The cross-section of two piston in hydraulic press are `2cm^(2) "and" 150 cm^(2)` respectivley . Calculate the minimum force required to support a weight of `2000"kg"` wt on the broader face of the press.

Answer

To solve the problem, we need to use the principle of hydraulic systems, which states that the pressure applied to a confined fluid is transmitted undiminished throughout the fluid.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Cross-sectional area of the smaller piston, \( A_1 = 2 \, \text{cm}^2 \)
- Cross-sectional area of the larger piston, \( A_2 = 150 \, \text{cm}^2 \)
- Weight supported on the larger piston, \( W = 2000 \, \text{kg} \)

2. **Convert the Weight to Force:**
- The force due to the weight can be calculated using the formula:
\[ F_{weight} = W \cdot g \]
- Where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).
- Thus,
\[ F_{weight} = 2000 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19620 \, \text{N} \]

3. **Calculate the Pressure on the Larger Piston:**
- Pressure is defined as force per unit area:
\[ P_2 = \frac{F_{weight}}{A_2} \]
- Convert \( A_2 \) from cm² to m²:
\[ A_2 = 150 \, \text{cm}^2 = 150 \times 10^{-4} \, \text{m}^2 = 0.015 \, \text{m}^2 \]
- Now calculate the pressure:
\[ P_2 = \frac{19620 \, \text{N}}{0.015 \, \text{m}^2} = 1308000 \, \text{Pa} \]

4. **Set Up the Pressure Equation for the Smaller Piston:**
- Since the pressure is the same on both sides, we can write:
\[ P_1 = P_2 \]
- Therefore,
\[ \frac{F_1}{A_1} = P_2 \]

5. **Calculate the Minimum Force Required on the Smaller Piston:**
- Rearranging the equation gives:
\[ F_1 = P_2 \cdot A_1 \]
- Convert \( A_1 \) from cm² to m²:
\[ A_1 = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 = 0.0002 \, \text{m}^2 \]
- Now substitute the values:
\[ F_1 = 1308000 \, \text{Pa} \cdot 0.0002 \, \text{m}^2 = 261.6 \, \text{N} \]

6. **Final Answer:**
- The minimum force required to support the weight of 2000 kg on the broader face of the press is approximately:
\[ F_1 \approx 261.6 \, \text{N} \]