# Ball on a wall

A ball is held up by a string with the string tangent to the ball. If the angle between the string and the wall is $\theta$, what is the minimum coefficient of static friction between the ball and the wall that keeps the ball from falling?

Problem Source : Introduction to Classical Mechanics

First, we identify the forces exerted on the ball :

To achieve a static state, we should have a balanced forces and balanced torque.

For balancing forces, we have

$\mu T \sin \theta + T \cos \theta \geq mg$

Noticed that we used larger or equal to instead of equal to. This is because the gravitational force may not overcome the sum of the friction force and tension force.

For balancing torques, we take the point where the ball is in contact with the wall as fulcrum.

\begin {aligned} mg(R) &= T(R+R\cos \theta) \\ mg &= T(1+\cos \theta) \end {aligned}

Substitute $mg$ into the previous equation,

\begin {aligned} T(\mu \sin \theta + \cos \theta) &\geq T(1+\cos \theta) \\ \mu &\geq \frac{1}{\sin \theta} \end {aligned}

## 3 thoughts on “Ball on a wall”

1. I’m not understanding where you got T [R+Rcos(theta)] from? We want force times distance. So force, T, times distance, R. (Would make sense if the center point were our fulcrum) AND force, T, times distance, Rcos(theta)? *lost*

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1. Hi Kaitlin,

You are right that we want force times distance. So force is T, but distance is not R. It should be the shortest (perpendicular) distance between fulcrum and the force. I used some geometry to get the distance :

Notice that the two segments add up to R + Rcos(theta). Hope this helps!

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1. Kaitlin says:

You’re awesome, thank you!

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