STATICS – Pushing a ball over a curb

[brotherbobby]

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?

 

[Orodruin]

(Note, this equation is valid whether the force is applied at the center or at the top.)

This is incorrect. The reaction force at the contact point will be in different directions depending on where you apply the force. Note that it consists of both a normal force and a frictional force. If there is not sufficient friction, the ball will slip.

 

[brotherbobby]

This is incorrect. The reaction force at the contact point will be in different directions depending on where you apply the force. Note that it consists of both a normal force and a frictional force. If there is not sufficient friction, the ball will slip.

Thank you for your quick reply. However, I have a couple of questions regarding the answer you gave.

1. I find that not only the value of the reaction force ($n_1$ in my diagram above) will be different depending where I apply the external force but also its direction. This comes out from the equation $n_1 \sin\theta = mg$ $-$ since $mg$ remains the same, a change in $n_1$ has to be accompanied by a change in $\theta$. But the surface in question is just a point. I am baffled as to how can a point apply a frictional force in the first place. And then, how can that frictional force change in direction. The force of friction has to be tangent to the surface of the ball above - meaning "along the surface" of the two bodies. Won't that constrain the normal reaction $n_1$ to be perpendicular to the tangent, along the normal?
2. Assuming that the frictional force is both value and direction dependent, am I right in the direction of the normal force $n_1$ above for the case (a)? Unaware back then that the tipping point would exert a frictional force on the ball, I drew the reaction normal to the surface of the ball ($\perp$ to tangent). If the external force $F_C$ acts at the centre, there would be no external torque about the center of the ball. Would a frictional force still come into effect, trying to offset slipping?

 

[Orodruin]

Thank you for your quick reply. However, I have a couple of questions regarding the answer you gave.

I find that not only the value of the reaction force ($n_1$ in my diagram above) will be different depending where I apply the external force but also its direction. This comes out from the equation $n_1 \sin\theta = mg$ $-$ since $mg$ remains the same, a change in $n_1$ has to be accompanied by a change in $\theta$. But the surface in question is just a point. I am baffled as to how can a point apply a frictional force in the first place. And then, how can that frictional force change in direction. The force of friction has to be tangent to the surface of the ball above - meaning "along the surface" of the two bodies. Won't that constrain the normal reaction $n_1$ to be perpendicular to the tangent, along the normal?

The angle $\theta$ is fixed by the problem. You even have an expression for it. The frictional force does not change direction, it is tangential to the ball. However, the normal + friction force will have to be what it has to be to ensure force equilibrium. That is why you can solve the problem through torque consideration only. Both of these forces act through the contact point and therefore give zero torque.

 

[brotherbobby]

Thank you for your response. It is clear that my approach has been incorrect, though the situation is far from trivial. I'd want to solve a similar problem here from Arfken's book. Please let me know if it's ok, in light of the above discussion.

Statement of the problem : A wheel of mass $m$ has a massless rope attached to it at the top as shown in the figure. The wheel rests against a curb whose height is half the radius of the wheel. Calculate the tension $T$ that will pull the barrel up the sidewalk. Additionally, calculate the normal force at the curb and the friction.

Let me draw the diagram with the forces marked.

Solution : Taking the curb as the pivot, $\Sigma \tau = 0\Rightarrow mgR\cos\theta = T\times \frac{3R}{2}\Rightarrow T = \frac{2}{3}mg \cos\theta = \frac{mg}{\sqrt{3}}$, as it can be shown from geometry that $\theta = 30^{\circ}$. Hence the tension in the rope $\boxed{T = \frac{mg}{\sqrt{3}} }$.

Using $\Sigma \vec F = 0 \Rightarrow mg+f\cos\theta = n \sin \theta\; \text{and}\; T = n \cos \theta + f \sin \theta$. Putting $\theta = 30^{\circ}$ and doing the algebra we get, $\boxed{n = mg}$ and $\boxed{f = -\frac{mg}{\sqrt{3}}}$.

Hence the direction of friction is opposite to the one I took. That makes sense on hindsight. If the wheel is to go forward, it has to roll "back" against the curb, making the friction acting forward.

 

[Orodruin]

It seems to me that this is exactly the same problem as your original problem when the rope is attached to the top of the ball. The only difference being that your original $h$ is fixed to $R/2$.

 

[brotherbobby]

Yes. The earlier problem does not ask for the normal force and the friction.

Am I correct in finding the two quantities?