Respuesta :
Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:
[tex]\tau_{net} = \tau + \tau'[/tex]
where
[tex]\tau[/tex] = Torque due to Force, F
[tex]\tau'[/tex] = Torque due to Force, F'
[tex]tau = F\times r[/tex]
[tex]tau' = F'\times r'[/tex]
Now,
[tex]\tau_{net} = - F\times r + F'\times r'[/tex]
[tex]\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm[/tex]
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:
[tex]|\tau_{net}| = 2.41\ Nm[/tex]
