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 \( \tan 37^\circ = 3/4 \))
To determine the mass \( m \) that causes block \( M = 10 \, \text{kg} \) to accelerate downwards at \( 2 \, \text{m/s}^2 \), we will analyze the forces on each block along their respective inclined planes.
Forces on block \( M \):
Component of gravity 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 equation for block \( M \) is:
\(M g \sin 53^\circ - \mu M g \cos 53^\circ - T = M a\)
Given:
\(\mu = 0.25\) (coefficient of friction)
\(a = 2 \, \text{m/s}^2\) (acceleration)
Forces on block \( m \) on its inclined plane:
Component of gravity 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 equation for block \( m \) is:
\(T - m g \sin 37^\circ - \mu m g \cos 37^\circ = m a\)
Solving these simultaneous equations will yield \( m \).
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 \( m \):
\(T - m g \sin 37^\circ - \mu m g \cos 37^\circ = m a\) \(45 - m \times 10 \times \frac{3}{5} - 0.25 \times m \times 10 \times \frac{4}{5} = m \times 2\) \(45 - 6m - 2m = 2m\) \(45 = 10m\) \(m = 4.5 \, \text{kg}\)
