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:
When a body is on an inclined plane, friction acts to oppose the direction of intended motion. To move a body "up," the applied force must overcome both the component of gravity acting down the slope and the maximum static friction acting down the slope. To "prevent sliding down," the applied force acts up the slope to support the body, while friction acts up the slope to assist the applied force.
Step 2: Key Formula or Approach:
1. Force to move up ($F_{up}$): \( F_{up} = mg(\sin\theta + \mu\cos\theta) \)
2. Force to prevent sliding ($F_{down}$): \( F_{down} = mg(\sin\theta - \mu\cos\theta) \)
3. Given condition: \( F_{up} = 3 \times F_{down} \)
Step 3: Detailed Explanation:
1. Write the equation based on the given condition: \[ mg(\sin\theta + \mu\cos\theta) = 3 \times mg(\sin\theta - \mu\cos\theta) \]
2. Cancel \(mg\) from both sides and substitute \(\theta = 45^\circ\): \[ \sin 45^\circ + \mu\cos 45^\circ = 3(\sin 45^\circ - \mu\cos 45^\circ) \]
3. Since \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\), the trigonometric terms cancel out: \[ 1 + \mu = 3(1 - \mu) \]
4. Expand and solve for \(\mu\): \[ 1 + \mu = 3 - 3\mu \] \[ \mu + 3\mu = 3 - 1 \] \[ 4\mu = 2 \implies \mu = \frac{2}{4} = 0.5 \]
Step 4: Final Answer
The coefficient of friction is 0.5.