A uniform rod of mass 3 kg and length 1 m is suspended from a fixed point by means of two strings of lengths 0.6m and 0.8 which are attached to the free ends of the rod. Find the tensions in the strings.

Answer

To solve the problem of finding the tensions in the strings supporting a uniform rod, we can follow these steps:

### Step 1: Understand the Setup
We have a uniform rod of mass \( m = 3 \, \text{kg} \) and length \( L = 1 \, \text{m} \), suspended by two strings of lengths \( L_1 = 0.6 \, \text{m} \) and \( L_2 = 0.8 \, \text{m} \). The rod is in equilibrium, so we need to balance the forces acting on it.

### Step 2: Determine the Weight of the Rod
The weight \( W \) of the rod can be calculated using the formula:
\[ W = mg \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)). Thus,
\[ W = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \]

### Step 3: Analyze the Geometry
The two strings create a right triangle with the rod. The lengths of the strings \( L_1 \) and \( L_2 \) can be used to find the angles. The vertical distance from the point of suspension to the center of the rod can be calculated using the Pythagorean theorem:
\[ L_1^2 + L_2^2 = (1 \, \text{m})^2 \]
This confirms that the triangle formed is a right triangle.

### Step 4: Set Up the Equations
Since the rod is in equilibrium, we can set up two equations based on the balance of vertical and horizontal forces.

1. **Vertical Forces**:
\[ T_1 \sin(\theta_1) + T_2 \sin(\theta_2) = W \]
where \( \sin(\theta_1) = \frac{L_1}{L} = \frac{0.6}{1} = 0.6 \) and \( \sin(\theta_2) = \frac{L_2}{L} = \frac{0.8}{1} = 0.8 \).

2. **Horizontal Forces**:
\[ T_1 \cos(\theta_1) = T_2 \cos(\theta_2) \]
where \( \cos(\theta_1) = \sqrt{1 - \sin^2(\theta_1)} \) and \( \cos(\theta_2) = \sqrt{1 - \sin^2(\theta_2)} \).

### Step 5: Solve the Equations
From the horizontal forces, we can express \( T_1 \) in terms of \( T_2 \):
\[ T_1 = \frac{T_2 \cos(\theta_2)}{\cos(\theta_1)} \]

Substituting \( \cos(\theta_1) \) and \( \cos(\theta_2) \):
\[ T_1 = \frac{T_2 \cdot 0.6}{0.8} \]

Now substitute this expression for \( T_1 \) into the vertical forces equation:
\[ \frac{T_2 \cdot 0.6}{0.8} \cdot 0.6 + T_2 \cdot 0.8 = 30 \]

### Step 6: Simplify and Solve for \( T_2 \)
Combining terms:
\[ \frac{0.36}{0.8} T_2 + 0.8 T_2 = 30 \]
\[ \left(\frac{0.36 + 0.64}{0.8}\right) T_2 = 30 \]
\[ T_2 = \frac{30 \cdot 0.8}{1} = 24 \, \text{N} \]

### Step 7: Find \( T_1 \)
Now substitute \( T_2 \) back into the equation for \( T_1 \):
\[ T_1 = \frac{24 \cdot 0.6}{0.8} = 18 \, \text{N} \]

### Final Answer
The tensions in the strings are:
- \( T_1 = 18 \, \text{N} \)
- \( T_2 = 24 \, \text{N} \)