A 5, m long pole of 3, kg mass is placed agai | vedclass

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(A) Let the length of the pole be $L = 5\, m$ and its mass be $m = 3\, kg$. The pole makes an angle $	heta = 37^{\circ}$ with the horizontal.The pole

Vedclass · Accepted Answer

$20\, N$

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(A) Let the length of the pole be $L = 5\, m$ and its mass be $m = 3\, kg$. The pole makes an angle $	heta = 37^{\circ}$ with the horizontal.The pole is in translational and rotational equilibrium.Let $N_1$ be the normal force from the horizontal surface,$f$ be the frictional force at the base,$N_2$ be the normal force from the smooth vertical wall,and $mg$ be the weight of the pole acting at its center of mass (midpoint).For translational equilibrium:Horizontal forces: $f = N_2$Vertical forces: $N_1 = mg = 3 	imes 10 = 30\, N$For rotational equilibrium,we take the torque about the base point (let it be $A$):$	au_{net} = 0$$mg 	imes (\frac{L}{2} \cos 37^{\circ}) - N_2 	imes (L \sin 37^{\circ}) = 0$$mg 	imes \frac{5}{2} 	imes \frac{4}{5} = N_2 	imes 5 	imes \frac{3}{5}$$30 	imes 2 = N_2 	imes 3$$60 = 3 N_2 \Rightarrow N_2 = 20\, N$Since $f = N_2$,the frictional force is $f = 20\, N$.

$A$ $5\, m$ long pole of $3\, kg$ mass is placed against a smooth vertical wall as shown in the figure. Under equilibrium condition,if the pole makes an angle of $37^{\circ}$ with the horizontal,the frictional force between the pole and the horizontal surface is

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