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 Question:
We need to compare the force required to push an object up an incline versus the force required to keep it from sliding down under the influence of friction.

Step 2: Key Formula or Approach:
1. Force to move up: \(F_1 = mg(\sin\theta + \mu\cos\theta)\).
2. Force to prevent sliding down: \(F_2 = mg(\sin\theta - \mu\cos\theta)\).

Step 3: Detailed Explanation:
Given: \(\theta = 45^\circ\) and \(F_1 = 3 F_2\).
\[ mg(\sin 45^\circ + \mu \cos 45^\circ) = 3 mg(\sin 45^\circ - \mu \cos 45^\circ) \] Since \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\), we can cancel these terms:
\[ \frac{1}{\sqrt{2}} (1 + \mu) = 3 \cdot \frac{1}{\sqrt{2}} (1 - \mu) \] \[ 1 + \mu = 3(1 - \mu) \] \[ 1 + \mu = 3 - 3\mu \] \[ 4\mu = 2 \] \[ \mu = \frac{2}{4} = 0.5 \]

Step 4: Final Answer:
The coefficient of friction is 0.5.