Problem Diagram

The question:

Find the range of value of $L/D$ such that the system remains in static equilibrium. And explain why such a range exists and not just a single value.

This is the question where I encountered a confusion. I know how friction acts on plane surfaces. But how do I analyze the friction in two places of the diagram marked in red. It's especially confusing to understand in which direction does the normal force act at a corner. Help anyone? It would be great if no one posted a complete solution to the problem (I would love to solve it myself), but instead if someone could provide an approach to solve such a case.

  • $\begingroup$ look for rotational equilibrium $\endgroup$ – Aditya Garg Feb 24 at 6:57

Let the lower contact point be A and the upper point be B. Also let the CG of the rod be at its centre and the rod has only the length dimension, no breath and width.

Then let the reaction at A be $R_A$ normal to the wall (point to the right), so the friction is pointing up $\mu R_A$ since the rod tends to fall down.

At point B the reaction $R_B$ is normal to the rod and the friction $\mu R_B$ is along the rod towards the right since the rod tends to move towards to the left.

Consider the horizontal forces we have $$R_A + \mu R_Bcos\theta = R_Bsin\theta$$

Vertically we have

$$\mu R_A + R_Bcos\theta + \mu R_Bsin\theta = mg$$

Solving for $R_B$ we have

$$R_B = \frac{mg}{2\mu sin\theta + (1 - \mu ^2)cos\theta}$$

Taking moment at A we have $$R_B\frac{d}{cos\theta} = mg \frac{L}{2}cos\theta$$

$$\implies \frac{L}{d} = \frac{2}{cos^2\theta[2\mu sin\theta + (1 - \mu ^2)cos\theta]}$$

Hence $\frac{L}{d}$ depends on $\mu$ and $\theta$. Also $\theta$ cannot be 90 degrees.


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