Physics - NLM Question with Solution | TestHub

Let $M_l$ be the mass of the left stick, and $M_r$ the mass of the right stick. Since the sticks have the same mass per unit length and meet at a right angle, with the left stick reaching just to the end of the right stick, the geometry gives

$$\frac{M_l}{M_r}=\text{tan}\theta$$

Let $N$ and $F_f$ be the normal and friction forces between the sticks. $F_f$ has a maximum value of $\mu N$. Balancing the torques on the left stick (around its contact point with the ground) gives

$$N=\frac{M_{l}g}{2}\text{sin}\theta$$

Balancing the torques on the right stick (around its contact point with the ground) gives

$$F_f=\frac{M_{r}g}{2}\text{cos}\theta$$

The condition $F_f\le \mu N$ is therefore

$$M_r\text{cos}\theta \le \mu M_l\text{sin}\theta \Rightarrow {\text{tan}}^2\theta \ge \frac{1}{\mu }$$

where we have used $M_l/M_r=\text{tan}\theta$. The minimum angle is $\theta =30^o$, so

$$\mu =\frac{1}{{\text{tan}}^{2}30^o}=\frac{1}{{\left(1/\sqrt{3}\right)}^2}=3$$