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

Answer 2

Step 1: Write down the given data
Mass of the brick, \( m = 2\,\text{kg} \)
Inclination angle, \( \theta = 45^\circ \)
The brick just begins to slide down the inclined plane. This means the limiting condition of static friction is reached, where the frictional force is at its maximum possible value.

Step 2: Analyze the forces acting on the brick
On an inclined surface, the forces acting on the brick are:
1) Weight \( mg \) acting vertically downward.
2) Normal reaction \( N \) acting perpendicular to the inclined plane.
3) Static friction \( f_s \) acting up the plane, opposing motion.

The components of weight are:
- Along the incline: \( mg \sin\theta \)
- Perpendicular to the incline: \( mg \cos\theta \)

Step 3: Condition for impending motion
At the instant the block is about to slide, static friction reaches its maximum value \( f_s = \mu_s N \).
In this limiting case, the component of weight down the plane just equals the maximum static friction:
\[ mg \sin\theta = \mu_s N \] But \( N = mg \cos\theta \). Substituting this in gives:
\[ mg \sin\theta = \mu_s (mg \cos\theta) \] Simplifying, we find:
\[ \mu_s = \tan\theta \]

Step 4: Substitute the given angle
Given \( \theta = 45^\circ \), we have:
\[ \mu_s = \tan 45^\circ = 1 \]

Step 5: Interpret the result
This means that the surface must have a coefficient of static friction equal to 1 for the brick to just start sliding at an incline of \( 45^\circ \). For angles smaller than \( 45^\circ \), the brick would remain stationary; for greater angles, it would slide freely.

Final Answer:

1