An object with a mass of #5 kg# is hanging from an axle with a radius of #15 m#. If the wheel attached to the axle has a radius of #3 m#, how much force must be applied to the wheel to keep the object from falling?

1 Answer
Mar 4, 2016

Answer:

Rotational Equilibrium : #\tau_w = \tau_F; \qquad r_w.mg = r_F.F#

#F = (r_w/r_F).mg = (15/3)\times(5\quad kg\times9.8 ms^{-2})=245\quad N#

Explanation:

Rotational Equilibrium : A system is in rotational equilibrium if the net torque acting on it is zero. A system in rotational equilibrium will either not rotate at all or rotate at a constant angular speed.

In this case since we want the object to not fall, we must apply a torque that balances the torque generated by the weight of the hanging mass.

In this problem the radius vector is perpendicular to the applied force and so #\vec{\tau_{}} = \vec{r_{}}\times\vec{F_{}}# has a magnitude that is simple the product of the magnitude of the two forces.

[1] Torque due to the weight of the hanging mass is : #\tau_w = r_w.mg#;
#r_w# is the radius of the axle, #mg# is the weight of the hanging mass.

[2] Torque due to the applied force : #\tau_F = r_F.F#

Rotational Equilibrium : #\tau_w = \tau_F; \qquad r_w.mg = r_F.F#

#F = (r_w/r_F).mg = (15/3)\times(5\quad kg\times9.8 ms^{-2})=245\quad N#