Question

Two uniform rods tied together with the help of a string are balanced as shown in the figure. The minimum coefficient of friction for which the system will remain in equilibrium in the position is $\frac{2}{x}$. The value of ${}^{\prime }{x}^{\prime }$ is

Answer 1

First balance the force in horizontal and vertical direction:
T = μNμN .... (i)
mg = N ......(ii)

From (i) and (ii) we get:

$T=\mu mg$

For the rods to be in rest the torque must be zero.

Taking torque about point 'P'.

$\tau =0\tau =mg\frac{l}{2}cos{37}^0-Tlsin{37}^0=0\Rightarrow mg\frac{l}{2}cos{37}^0=Tlsin{37}^0$

By putting value of T:

$T=\mu mg$

We get:

$mg\frac{l}{2}cos{37}^0=\mu mglsin{37}^0\frac{1}{2}cos{37}^0=\mu sin{37}^0\Rightarrow \mu =\frac{2}{3}$

$\mu =\frac{2}{3}=\frac{2}{x}$

Therefore, x=3