.57 m

Force on the cylinder in the downward direction, F' = 7.58 N

Now, the magnitude of the net torque is given by:

`\tau_(net) = \tau + \tau'`

where

`\tau` = Torque due to Force, F

`\tau'` = Torque due to Force, F'

`\tau = F* r\tau' = F'* r'`

Now,

`\tau_(net) = - F* r + F'* r'\tau_(net) = - 4.56* 1.24 + 7.58* 0.57 \\\\= - 1.3338\ Nm`

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

`|\tau_(net)| = 1.3338\ Nm`