Question

If the minimum force required to move a body up an inclined plane of inclination 45\(^\circ\) is 3 times the force required to just prevent it sliding down, the coefficient of friction is

Hint:

For an incline with angle \(\theta\), if the ratio of force up to force down is \(k\), then \(\mu = \tan\theta \cdot \frac{k-1}{k+1}\). Here, \(\mu = \tan(45^\circ) \cdot \frac{3-1}{3+1} = 1 \cdot \frac{2}{4} = 0.5\).

Answer 1

Step 1: Understanding the Concept:
On an inclined plane, the force to move a body "up" must overcome both gravity (\( mg\sin\theta \)) and friction (\( \mu mg\cos\theta \)). The force to "prevent sliding down" only needs to bridge the gap between gravity and the maximum available friction helping the body stay up.
Step 2: Key Formula or Approach:
1. Force Up (\( F_1 \)) = \( mg(\sin\theta + \mu\cos\theta) \)
2. Force to prevent slide (\( F_2 \)) = \( mg(\sin\theta - \mu\cos\theta) \)
3. Given: \( F_1 = 3F_2 \)
Step 3: Detailed Explanation:
1. Set up the ratio based on the problem statement: \[ mg(\sin 45^\circ + \mu\cos 45^\circ) = 3 \cdot mg(\sin 45^\circ - \mu\cos 45^\circ) \]
2. Cancel \( mg \) and note that \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \): \[ \frac{1}{\sqrt{2}} + \frac{\mu}{\sqrt{2}} = 3 \left( \frac{1}{\sqrt{2}} - \frac{\mu}{\sqrt{2}} \right) \]
3. Multiply the entire equation by \( \sqrt{2} \): \[ 1 + \mu = 3(1 - \mu) \] \[ 1 + \mu = 3 - 3\mu \]
4. Group the \( \mu \) terms: \[ 4\mu = 2 \implies \mu = 0.5 \]
Step 4: Final Answer
The coefficient of friction is 0.5.