A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm^(2). What maximum pressure would the small piston have to bear?

Explanation

To solve the problem of determining the maximum pressure that the small piston of a hydraulic lift must bear, we can follow these steps:

### Step 1: Understand the relationship between force, area, and pressure.
The formula for pressure (P) is given by:
\[ P = \frac{F}{A} \]
where:
- \( P \) is the pressure,
- \( F \) is the force applied,
- \( A \) is the area over which the force is applied.

### Step 2: Calculate the force (F) acting on the piston.
The force exerted by the car can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m \) is the mass of the car (3000 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

Substituting the values:
\[ F = 3000 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29400 \, \text{N} \]

### Step 3: Convert the area from cm² to m².
The area of the piston is given as:
\[ A = 425 \, \text{cm}^2 \]
To convert this to square meters:
\[ A = 425 \, \text{cm}^2 \times \frac{1 \, \text{m}^2}{10000 \, \text{cm}^2} = 0.0425 \, \text{m}^2 \]

### Step 4: Calculate the pressure (P) using the force and area.
Now, we can substitute the values of force and area into the pressure formula:
\[ P = \frac{F}{A} = \frac{29400 \, \text{N}}{0.0425 \, \text{m}^2} \]

Calculating this gives:
\[ P = 691764.71 \, \text{Pa} \]
or approximately:
\[ P \approx 6.92 \times 10^5 \, \text{Pa} \]

### Final Answer:
The maximum pressure that the small piston would have to bear is approximately \( 6.92 \times 10^5 \, \text{Pa} \).

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