For the equilibrium of a body on an inclined plane of inclination `45^(@)`, the coefficient of static friction will be

Options

A. greater than one
B. less than one
C. zero
D. less than zero

Answer

To find the coefficient of static friction (\( \mu \)) for a body in equilibrium on an inclined plane with an inclination of \( 45^\circ \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Body
On an inclined plane, the forces acting on the body include:
- The gravitational force (\( mg \)), which can be resolved into two components:
- Perpendicular to the incline: \( mg \cos(45^\circ) \)
- Parallel to the incline: \( mg \sin(45^\circ) \)
- The normal force (\( N \)), which balances the perpendicular component of the weight.
- The frictional force (\( f \)), which opposes the motion down the incline and is given by \( f = \mu N \).

### Step 2: Write the Equilibrium Conditions
For the body to be in equilibrium, the net force acting parallel to the incline must be zero. Therefore, we can set up the following equation:
\[ mg \sin(45^\circ) = f \]
Since the frictional force \( f \) is given by \( f = \mu N \), we can substitute this into the equation:
\[ mg \sin(45^\circ) = \mu N \]

### Step 3: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = mg \cos(45^\circ) \]

### Step 4: Substitute the Normal Force into the Equation
Now, we can substitute \( N \) into the equilibrium equation:
\[ mg \sin(45^\circ) = \mu (mg \cos(45^\circ)) \]

### Step 5: Simplify the Equation
We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \sin(45^\circ) = \mu \cos(45^\circ) \]

### Step 6: Solve for the Coefficient of Static Friction
Now, we can solve for \( \mu \):
\[ \mu = \frac{\sin(45^\circ)}{\cos(45^\circ)} \]
Using the values of sine and cosine for \( 45^\circ \):
\[ \sin(45^\circ) = \cos(45^\circ) = \frac{1}{\sqrt{2}} \]
Thus,
\[ \mu = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1 \]

### Conclusion
The coefficient of static friction required for equilibrium on a \( 45^\circ \) incline is \( \mu = 1 \).

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