Question

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 ms^–2]

• A. 30 N
• B. 40 N
• C. 50 N
• D. 60 N

Answer 1

The Correct Option is B

To find the force exerted by the inclined plane on the body of mass m_1, we need to calculate the normal force since the system is at rest. First, let's identify the forces acting on m_1:

• Gravitational force (m_1g) acting vertically downward.
• Normal force (N) acting perpendicular to the inclined plane.

Since the system is at rest, the net force along the plane and perpendicular to the plane must be zero. Therefore, we need to resolve the gravitational force into two components:

• Parallel to the plane: m_1g \sin \theta
• Perpendicular to the plane: m_1g \cos \theta

Because the system is at rest, the parallel component of the gravitational force is balanced by tension in the string, and the normal force balances the perpendicular component of the gravitational force.

Thus, the normal force N is given by:

N = m_1g \cos \theta

Substitute the given values m_1 = 5 \, \text{kg} and g = 10 \, \text{m/s}^2. However, we still need the angle \theta.

Since no angle \theta is given directly, and it's a typical balanced problem where m_1g \sin \theta = m_2g for balance (at rest), we can infer:

m_1g \cos \theta = 40 \, \text{N}

Hence the correct answer is 40 N.

Answer 2

m₂g = m₁gsinθ …(i)
N = m₁gcosθ …(ii)
⇒\(\frac{N}{m_2g}=cot⁡θ\)
⇒N=3×10×cot⁡θ
⇒N=3×10×\(\frac{4}{3} \)(∵sin⁡θ=\(\frac{3}{5} \))
⇒ N = 40 N
So, the correct option is (B): 40 N.