A block is in limiting equilibrium on a rough horizontal surface. If the net contact force is `sqrt(3)` times the normal force, the coefficient of static friction is

Options

A. `sqrt(2)`
B. `1/sqrt(2)`
C. 0.5
D. `1/sqrt(3)`

Answer

To solve the problem, we need to find the coefficient of static friction (μ) given that the net contact force is \(\sqrt{3}\) times the normal force (N).

### Step-by-step Solution:

1. **Understanding the Forces**:
- The block is in limiting equilibrium on a rough horizontal surface.
- The normal force (N) acts perpendicular to the surface, and the frictional force (F) acts parallel to the surface.

2. **Expression for Frictional Force**:
- The frictional force can be expressed as:
\[ F = \mu N \]
where μ is the coefficient of static friction.

3. **Net Contact Force**:
- The net contact force (R) is the resultant of the normal force and the frictional force. Since the friction is acting horizontally and the normal force is vertical, we can use the Pythagorean theorem:
\[ R = \sqrt{N^2 + F^2} \]
Substituting the expression for F:
\[ R = \sqrt{N^2 + (\mu N)^2} = \sqrt{N^2(1 + \mu^2)} = N\sqrt{1 + \mu^2} \]

4. **Setting Up the Equation**:
- According to the problem, the net contact force is given as:
\[ R = \sqrt{3} N \]
- Equating the two expressions for R:
\[ N\sqrt{1 + \mu^2} = \sqrt{3} N \]

5. **Cancelling N**:
- Since N is not zero, we can divide both sides by N:
\[ \sqrt{1 + \mu^2} = \sqrt{3} \]

6. **Squaring Both Sides**:
- To eliminate the square root, we square both sides:
\[ 1 + \mu^2 = 3 \]

7. **Solving for μ**:
- Rearranging the equation gives:
\[ \mu^2 = 3 - 1 = 2 \]
- Taking the square root of both sides:
\[ \mu = \sqrt{2} \]

### Final Answer:
The coefficient of static friction (μ) is \(\sqrt{2}\).

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