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 the problem, we analyze forces on the body on the inclined plane.
Step 1: Forces Forces acting on a body on an inclined plane with friction include: 1. Normal Force (\( N \)): Perpendicular force from the plane. 2. Frictional Force (\( F_{\text{friction}} \)): Opposes motion, given by: \[ F_{\text{friction}} = \mu N \] 3. Gravitational Force (\( mg \)): Downward force due to gravity.
Step 2: Minimum Force to Start Moving (\( F_1 \)) The minimum force to push the body up the plane overcomes friction and the gravitational component parallel to the plane. - Gravitational component parallel to incline: \( mg \sin\theta \). - Frictional force opposing motion: \( \mu mg \cos\theta \). Therefore, the minimum force \( F_1 \) is: \[ F_1 = mg \sin\theta + \mu mg \cos\theta \]
Step 3: Minimum Force to Prevent Sliding (\( F_2 \)) The minimum force to prevent sliding down the incline balances the gravitational component pulling the body down. - Gravitational component parallel to incline: \( mg \sin\theta \). - Frictional force opposes sliding, with a maximum value of \( \mu mg \cos\theta \). Therefore, the minimum force \( F_2 \) is: \[ F_2 = mg \sin\theta - \mu mg \cos\theta \]
Step 4: Calculating the Ratio \( \frac{F_1}{F_2} \) Calculate the ratio \( \frac{F_1}{F_2} \): \[ \frac{F_1}{F_2} = \frac{mg \sin\theta + \mu mg \cos\theta}{mg \sin\theta - \mu mg \cos\theta} \] 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 \( \tan\theta = 2\mu \), use \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) to express \( \sin\theta \) and \( \cos\theta \) in terms of \( \mu \). From \( \tan\theta = 2\mu \): \[ \sin\theta = 2\mu \cos\theta \] Substitute 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. The correct answer is: \[ \boxed{(D) 3} \]