A steel wire can withstand a load up to 2940 | vedclass

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(B) Let the wire be displaced by an angle $	heta$ from the vertical. When the load passes through the equilibrium position (the lowest point),the ten

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$60$

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(B) Let the wire be displaced by an angle $	heta$ from the vertical. When the load passes through the equilibrium position (the lowest point),the tension $T$ in the wire must balance the weight $mg$ and provide the necessary centripetal force.However,the question asks for the maximum angle $	heta$ such that the wire does not break. The tension $T$ is maximum at the lowest point of the swing.At the lowest point,$T = mg + \frac{mv^2}{R}$.If we consider the energy conservation from the release point at angle $	heta$ to the lowest point,the potential energy lost equals the kinetic energy gained: $mgR(1 - \cos 	heta) = \frac{1}{2}mv^2$,so $mv^2 = 2mgR(1 - \cos 	heta)$.Substituting this into the tension equation: $T = mg + 2mg(1 - \cos 	heta) = mg(3 - 2 \cos 	heta)$.Given $T_{max} = 2940 ~N$ and $m = 150 ~kg$,$g = 9.8 ~m/s^2$,so $mg = 150 	imes 9.8 = 1470 ~N$.$2940 = 1470(3 - 2 \cos 	heta)$.$2 = 3 - 2 \cos 	heta$.$2 \cos 	heta = 1$.$\cos 	heta = 0.5$.$	heta = 60^{\circ}$.

$A$ steel wire can withstand a load up to $2940 ~N$. $A$ load of $150 ~kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position,so that the wire does not break when the load passes through the position of equilibrium,is (in $^{\circ}$)

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