Two bodies of masses m₁=5 kg and m₂=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 m₁ will be [Take g=10 m s⁻²]

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(2) 40 N. Idea: the system being at rest fixes the angle, and the angle then fixes the normal force. The plane is smooth, so the force it exerts is purely normal. Along the incline, at rest, the tension equals the hanging weight and balances m₁'s component: m₁g sin θ=m₂g sin θ=(m₂)/(m₁)=3/5, so cos θ=4/5. Perpendicular to the incline: N=m₁g cos θ=5×10×4/5=40 N. The 3-4-5 triangle hidden in the mass ratio is what makes the answer come out whole. Note the question asks for the force from the plane, not the tension: on a smooth surface that is the normal force alone.

Two bodies of masses m₁=5 kg and m₂=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 m₁ will be [Take g=10 m s⁻²]

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