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 1

Road has a length $=L$
It makes an angle $\theta$ with the floor
The verticle wall has a height $=h$
$R_2=mg-R_1\cos \theta .....\left(1\right)$
$R_1\sin \theta =\mu R_2....\left(2\right)$
$R_1\cos \theta \times \left(h/\tan \theta \right)+R_1\sin \theta \times h=mg\times 1/2\cos \theta$
$\Rightarrow R_1\left({\cos }^2\theta /\sin \theta \right)h+R_1\sin \theta h=mg\times 1/2\cos \theta$
$\Rightarrow R_1=\frac{mg\times L/2\cos \theta }{\left\{\right({\cos }^2\theta /\sin \theta )h+\sin \theta h\}}$
$\Rightarrow R_1\cos \theta =\frac{mgL/2{\cos }^2\theta \sin \theta }{\left\{\right({\cos }^2\theta /\sin \theta )h+\sin \theta h\}}$
$\Rightarrow \mu =R_1\sin \theta /R_2=\frac{mgL/2\cos \theta .\sin \theta }{\left\{\right({\cos }^2\theta /\sin \theta )mg-mg1/2{\cos }^2\theta \}}$
$\Rightarrow \mu =\frac{L/2\cos \theta .\sin \theta \times 2\sin \theta }{2({\cos }^2\theta h+{\sin }^2\theta h)-L{\cos }^2\theta \sin \theta }$
$\Rightarrow \mu =\frac{L\cos \theta {\sin }^2\theta }{2h-L{\cos }^2\theta \sin \theta }$