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 : [Take g = 10 m s - 2 ]

Question

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 : [Take g = 10 m s - 2 ] ​​​​​​​

Aicharya · Accepted Answer

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.

For equilibrium condition, forces should add to zero.
Thus, equating, $m_{2} g = m_{1} g \sin θ$
$\Rightarrow \sin θ = \frac{m_{2}}{m_{1}} = \frac{3}{5}$ and $\cos θ = \frac{4}{5}$
Normal force on $m_{1}$ is $N = m_{1} g \cos θ = 5 g \cos θ$
$= 5 \times 10 \times \frac{4}{5} = 40 N$

Two bodies of masses m 1 = 5 kg and m 2 = 3 kg are connected by a light string going over a smooth light…

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