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A uniform rod of length 5 m is placed against the wall as shown. If coefficient of friction $\mu$ is the same for both the walls, the minimum value for $\mu$ not to slip.

\[\begin {array} {1 1} (a)\;\mu=\frac{1}{2} & \quad (b)\;\mu=\frac{1}{4} \\ (c)\;\mu=\frac{1}{3} & \quad  (d)\;\mu=\frac{1}{5} \end {array}\]

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For translation equations :
$\sum F_x =0 => N_2= f_1$
$f_1=N_2$
$\sum F_y =0 => mg=f_2+N_1$
limiting equation
$f_1= \mu N_1$
$f_2= \mu N_2$
$f_1=N_2=\mu N_1$
$\therefore f_2=\mu ^2 N_1$
For rotational equation:
Taking torques about A:
Clock wise moments= Anti clock wise moments
$mg \times \frac{l}{2} \cos \theta+ f_1 \times l \sin \theta= N_1 \times l cos \theta$
Solving above equations:
$\mu =\large\frac{1}{3}$
answered Nov 27, 2013 by meena.p
 

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