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 \(m_1 = 5 \, \text{kg}\), we can proceed as follows:

Since the system is at rest, the net force on each body must be zero. For the body \(m_1\) on the inclined plane, the forces acting are:

• The gravitational force \(m_1 g\) acting vertically downward.
• The normal force \(N\) exerted by the inclined plane.
• The tension \(T\) in the string.

Resolve the gravitational force into components parallel and perpendicular to the incline:

Component parallel to the incline: \(m_1 g \sin \theta\)

Component perpendicular to the incline: \(m_1 g \cos \theta\)

Since the system is at rest, the parallel component is balanced by the tension \(T\), and the perpendicular component is balanced by the normal force \(N\):

1. \(T = m_1 g \sin \theta\) 
   \(N = m_1 g \cos \theta\)

For \(m_2 = 3 \, \text{kg}\), the tension is balanced by its weight, so:

1. \(T = m_2 g\)

Equating the two expressions for tension gives:

1. \(m_1 g \sin \theta = m_2 g\) 
   \(m_1 \sin \theta = m_2\)

Thus, \(\sin \theta = \frac{m_2}{m_1} = \frac{3}{5}\). Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\cos \theta\):

\(\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}\)

Substitute into the expression for \(N\):

1. \(N = m_1 g \cos \theta\) 
   \(N = 5 \times 10 \times \frac{4}{5} = 40 \, \text{N}\)

The force exerted by the inclined plane on the body of mass \(m_1\) is 40 N.