An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of frictio

Q: Why do we resolve gravity instead of the normal force?

Because the motion (or tendency of motion) is along the curved surface. It is always mathematically simpler to set our coordinate axes parallel and perpendicular to the surface at the point of contact. This makes the normal force and friction align perfectly with our axes, leaving only gravity to be resolved.

Q: Why is the normal force not equal to $mg$?

The normal force only balances the forces pressing directly into the surface. Since the surface is inclined at an angle $\alpha$ to the horizontal (at the insect's position), only a component of gravity ($mg \cos \alpha$) presses into the surface. Thus, $N = mg \cos \alpha$, which is less than $mg$.

Q: What happens if the insect's mass is doubled?

Nothing changes regarding the maximum angle! As we saw in the derivation, the mass $m$ cancels out from both sides of the equation $\mu (mg \cos \alpha) = mg \sin \alpha$. The maximum angle depends entirely on the coefficient of friction, not the mass.

Q: Why did we use $\cot \alpha$ instead of $\tan \alpha$?

You can absolutely use $\tan \alpha$! The equation simplifies to $\tan \alpha = \mu$. Since $\mu = 1/3$, we get $\tan \alpha = 1/3$. The options in the question were given in terms of $\cot \alpha$, $\tan \alpha$, $\sec \alpha$, etc. Since $\cot \alpha = 1/\tan \alpha$, we get $\cot \alpha = 3$, which perfectly matches option (a).

Q: What does "crawls up very slowly" mean in physics terms?

"Very slowly" is a classic physics code phrase for "quasi-static process." It means the insect's velocity is practically zero, and more importantly, its acceleration is zero. This allows us to treat the problem as a pure statics (equilibrium) problem without worrying about centripetal acceleration or pseudo forces.

Question

An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is $\frac{1}{3}$. If the line joining the centre of the hemispherical surface to the insect makes an angle $\alpha$ with the vertical, the maximum possible value of $\alpha$ is given

Options:

(A) $\cot \alpha = 3$

(B) $	an \alpha = 3$

(C) $\sec \alpha = 3$

(D) $\operatorname{cosec} \alpha = 3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an insect, and you have decided to conquer a smooth, hemispherical bowl. As you start crawling from the bottom, the surface is relatively flat. But as you climb higher, the slope becomes steeper and steeper.

The problem states that the insect crawls "very slowly." In physics, this is a crucial hint. It means the insect is moving quasi-statically, with zero acceleration. We can treat this entirely as a problem of static equilibrium.

Let the radius of the hemisphere be $R$. The position of the insect is defined by the angle $\alpha$, which is the angle the radius vector makes with the vertical axis. Our goal is to find the maximum possible value of this angle $\alpha$ before the insect loses its grip and slides down.

### The Forces at Play

To understand why the insect eventually slips, we must analyze the forces acting on it. First, there is the force of gravity, pulling the insect straight down towards the center of the Earth. This weight is given by $mg$.

However, because the insect is on a curved surface, gravity doesn't just pull it down; it does two things simultaneously. To see this, we resolve the gravitational force into two perpendicular components.

The first component acts radially inwards, pressing the insect against the surface. By simple geometry, the angle between the vertical and the radial line is $\alpha$, so this inward pressing component is $mg \cos \alpha$.

The second component acts tangentially along the surface, trying to drag the insect back down to the bottom of the bowl. This sliding component is $mg \sin \alpha$.

### The Master Equations

Since the insect is not sinking into the bowl, the surface must push back with an equal and opposite force. This is the normal reaction, $N$. Because there is no motion in the radial direction, we can write our first equation of equilibrium:

$$ N = mg \cos \alpha $$

Now, what prevents the insect from sliding down under the influence of $mg \sin \alpha$? It is the static friction between the insect's feet and the surface. This frictional force, $f$, acts tangentially upwards.

As the insect climbs higher, $\alpha$ increases. This means the sliding force ($mg \sin \alpha$) increases, while the normal force ($mg \cos \alpha$) decreases. Consequently, the required friction increases.

At the maximum possible height, the insect is on the absolute verge of slipping. At this critical moment, the static friction reaches its maximum limit, known as limiting friction. The formula for limiting friction is:

$$ f_{	ext{max}} = \mu N $$

Equating this maximum upward friction to the downward sliding force gives us our second master equation:

$$ \mu N = mg \sin \alpha $$

### The Elegant Cancellation

We now have a system of two equations. Let's substitute the expression for the normal force from the first equation into our friction equation. This yields:

$$ \mu (mg \cos \alpha) = mg \sin \alpha $$

Look closely at this equation. The term $mg$ appears on both sides. This means we can completely cancel out the mass of the insect and the acceleration due to gravity!

This is a profound physical insight. It tells us that the maximum angle of climb is entirely independent of the insect's mass. A heavy beetle and a tiny ant, assuming they have the same coefficient of friction with the surface, will slip at the exact same height.

### Final Calculation

After canceling $mg$, our equation simplifies beautifully to:

$$ \mu \cos \alpha = \sin \alpha $$

We want to find a trigonometric ratio for $\alpha$. Let's divide both sides by $\sin \alpha$ and divide by $\mu$:

$$ \frac{\cos \alpha}{\sin \alpha} = \frac{1}{\mu} $$

We know from trigonometry that cosine divided by sine is the cotangent function. Therefore:

$$ \cot \alpha = \frac{1}{\mu} $$

The problem provides the coefficient of friction, $\mu = \frac{1}{3}$. Substituting this value into our equation, we get:

$$ \cot \alpha = \frac{1}{1/3} $$

$$ \cot \alpha = 3 $$

This perfectly matches option (a). The physics reveals that the maximum angle is purely a geometric consequence of the friction coefficient.

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Visualized Solution

$The Physical Setup$

$The Force of Gravity$

$Resolving Gravity$

$The Normal Reaction$

$Radial Equilibrium$

$The Role of Friction$

$Limiting Friction$

$Tangential Equilibrium$

$Combining the Equations$

$Mass Independence$

$Trigonometric Rearrangement$

$The Final Calculation$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Forces at Play

The Master Equations

The Elegant Cancellation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Force of Gravity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving Gravity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Normal Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radial Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Role of Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangential Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combining the Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Independence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Trigonometric Rearrangement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Final Calculation