A body of mass m is suspended by two strings | vedclass

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(B) For the body to be in equilibrium,the horizontal and vertical components of the tensions must balance the forces.Horizontal equilibrium: $T_1 \cos

Vedclass · Accepted Answer

$	heta_1 = 60^{\circ}, 	heta_2 = 30^{\circ}$ with $T_2 = \frac{mg}{2}$

Vedclass · Answer

(B) For the body to be in equilibrium,the horizontal and vertical components of the tensions must balance the forces.Horizontal equilibrium: $T_1 \cos 	heta_1 = T_2 \cos 	heta_2$.Given $T_1 = \sqrt{3} T_2$,we have $\sqrt{3} T_2 \cos 	heta_1 = T_2 \cos 	heta_2$,which simplifies to $\sqrt{3} \cos 	heta_1 = \cos 	heta_2$.Vertical equilibrium: $T_1 \sin 	heta_1 + T_2 \sin 	heta_2 = mg$.Substituting $T_1 = \sqrt{3} T_2$: $T_2 (\sqrt{3} \sin 	heta_1 + \sin 	heta_2) = mg$.Testing option $B$: $	heta_1 = 60^{\circ}$ and $	heta_2 = 30^{\circ}$.Check horizontal: $\sqrt{3} \cos 60^{\circ} = \sqrt{3} 	imes \frac{1}{2} = \frac{\sqrt{3}}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$. This matches.Check vertical: $T_2 (\sqrt{3} \sin 60^{\circ} + \sin 30^{\circ}) = T_2 (\sqrt{3} 	imes \frac{\sqrt{3}}{2} + \frac{1}{2}) = T_2 (\frac{3}{2} + \frac{1}{2}) = T_2 (2) = mg$.Thus,$T_2 = \frac{mg}{2}$.

$A$ body of mass $m$ is suspended by two strings making angles $\theta_1$ and $\theta_2$ with the horizontal ceiling with tensions $T_1$ and $T_2$ respectively. If $T_1 = \sqrt{3} T_2$,then the angles $\theta_1$ and $\theta_2$ are:

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