A uniform metre stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.

Explanation

To solve the problem of finding the tensions in the two strings supporting a uniform meter stick with an additional mass, we can follow these steps:

### Step 1: Identify the Forces Acting on the System
- The meter stick has a mass \( M = 200 \, \text{g} = 0.2 \, \text{kg} \).
- The small object has a mass \( m = 20 \, \text{g} = 0.02 \, \text{kg} \).
- The weight of the meter stick acts downwards at its center of mass (50 cm from the left end).
- The weight of the small object acts downwards at 70 cm from the left end.
- The tensions in the strings are \( T_1 \) (left string) and \( T_2 \) (right string).

### Step 2: Write the Equilibrium Conditions
Since the system is in equilibrium, the sum of the vertical forces must equal zero:
\[ T_1 + T_2 = Mg + mg \]
Substituting the values:
\[ T_1 + T_2 = (0.2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) + (0.02 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) \]
Calculating the weights:
\[ T_1 + T_2 = 1.96 \, \text{N} + 0.196 \, \text{N} = 2.156 \, \text{N} \quad \text{(Equation 1)} \]

### Step 3: Calculate the Torque About One End
To find the tensions, we can take moments about the left end (where \( T_1 \) acts). The sum of the torques must also equal zero:
\[ T_2 \cdot 100 \, \text{cm} = Mg \cdot 50 \, \text{cm} + mg \cdot 70 \, \text{cm} \]
Substituting the values:
\[ T_2 \cdot 100 = (0.2 \cdot 9.8 \cdot 50) + (0.02 \cdot 9.8 \cdot 70) \]
Calculating the right-hand side:
\[ T_2 \cdot 100 = (0.2 \cdot 9.8 \cdot 50) + (0.02 \cdot 9.8 \cdot 70) = 0.2 \cdot 9.8 \cdot 50 + 0.02 \cdot 9.8 \cdot 70 \]
\[ = 0.2 \cdot 490 + 0.02 \cdot 686 = 98 + 13.72 = 111.72 \, \text{N \cdot cm} \]
Now, solving for \( T_2 \):
\[ T_2 = \frac{111.72}{100} = 1.1172 \, \text{N} \approx 1.12 \, \text{N} \]

### Step 4: Substitute \( T_2 \) Back into Equation 1
Now we can substitute \( T_2 \) back into Equation 1 to find \( T_1 \):
\[ T_1 + 1.12 = 2.156 \]
Solving for \( T_1 \):
\[ T_1 = 2.156 - 1.12 = 1.036 \, \text{N} \approx 1.04 \, \text{N} \]

### Final Answer
Thus, the tensions in the two strings are:
- \( T_1 \approx 1.04 \, \text{N} \)
- \( T_2 \approx 1.12 \, \text{N} \)