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

Step 1: Identify the forces acting on the insect.

Let $m$ be the mass of the insect. The forces acting on the insect at an angle $\alpha$ with the vertical are:

• Weight $mg$ acting vertically downwards
• Normal force $N$ acting radially inwards towards the center of the hemisphere
• Friction force $f$ acting tangentially upwards along the surface

Step 2: Resolve the forces in radial and tangential directions.

Since the insect crawls very slowly, it is in quasi-static equilibrium.

$$N = mg \cos\alpha$$ $$f = mg \sin\alpha$$

Step 3: Apply the condition for no slipping.

The static friction force must satisfy the inequality $f \le \mu N$.

$$mg \sin\alpha \le \mu mg \cos\alpha$$ $$\tan\alpha \le \mu$$

Step 4: Calculate the maximum angle $\alpha$.

Substitute the given coefficient of friction $\mu = \frac{1}{3}$ to find the maximum angle.

$$\tan\alpha_{\max} = \frac{1}{3}$$ $$\cot\alpha_{\max} = 3$$

Answer: The maximum possible value of $\alpha$ is given by $\cot\alpha = 3$.