A football of radius R is kept on a hole of radius r (r

Q: Why does the normal force at the upper edge become zero?

As the plank tilts, the line of action of the gravitational force shifts towards the lower edge. To maintain equilibrium, the lower edge must provide more normal force. At the critical angle, the entire weight is supported by the lower edge, and the football is on the verge of losing contact with the upper edge.

Q: Why do we take torque about the lower edge?

Taking torque about the lower edge is convenient because the normal force at that point passes through the axis of rotation, creating zero torque. This allows us to directly relate the gravitational force to the tipping condition without needing to calculate the exact value of the normal force.

Q: Can we solve this using forces instead of torque?

Yes, you can resolve the gravitational force into components parallel and perpendicular to the plank and balance them with the normal forces. However, using torque about the pivot point (lower edge) is much faster and more elegant.

Q: What if the hole radius $r$ is equal to the football radius $R$?

If $r = R$, the football would fall through the hole! Mathematically, $\sin\theta_{\max} = 1$, which means $\theta_{\max} = 90^\circ$. The football would be perfectly wedged and would only fall out if the plank was completely vertical.

Q: Does the mass of the football affect the maximum angle?

No, the mass $M$ cancels out in the torque equation. The tipping condition depends purely on the geometry of the setup, specifically the ratio of the hole radius to the football radius.

Question

A football of radius R is kept on a hole of radius r ($r < R$) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle $	heta$ from the horizontal as shown in the figure below. The maximum value of $	heta$ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale) -

Options:

(A) $\sin	heta = \frac{r}{R}$

(B) $	an	heta = \frac{r}{R}$

(C) $\sin	heta = \frac{r}{2R}$

(D) $\cos	heta = \frac{r}{2R}$

Sigma Percentile · Accepted Answer

The problem of the football resting in a hole on a tilted plank is a classic example of rotational equilibrium and tipping points. It beautifully combines static forces with geometric constraints. Let's break down the physics and the math behind this elegant setup.

### Analyzing the Setup

Imagine the football resting peacefully in the hole while the plank is perfectly horizontal. Gravity is pulling the football straight down from its center of mass. To counteract this, the two edges of the hole push back with normal forces, let's call them $N_1$ (at the lower edge) and $N_2$ (at the upper edge).

Because the setup is symmetric when horizontal, these two normal forces are equal. The football is in a state of perfect static equilibrium. But what happens when we start to lift one end of the plank?

### The Tipping Point

As the tilt angle $	heta$ increases, the symmetry is broken. The line of action of the gravitational force $Mg$ begins to shift towards the lower edge of the hole. Consequently, the lower edge has to support more of the football's weight, meaning $N_1$ increases. Conversely, the upper edge supports less weight, so $N_2$ decreases.

The critical moment—the tipping point—occurs when the football is just about to roll out of the hole. At this exact maximum angle, $	heta_{\max}$, the football completely loses contact with the upper edge. Mathematically, this means the normal force $N_2$ becomes exactly zero.

### The Master Equation

For the football to remain in equilibrium just before rolling, the net torque about any point must be zero. The smartest point to choose as our pivot for the torque calculation is the lower contact edge. Why? Because the normal force $N_1$ passes directly through this point, creating zero torque.

With $N_2 = 0$, the only force that could cause a torque is gravity, $Mg$. For the net torque to be zero, the line of action of $Mg$ must also pass perfectly through the lower contact edge. Geometrically, this means the line connecting the center of the football to the lower edge must be perfectly vertical.

### Final Calculation

Let's extract the geometry from this critical state. The angle between the vertical gravity line and the normal to the plank is exactly $	heta_{\max}$.

Now, consider the right-angled triangle formed by three points:
1. The center of the football.
2. The center of the hole.
3. The lower contact edge.

In this triangle, the hypotenuse is the radius of the football, $R$. The side opposite to the angle $	heta_{\max}$ is half the width of the hole, which is $r$.

Using basic trigonometry, we can write:
$$ \sin	heta_{\max} = \frac{	ext{Opposite}}{	ext{Hypotenuse}} $$

Substituting our values, we get the final, elegant condition:
$$ \sin	heta_{\max} = \frac{r}{R} $$

This simple yet profound result shows that the maximum tilt angle depends entirely on the geometric ratio of the hole's radius to the football's radius, completely independent of the football's mass!

Select Answer:

Visualized Solution

$M g, N_{1}, N_{2}$

$N_{2} \to 0$

$τ_{lower} = 0$

$θ_{m a x} = α$

$sin θ_{m a x} = \frac{r}{R}$

$r \to R ⟹ θ_{m a x} \to 9 0^{\circ}$

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Tipping Point

The Master Equation

Final Calculation

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