Two masses m and M are attached to the string | vedclass

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(A) Let $T$ be the tension in the string between mass $m$ and mass $M$. For mass $m$ in equilibrium:$2T \sin 45^{\circ} = mg$$2T \left(\frac{1}{\sqrt{

Vedclass · Accepted Answer

$	an	heta = 1 + \frac{2M}{m}$

Vedclass · Answer

(A) Let $T$ be the tension in the string between mass $m$ and mass $M$. For mass $m$ in equilibrium:$2T \sin 45^{\circ} = mg$$2T \left(\frac{1}{\sqrt{2}}ight) = mg \implies T = \frac{mg}{\sqrt{2}}$Let $T_1$ be the tension in the string above mass $M$. Resolving forces at the junction of mass $M$:Horizontal component: $T_1 \cos 	heta = T \cos 45^{\circ} = \left(\frac{mg}{\sqrt{2}}ight) \left(\frac{1}{\sqrt{2}}ight) = \frac{mg}{2}$Vertical component: $T_1 \sin 	heta = Mg + T \sin 45^{\circ} = Mg + \left(\frac{mg}{\sqrt{2}}ight) \left(\frac{1}{\sqrt{2}}ight) = Mg + \frac{mg}{2}$Dividing the vertical component by the horizontal component:$	an 	heta = \frac{Mg + \frac{mg}{2}}{\frac{mg}{2}} = \frac{Mg}{\frac{mg}{2}} + 1 = \frac{2M}{m} + 1$Thus,$	an 	heta = 1 + \frac{2M}{m}$.

Two masses $m$ and $M$ are attached to the strings as shown in the figure. If the system is in equilibrium,then

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