Question

A body weighing 125 kg just slides down a rough inclined plane that rises 1 m in every 2 m. What is the coefficient of friction?

• A. \( \frac{\pi}{8} \)
• B. \( \frac{\pi}{5} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{6} \)

Hint:

For limiting equilibrium on incline, always use \( \mu = \tan\theta \).

Answer 1

The Correct Option is D

Step 1: Condition of just sliding.
When body just slides:
\[ \mu = \tan\theta \]

Step 2: Interpret incline data.
Rise = 1 m, base = 2 m → slope angle:
\[ \tan\theta = \frac{1}{2} \]

Step 3: Find angle.
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

Step 4: Approximate angle.
\[ \theta \approx 26.5^\circ \]

Step 5: Match with options.
Closest standard angle:
\[ \frac{\pi}{6} = 30^\circ \]

Step 6: Hence coefficient.
\[ \mu \approx \tan\left(\frac{\pi}{6}\right) \]

Step 7: Final Answer.
\[ \boxed{\frac{\pi}{6}} \]