Question

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?

Answer 1

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`