Question

A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make the horizontal is θ.

Figure

Answer 1

Given:
Length of rod = L
It makes an angle θ with the floor.
Height of the wall is h.



The system is in translational equilibrium; therefore, we get:

$$\begin{gathered} N_2=mg-N_1\cos \theta ...\left(i\right) \\ {N}_1\sin \mathrm{\theta }=\mu N_2...\left(ii\right) \\ \mathrm{System}\mathrm{is}\mathrm{in}\mathrm{rotational}\mathrm{equilibrium};\mathrm{therefore},\mathrm{we}\mathrm{have}: \\ {N}_1\cos \theta \times \frac{h}{\tan \theta }+N_1\sin \theta \times h=mg\times \frac{L}{2}\cos \theta \\ \Rightarrow N_1=\frac{mg\times {\displaystyle \frac{L}{2}}\cos \theta }{\left({\displaystyle \frac{{\cos }^2\theta }{\sin \theta }}\right)h+h\sin \theta } \\ \Rightarrow N_1\mathrm{cos\theta }=\frac{{\displaystyle \frac{1}{2}}mgL{\cos }^2\theta }{\left\{\left({\displaystyle \frac{{\cos }^2\theta }{\sin \theta }}\right)h+h\sin \theta \right\}} \end{gathered}$$
From equation (ii), we have:
$$\begin{gathered} \mu =\frac{N_1\sin \theta }{N_2} \\ =\frac{N_1\sin \theta }{mg-N_1\cos \theta } \\ \Rightarrow \mu =\frac{L/2\cos \theta \sin \theta \times 2\sin \theta }{2\left({\cos }^2\theta +{\sin }^2\theta \right)h-L{\cos }^2\theta \sin \theta } \\ \Rightarrow \mu =\frac{L\cos \theta {\sin }^2\theta }{2h-L{\cos }^2\theta \sin \theta } \end{gathered}$$