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$A$ uniform ladder of length $5m$ is placed against the wall as shown in the figure. If coefficient of friction $\mu$ is the same for both the walls, what is the minimum value of $\mu$ for it not to slip?
$\mu = \frac{1}{2}$
$\mu = \frac{1}{4}$
$\mu = \frac{1}{3}$
$\mu = \frac{1}{5}$
Solution
For translation equations$:$
$\sum F_{x}=0 \Rightarrow N_{2}=f_{1}$
$f_{1}=N_{2}$
$\sum F_{y}=0 \quad \Rightarrow m g=f_{2}+N$
limiting equation
$f_{1}=\mu N_{1}$
$f_{2}=\mu N_{2}$
$f_{1}=N_{2}=\mu N_{1}$
$\therefore \quad f_{2}=\mu^{2} N_{1}$
For rotational equation$:-$
Taking torques about $A:$
Clockwise moments $=$ Anti clockwise $-$ Clockwise moments
$m g \times \frac{l}{2} \cos \theta+f_{1} \times l \sin \theta=N_{1} \times l \cos \theta$
Solving above equation
$\mu=\frac{1}{3}$
