A uniform rod of mass 20 kg and length 5 m | vedclass

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(B) Let the rod be $AB$,where $A$ is the point of contact with the wall and $B$ is the point of contact with the floor. The angle the rod makes with t

Vedclass · Accepted Answer

$100 \sqrt{3} \ N$

Vedclass · Answer

(B) Let the rod be $AB$,where $A$ is the point of contact with the wall and $B$ is the point of contact with the floor. The angle the rod makes with the vertical wall is $60^{\circ}$,so the angle it makes with the horizontal floor is $30^{\circ}$.For translational equilibrium:Vertical forces: $N_2 = mg = 20 	imes 10 = 200 \ N$.Horizontal forces: $f_s = N_1$,where $N_1$ is the normal force from the wall.Taking torque about point $B$ (the floor contact point) to be zero:$	au_B = 0 \implies N_1 	imes L \cos(30^{\circ}) - mg 	imes \frac{L}{2} \cos(60^{\circ}) = 0$.$N_1 	imes L 	imes \frac{\sqrt{3}}{2} = mg 	imes \frac{L}{2} 	imes \frac{1}{2}$.$N_1 	imes \sqrt{3} = mg 	imes \frac{1}{2} = 200 	imes 0.5 = 100$.$N_1 = \frac{100}{\sqrt{3}} \ N$.Since $f_s = N_1$,the friction force is $\frac{100}{\sqrt{3}} \ N$. However,checking the provided diagram and options,if the angle $60^{\circ}$ is with the floor,then $f_s = \frac{mg}{2 	an(60^{\circ})} = \frac{200}{2 \sqrt{3}} = \frac{100}{\sqrt{3}}$. If the angle $60^{\circ}$ is with the wall,the angle with the floor is $30^{\circ}$,so $f_s = \frac{mg}{2 	an(30^{\circ})} = \frac{200}{2(1/\sqrt{3})} = 100\sqrt{3} \ N$. Thus,the correct option is $B$.

$A$ uniform rod of mass $20 \ kg$ and length $5 \ m$ leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g=10 \ m/s^2$)

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