Question

A uniform rod of length $L$ rests against a smooth wall 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 $\theta$.

• A. $\frac{L{\cos }^2\theta \sin \theta }{2h-l\cos \theta {\sin }^2\theta }$
• B. $\frac{L\cos \theta {\sin }^2\theta }{2h-l\cos \theta {\sin }^2\theta }$
• C. $\frac{L{\cos }^2\theta \sin \theta }{2h-l{\cos }^2\theta \sin \theta }$
• D. $\frac{L\cos \theta {\sin }^2\theta }{2h-l{\cos }^2\theta \sin \theta }$

Answer 1

The correct option is D $\frac{L\cos \theta {\sin }^2\theta }{2h-l{\cos }^2\theta \sin \theta }$

Figure(b) gives the forces .

$R_2=$Reactional force on rod by ground

$R_1=$ Reactional force on rod by wall

$f=$ frictional force $=\mu R_2$

$\theta$ is the maximum angle the rod can make with horizontal.

Equating the forces on verticle direction we get

$R_2=mg-R_{1}cos\theta --------\left(1\right)$

Equating the forces on horizontal direction we get

$R_{1}sin\theta =f=\mu R_2$

Now the torque about point B should be balanced

$R_{1}cos\theta \times AB+R_{1}Sin\theta \times h=mg\frac{l}{2}cos\theta$

$⟹R_{1}h\left[\frac{co{s}^2\theta +si{n}^2\theta }{sin\theta }\right]=\frac{mgl}{2}cos\theta$

$⟹R_1=\frac{mgl}{2h}cos\theta sin\theta$

$⟹R_{1}cos\theta =\frac{mgl}{2h}co{s}^2\theta sin\theta$

So, $R_2=mg-\frac{mgl}{2h}co{s}^2\theta sin\theta =\frac{mg}{2h}(2h-lco{s}^2\theta sin\theta )$

And $\mu =\frac{R_{1}sin\theta }{R_2}=\frac{\frac{mgh}{2h}cos\theta si{n}^2\theta }{\frac{mg}{2h}(2h-lco{s}^2\theta si{n}^2\theta )}$

$\mu =\frac{lcos\theta si{n}^2\theta }{2h-lco{s}^2\theta sin\theta }$