A body weighing 20kg just slides down a rough inclined plane that rises 5 in 12. What is the coefficient of friction ?

Explanation

To solve the problem of finding the coefficient of friction for a body weighing 20 kg that just slides down a rough inclined plane, we can follow these steps:

### Step 1: Identify the Given Values
- Mass of the body, \( m = 20 \, \text{kg} \)
- The incline rises 5 in 12, which gives us \( \sin \theta = \frac{5}{12} \).

### Step 2: Resolve the Forces
The weight of the body can be resolved into two components:
- The component parallel to the incline: \( mg \sin \theta \)
- The component perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Write the Equations for Normal Force and Friction
1. The normal force \( R \) acting on the body is given by:
\[ R = mg \cos \theta \]

2. The frictional force \( F \) acting on the body is given by:
\[ F = mg \sin \theta \]

### Step 4: Relate Frictional Force and Normal Force
The coefficient of friction \( \mu \) can be expressed as:
\[ \mu = \frac{F}{R} \]
Substituting the expressions for \( F \) and \( R \):
\[ \mu = \frac{mg \sin \theta}{mg \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

### Step 5: Calculate \( \cos \theta \)
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{5}{12}\right)^2 = 1 - \frac{25}{144} = \frac{119}{144} \]
Thus,
\[ \cos \theta = \sqrt{\frac{119}{144}} = \frac{\sqrt{119}}{12} \]

### Step 6: Substitute Values to Find \( \mu \)
Now substituting \( \sin \theta \) and \( \cos \theta \) into the equation for \( \mu \):
\[ \mu = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{12}}{\frac{\sqrt{119}}{12}} = \frac{5}{\sqrt{119}} \]

### Step 7: Calculate the Numerical Value of \( \mu \)
To find the numerical value of \( \mu \):
\[ \mu \approx \frac{5}{\sqrt{119}} \approx 0.46 \]

### Final Answer
The coefficient of friction \( \mu \) is approximately \( 0.46 \).

---