Question 90 of 210

Two bodies of masses ${m}_{1}=5kg$ and ${m}_{2}=3kg$ 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 : [Take $g=10m{s}^{-2}$]
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A$30N$B$40N$C$50N$D$60N$

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Answer: (B) $40N$

The forces acting on two masses are shown below.
Here, $N$ is normal to the incline and tension $T$ along the string up the incline. As the string and the pulley are all light and smooth, the tension in the string is uniform everywhere.
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For equilibrium condition, forces should add to zero.
Thus, equating, ${m}_{2}g={m}_{1}g\sin \theta$
$\Rightarrow \sin \theta =\dfrac{{m}_{2}}{{m}_{1}}=\dfrac{3}{5}$ and $\cos \theta =\dfrac{4}{5}$
Normal force on ${m}_{1}$ is $N={m}_{1}g\cos \theta =5g\cos \theta$
$=5\times 10\times \dfrac{4}{5}=40N$