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

Answer:

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

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'

$\tau = F\times r\tau' = F'\times r'$

Now,

$\tau_{net} = - F\times r + F'\times r'\tau_{net} = - 4.56\times 1.24 + 7.58\times 0.57 \\\\= - 1.3338\ Nm$

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:

$|\tau_{net}| = 1.3338\ Nm$