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(C) Let the ladder have length $L = 5 \ m$. From the geometry,$\cos 	heta = \frac{3}{5}$ and $\sin 	heta = \frac{4}{5}$.For translational equilibriu

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the ladder have length $L = 5 \ m$. From the geometry,$\cos 	heta = \frac{3}{5}$ and $\sin 	heta = \frac{4}{5}$.For translational equilibrium:$\sum F_x = 0 \Rightarrow N_2 = f_1 = \mu N_1$$\sum F_y = 0 \Rightarrow N_1 + f_2 = mg$Since $f_2 = \mu N_2$,we have $N_1 + \mu(\mu N_1) = mg \Rightarrow N_1(1 + \mu^2) = mg \Rightarrow N_1 = \frac{mg}{1 + \mu^2}$.For rotational equilibrium,taking torques about the base $B$:$mg 	imes (\frac{L}{2} \cos 	heta) = N_2 	imes (L \sin 	heta) + f_2 	imes (L \cos 	heta)$$mg \frac{\cos 	heta}{2} = (\mu N_1) \sin 	heta + (\mu^2 N_1) \cos 	heta$Substituting $N_1 = \frac{mg}{1 + \mu^2}$:$\frac{mg \cos 	heta}{2} = \frac{mg}{1 + \mu^2} (\mu \sin 	heta + \mu^2 \cos 	heta)$$\frac{\cos 	heta}{2} = \frac{\mu \sin 	heta + \mu^2 \cos 	heta}{1 + \mu^2}$$(1 + \mu^2) \cos 	heta = 2\mu \sin 	heta + 2\mu^2 \cos 	heta$$\cos 	heta = 2\mu \sin 	heta + \mu^2 \cos 	heta$Dividing by $\cos 	heta$: $1 = 2\mu 	an 	heta + \mu^2$Given $	an 	heta = \frac{4}{3}$,so $1 = 2\mu(\frac{4}{3}) + \mu^2 \Rightarrow \mu^2 + \frac{8}{3}\mu - 1 = 0$.Solving the quadratic equation: $\mu = \frac{-\frac{8}{3} + \sqrt{(\frac{8}{3})^2 - 4(1)(-1)}}{2} = \frac{-\frac{8}{3} + \sqrt{\frac{64}{9} + 4}}{2} = \frac{-\frac{8}{3} + \sqrt{\frac{100}{9}}}{2} = \frac{-\frac{8}{3} + \frac{10}{3}}{2} = \frac{2/3}{2} = \frac{1}{3}$.

$A$ uniform ladder of length $5 \ m$ is placed against a wall as shown in the figure. If the coefficient of friction $\mu$ is the same for both the wall and the floor,what is the minimum value of $\mu$ for it not to slip?

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