Question

In the given arrangement of a doubly inclined plane, two blocks of masses \( M \) and \( m \) are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of \( m \), for which \( M = 10 \, \text{kg} \) will move down with an acceleration of \( 2 \, \text{m/s}^2 \), is: (take \( g = 10 \, \text{m/s}^2 \) and \( \tan 37^\circ = 3/4 \))

• A. 9 kg
• B. 4.5 kg
• C. 6.5 kg
• D. 2.25 kg

Answer 1

The Correct Option is B

To find the mass \( m \) that causes block \( M = 10 \, \text{kg} \) to move down with an acceleration of \( 2 \, \text{m/s}^2 \), we analyze the forces acting on both blocks along the inclined planes.

Consider the forces acting on block \( M \): 

• Gravitational force down the plane: \( M g \sin 53^\circ \)
• Frictional force opposing motion: \( \mu M g \cos 53^\circ \)
• Tension in the string: \( T \)

The net force on block \( M \) is given by:

\(M g \sin 53^\circ - \mu M g \cos 53^\circ - T = M a\)

Where:

• \(\mu = 0.25\) is the coefficient of friction.
• \(a = 2 \, \text{m/s}^2\) is the acceleration.

For block \( m \) on the inclined plane:

• Gravitational force down the plane: \( m g \sin 37^\circ \)
• Frictional force opposing motion: \( \mu m g \cos 37^\circ \)
• Tension in the string: \( T \)

The net force on block \( m \) is given by:

\(T - m g \sin 37^\circ - \mu m g \cos 37^\circ = m a\)

We solve these equations simultaneously to find \( m \).

Calculations:

For block \( M \):

\(M g \sin 53^\circ - \mu M g \cos 53^\circ - T = M a\)
\(10 \times 10 \times \sin 53^\circ - 0.25 \times 10 \times 10 \times \cos 53^\circ - T = 10 \times 2\)
\(100 \times \frac{4}{5} - 0.25 \times 100 \times \frac{3}{5} - T = 20\)
\(80 - 15 - T = 20\)
\(T = 45 \, \text{N}\)

For block \( m \):

\(T - m g \sin 37^\circ - \mu m g \cos 37^\circ = m a\)
\(45 - m \times 10 \times \frac{3}{5} - 0.25 \times m \times 10 \times \frac{4}{5} = m \times 2\)
\(45 - 6m - 2m = 2m\)
\(45 = 10m\)
\(m = 4.5 \, \text{kg}\)

Thus, the mass \( m \) required is 4.5 kg.

Answer 2

Considering the forces acting on block \( M \) on the inclined plane:

\[ 10g \sin 53^\circ - \mu (10g) \cos 53^\circ - T = 10 \times 2 \]

Substituting values:

\[ T = 80 - 15 - 20 = 45 \, \text{N} \]

For block \( m \) on the other inclined plane:

\[ T - mg \sin 37^\circ - \mu mg \cos 37^\circ = m \times 2 \]

Substituting values:

\[ 45 = 10m \] \[ m = 4.5 \, \text{kg} \]