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

Answer: (A)

1

Answer 2

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