Question

Consider a block kept on an inclined plane (inclined at $45^{\circ}$ ) as shown in the figure If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu)$ is equal to :

• A. $0.50$
• B. $0.33$
• C. $0.25$
• D. $0.60$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the coefficient of friction \(\mu\) for a block on an inclined plane. The angle of inclination is \(45^\circ\).

First, consider the forces acting on the block:

• The gravitational force \(mg\).
• The normal force \(N = mg \cos \theta\).
• The frictional force, which acts against motion.

When the block is just about to slide down, the force required to prevent it from sliding is equal to the component of gravitational force acting down the incline minus the frictional force. The frictional force at this point is maximal, given by \(\mu N\).

The force equation for preventing the block from sliding down is:

\(F_1 = mg \sin \theta - \mu mg \cos \theta\)

When pushing the block up, the force required is greater due to the frictional force acting in the opposite direction:

\(F_2 = mg \sin \theta + \mu mg \cos \theta\)

According to the problem, \(F_2\) is twice \(F_1\):

\(F_2 = 2F_1\)

Substituting the expressions for \(F_1\) and \(F_2\) gives:

\(mg \sin \theta + \mu mg \cos \theta = 2(mg \sin \theta - \mu mg \cos \theta)\)

Simplifying by canceling \(mg\) from both sides, and substituting \(\theta = 45^\circ\):

\(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\)

The equation becomes:

\(\frac{1}{\sqrt{2}} + \mu \frac{1}{\sqrt{2}} = 2 \left( \frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}} \right)\)

Simplifying further:

\(1 + \mu = 2(1 - \mu )\)

Solving for \(\mu\):

\(1 + \mu = 2 - 2\mu\)
\(3\mu = 1\)
\(\mu = \frac{1}{3} = 0.33\)

Thus, the coefficient of friction is \(0.33\).