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?

Mar 4, 2016

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

$F = \left({r}_{w} / {r}_{F}\right) . m g = \left(\frac{15}{3}\right) \setminus \times \left(5 \setminus \quad k g \setminus \times 9.8 m {s}^{- 2}\right) = 245 \setminus \quad N$

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 $\setminus \vec{\setminus {\tau}_{}} = \setminus \vec{{r}_{}} \setminus \times \setminus \vec{{F}_{}}$ 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 : $\setminus {\tau}_{w} = {r}_{w} . m g$;
${r}_{w}$ is the radius of the axle, $m g$ is the weight of the hanging mass.

[2] Torque due to the applied force : $\setminus {\tau}_{F} = {r}_{F} . F$

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

$F = \left({r}_{w} / {r}_{F}\right) . m g = \left(\frac{15}{3}\right) \setminus \times \left(5 \setminus \quad k g \setminus \times 9.8 m {s}^{- 2}\right) = 245 \setminus \quad N$