The average mass that must be lifted by a hydraulic press is `80 "kg"`.If the radius of the larger piston is five times that of the smaller piston, what is the minimum force that must be applied?

Answer

To solve the problem, we will use the principle of hydraulic systems, which states that the pressure applied on a fluid in a closed system is transmitted undiminished throughout the fluid.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass to be lifted (m) = 80 kg
- Radius of the larger piston (R) = 5 * radius of the smaller piston (r)

2. **Calculate the Weight of the Mass**:
- The weight (W) of the mass can be calculated using the formula:
\[ W = m \cdot g \]
- Where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).
- Therefore,
\[ W = 80 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 784.8 \, \text{N} \]

3. **Determine the Areas of the Pistons**:
- The area \( A \) of a piston is given by the formula:
\[ A = \pi r^2 \]
- For the smaller piston (area \( A_1 \)):
\[ A_1 = \pi r^2 \]
- For the larger piston (area \( A_2 \)):
\[ A_2 = \pi (5r)^2 = \pi (25r^2) = 25\pi r^2 \]

4. **Apply the Hydraulic Pressure Principle**:
- According to Pascal's principle, the pressure applied on the smaller piston (P1) is equal to the pressure on the larger piston (P2):
\[ P_1 = P_2 \]
- This can be expressed as:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]
- Where \( F_1 \) is the force applied on the smaller piston and \( F_2 \) is the force exerted by the larger piston (which is equal to the weight W).

5. **Substituting the Areas**:
- Substitute \( A_1 \) and \( A_2 \) into the pressure equation:
\[ \frac{F_1}{\pi r^2} = \frac{784.8 \, \text{N}}{25\pi r^2} \]

6. **Simplifying the Equation**:
- Cancel \( \pi r^2 \) from both sides:
\[ F_1 = \frac{784.8 \, \text{N}}{25} \]
- Calculate \( F_1 \):
\[ F_1 = 31.392 \, \text{N} \]

### Final Answer:
The minimum force that must be applied to the smaller piston is approximately **31.39 N**.

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