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(D) Let the length of the rod be $L$ and the angle it makes with the horizontal be $	heta$. The distance between the walls is $d = 2 \ m$. Thus,$L \c

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$\frac{\sqrt{17}}{2} \ m$

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(D) Let the length of the rod be $L$ and the angle it makes with the horizontal be $	heta$. The distance between the walls is $d = 2 \ m$. Thus,$L \cos 	heta = d = 2 \ m$,which implies $\cos 	heta = \frac{2}{L}$.For the rod to be in equilibrium,the net torque about the point of contact with wall $B$ must be zero. Let $N$ be the normal force from wall $A$. The forces acting on the rod are: normal force $N$ at wall $A$,normal force $N$ at wall $B$,frictional force $f = \mu N$ at wall $B$,and weight $Mg$ acting at the center of mass.Taking torque about the contact point at wall $B$:$N \cdot L \sin 	heta = Mg \cdot \frac{L}{2} \cos 	heta$Also,for vertical equilibrium: $f = Mg \Rightarrow \mu N = Mg$.Substituting $Mg = \mu N$ into the torque equation:$N \cdot L \sin 	heta = (\mu N) \cdot \frac{L}{2} \cos 	heta$$\sin 	heta = \frac{\mu}{2} \cos 	heta \Rightarrow 	an 	heta = \frac{\mu}{2}$.Given $\mu = 0.5$,we have $	an 	heta = \frac{0.5}{2} = 0.25 = \frac{1}{4}$.Since $	an 	heta = \frac{1}{4}$,we have $\cos 	heta = \frac{4}{\sqrt{1^2 + 4^2}} = \frac{4}{\sqrt{17}}$.Since $L \cos 	heta = 2$,we have $L \cdot \frac{4}{\sqrt{17}} = 2 \Rightarrow L = \frac{2\sqrt{17}}{4} = \frac{\sqrt{17}}{2} \ m$.

Two vertical walls are separated by a distance of $2 \ m$. Wall $A$ is smooth while wall $B$ is rough with a coefficient of friction $\mu = 0.5$. $A$ uniform rod is placed between them as shown. The length of the longest rod that can be placed between the walls in equilibrium is equal to

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