- #1

brotherbobby

- 699

- 163

- Homework Statement
- A ball of mass m and radius ##R## stands against a curb of height ##h## as shown in the figure alongside. In order to push the ball over the curb, we apply a horizontal force ##F##. Find the minimum force required to do so when the force F is applied at (a) at the center of the wheel and (b) at the top of the wheel. Argue which of the two forces is less and why.

- Relevant Equations
- ##\Sigma F = 0## (translational equilibrium) and ##\Sigma \tau = 0## about any point (rotational equilibrium)

I am able to solve the problem for both cases (a) and (b) using ##\Sigma \tau = 0##. It is only when, stranglely, that I use ##\Sigma F=0## for the ball that I run into trouble.

Let me provide the solutions.

(a) Using ##\Sigma \vec \tau = 0## about the tipping point, ##mg R\cos\theta = F_C R \sin \theta \Rightarrow F_C = \frac{mg}{\tan \theta}## (calling the force as ##F_C## when applied at the center ). From the figure ##\tan \theta = \frac{R-h}{\sqrt{R^2-(R-h)^2}} = \frac{R-h}{\sqrt{2hR-h^2}}##. Hence, we have ##\boxed{F_C = \frac{\sqrt{2hR-R^2}}{R-h} mg}##, which agrees with the answer.

(b) Calling the force ##F_T## when applied to the top of the ball and using ##\Sigma \vec \tau = 0## about the same tipping point, we get : ##F_T (2R-h) = mgR \cos\theta\Rightarrow F_T(2R-h) = mg\sqrt{2hR-h^2}\Rightarrow \boxed{F_T = \frac{\sqrt{2hR-h^2}}{2R-h}mg}##, which is correct.

Clearly, using ##\frac{F_T}{F_C} = \frac{R-h}{2R-h}<1 \Rightarrow \boxed{F_T < F_C}##. The force one needs to apply at the top is less than the force one needs to apply at the center is because the top force has a longer lever arm about the center of rotation which is the point of tipping.

All well and good.

**The confusion**:

Using ##\Sigma \vec F=0## in both horizontal and vertical directions and remembering that the normal reaction from the ground (##n_2##) is 0 at tipping point, we have : ##n_1 \sin \theta = mg## and ##n_1 \cos\theta = F_C##. (Note, this equation is valid whether the force is applied at the center or at the top.) Hence, the force applied should be the same in both cases. That common value of force turns out to be that found in (a) : ##F_C = \frac{\sqrt{2hR-R^2}}{R-h} mg##. Clearly, using ##\Sigma F = 0## is mistaken, as the force when applied on the top is less than that applied at the center, due to a larger lever arm for the former. And yet, ##\Sigma F = 0## is valid for the ball as a whole.

**What is going on?**