Question
A uniform rod of length L rests against a smooth roller as shown in figure. Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is $\theta.$
✓
Answer
Rod has a length = LIt makes an angle $\theta$ with the floor
The vertical wall has a height = h
$\text{R}_2=\text{mg}-\text{R}_1\cos\theta\ \dots(1)$
$\text{R}_1\sin\theta=\mu\text{R}_2\ \dots(2)$
$\text{R}_1\cos\theta\times\Big(\frac{\text{h}}{\tan\theta}\Big)+\text{R}_1\sin\theta\times\text{h}=\text{mg}\times\frac{1}{2}\cos\theta$
$\Rightarrow\text{R}_1\times\Big(\frac{\cos^2\theta}{\tan\theta}\Big)+\text{R}_1\sin\theta\times\text{h}=\text{mg}\times\frac{1}{2}\cos\theta$
$\Rightarrow\text{R}_1=\frac{\text{mg}\times\frac{\text{L}}{2\cos\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}}$
$\Rightarrow\text{R}_1\cos\theta=\frac{\frac{\text{mg}\text{L}}{2\cos^2\theta\sin\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}}$
$\Rightarrow\mu=\frac{\text{R}_1\sin\theta}{\text{R}_2}=\frac{\frac{\text{mgL}}{2\cos\theta\sin\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}\text{mg}-\text{mg}\frac{1}{2}\cos^2\theta}$
$\Rightarrow\mu=\frac{\frac{\text{L}}{2\cos\theta.\sin\theta}\times2\sin\theta}{2\big(\cos^2\theta\text{h}+\sin^2\theta\text{h}\big)-\text{L}\cos^2\theta\sin\theta}$
$\Rightarrow\mu=\frac{\text{L}\cos\theta\sin^2\theta}{\text{2h}-\text{L}\cos^2\theta\sin\theta}$
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