Question

A 2 kg brick begins to slide over a surface which is inclined at an angle of \( 45^\circ \) with respect to the horizontal axis. The coefficient of static friction between their surfaces is:

• A. 1
• B. \( \frac{1}{\sqrt{3}} \)
• C. 0.5
• D. 1.7

Answer 1

The Correct Option is A

1. **Analyze the Forces:**
For a body on an inclined plane with angle \( \theta = 45^\circ \):
- The component of gravitational force parallel to the incline is \( f_L = mg \sin \theta \).
- The normal force perpendicular to the incline is \( N = mg \cos \theta \).

2. **Apply the Condition for Motion:**
Since the brick just begins to slide, the frictional force \( f_L \) is equal to the maximum static friction force, \( \mu_s N \). Thus:
\[ mg \sin 45^\circ = \mu_s mg \cos 45^\circ. \]
Simplifying, we get:
\[ \mu_s = \tan 45^\circ = 1. \]

3. **Conclusion:**
Therefore, the coefficient of static friction \( \mu_s \) is 1.

Answer: 1