Two bodies of masses $m_1=5 \mathrm{kg}$ and $m_2=3 \mathrm{kg}$ 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 : [Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$]

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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, $m_{2}g=m_{1}g\sin \theta$

$\Rightarrow \sin \theta =\frac{m_2}{m_1}=\frac{3}{5}$ and $\cos \theta =\frac{4}{5}$

Normal force on $m_1$ is $N=m_{1}g\cos \theta =5g\cos \theta$

$=5\times 10\times \frac{4}{5}=40 \mathrm{N}$