Question

The minimum force required to start pushing a body up a rough (having coefficient of friction \( \mu \)) inclined plane is \( F_1 \), while the minimum force needed to prevent it from sliding is \( F_2 \). If the inclined plane makes an angle \( \theta \) with the horizontal such that \( \tan\theta = 2\mu \), then the ratio \( \frac{F_1}{F_2} \) is:

• A. 4
• B. 1
• C. 2
• D. 3

Hint:

When dealing with friction and inclined planes, remember the components of gravitational force acting along and perpendicular to the incline, and use the frictional force \( F_{\text{friction}} = \mu N \) to solve problems.

Answer 1

The Correct Option is D

To solve this problem, we need to consider the forces acting on the body when it is on the inclined plane.
Step 1: Forces Acting on the Body When a body is on an inclined plane with friction, the forces that act on the body include: 1. Normal Force (\( N \)): The perpendicular force from the surface of the plane. 2. Frictional Force (\( F_{\text{friction}} \)): This opposes the motion and is given by: \[ F_{\text{friction}} = \mu N \] 3. Gravitational Force (\( mg \)): The force due to gravity acting vertically downwards.
Step 2: Minimum Force to Start Moving (\( F_1 \)) The minimum force required to start pushing the body up the inclined plane is the force that overcomes both the frictional force and the component of the gravitational force parallel to the plane. - The component of the gravitational force parallel to the incline is \( mg \sin\theta \). - The frictional force opposing the motion is \( \mu mg \cos\theta \). Thus, the minimum force \( F_1 \) to move the body up the plane is: \[ F_1 = mg \sin\theta + \mu mg \cos\theta \]
Step 3: Minimum Force to Prevent Sliding (\( F_2 \)) The minimum force required to prevent the body from sliding down the incline is the force that exactly balances the component of the gravitational force pulling the body down the plane. - The component of the gravitational force parallel to the incline is \( mg \sin\theta \). - The frictional force opposes sliding, and it has a maximum value of \( \mu mg \cos\theta \). Thus, the minimum force \( F_2 \) required to prevent sliding is: \[ F_2 = mg \sin\theta - \mu mg \cos\theta \]
Step 4: Calculating the Ratio \( \frac{F_1}{F_2} \) Now, we can calculate the ratio \( \frac{F_1}{F_2} \): \[ \frac{F_1}{F_2} = \frac{mg \sin\theta + \mu mg \cos\theta}{mg \sin\theta - \mu mg \cos\theta} \] We can simplify the expression: \[ \frac{F_1}{F_2} = \frac{\sin\theta + \mu \cos\theta}{\sin\theta - \mu \cos\theta} \]
Step 5: Substituting \( \tan\theta = 2\mu \) Given that \( \tan\theta = 2\mu \), we can use the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) to express \( \sin\theta \) and \( \cos\theta \) in terms of \( \mu \). From \( \tan\theta = 2\mu \), we get: \[ \sin\theta = 2\mu \cos\theta \] Substitute this into the equation for \( \frac{F_1}{F_2} \): \[ \frac{F_1}{F_2} = \frac{2\mu \cos\theta + \mu \cos\theta}{2\mu \cos\theta - \mu \cos\theta} \] Simplifying: \[ \frac{F_1}{F_2} = \frac{3\mu \cos\theta}{\mu \cos\theta} = 3 \]
Step 6: Conclusion The ratio \( \frac{F_1}{F_2} \) is 3. Therefore, the correct answer is: \[ \boxed{(D) 3} \]