Two bodies of masses m_{1}=5,kg and m_{2}=3,k | vedclass

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(B) For the system to be at rest (equilibrium),the tension $T$ in the string must balance the weight of $m_{2}$ and the component of weight of $m_{1}$

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) For the system to be at rest (equilibrium),the tension $T$ in the string must balance the weight of $m_{2}$ and the component of weight of $m_{1}$ along the incline.$T = m_{2}g$$T = m_{1}g \sin 	heta$Equating the two,we get $m_{2}g = m_{1}g \sin 	heta$,which implies $\sin 	heta = \frac{m_{2}}{m_{1}} = \frac{3}{5}$.Since $\sin 	heta = \frac{3}{5}$,we have $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.The force exerted by the inclined plane on the body of mass $m_{1}$ is the normal force $N$,which balances the component of weight perpendicular to the incline:$N = m_{1}g \cos 	heta$Substituting the values: $N = 5 	imes 10 	imes \frac{4}{5} = 40\,N$.

Two bodies of masses $m_{1}=5\,kg$ and $m_{2}=3\,kg$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass $m_{1}$ will be $....N$ [Take $g=10\,ms^{-2}$]

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