A uniform ladder of length 5m is placed again | vedclass

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Question

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 eq

Vedclass · Accepted Answer

$\mu = \frac{1}{3}$

Vedclass · Answer

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

$f_{1}=\mu N_{1}$

$f_{2}=\mu N_{2}$

$f_{1}=N_{2}=\mu N_{1}$

$	herefore \quad f_{2}=\mu^{2} N_{1}$

For rotational equation$:-$

Taking torques about $A:$

Clockwise moments $=$ Anti clockwise $-$ Clockwise moments

$m g 	imes \frac{l}{2} \cos 	heta+f_{1} 	imes l \sin 	heta=N_{1} 	imes l \cos 	heta$

Solving above equation

$\mu=\frac{1}{3}$

$A$ 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?

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