A ladder rests against a frictionless vertica | vedclass

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Question

(A) Let the length of the ladder be $L$. From the geometry of the right-angled triangle formed by the ladder,the wall,and the ground,$L = \sqrt{4^2 +

Vedclass · Accepted Answer

$111$

Vedclass · Answer

(A) Let the length of the ladder be $L$. From the geometry of the right-angled triangle formed by the ladder,the wall,and the ground,$L = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \, m$.Let $	heta$ be the angle the ladder makes with the ground. Then $\cos 	heta = \frac{4}{L} = \frac{4}{\sqrt{52}}$ and $\sin 	heta = \frac{6}{L} = \frac{6}{\sqrt{52}}$.Let $N_1$ be the normal reaction from the ground,$f$ be the friction from the ground,and $N_2$ be the reaction from the frictionless wall.For vertical equilibrium: $N_1 = 500 \, N$.For horizontal equilibrium: $f = N_2$.Taking torque about the lower end (point $A$): $\sum 	au_A = 0$.The weight acts at $L/3$ from the lower end. The torque due to weight is $500 	imes (L/3) \cos 	heta$.The torque due to the wall reaction $N_2$ is $N_2 	imes (6) = N_2 	imes (L \sin 	heta)$.Equating torques: $N_2 (L \sin 	heta) = 500 (L/3) \cos 	heta$.$N_2 = \frac{500}{3} \frac{\cos 	heta}{\sin 	heta} = \frac{500}{3} 	imes \frac{4/L}{6/L} = \frac{500}{3} 	imes \frac{4}{6} = \frac{2000}{18} = \frac{1000}{9} \approx 111.11 \, N$.

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