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 cm2 . What maximum pressure would the smaller piston have to bear ?
| Parameter | Value |
|---|---|
| Maximum car mass | 3000 kg |
| Piston area (\(A\)) | 425 cm² = \(4.25 \times 10^{-3}\) m² |
| \(g\) | 9.8 m/s² (or 10 m/s² approx) |
Large Piston: Car (3000 kg)
↓ Pressure transmitted equally ↓
Small Piston: Maximum pressure?
Hydraulic lift: pressure uniform throughout incompressible fluid.
$$P = P_1 = P_2 = \frac{F_1}{A_1} = \frac{F_2}{A_2}$$
Load piston pressure:
$$P = \frac{F_\text{load}}{A_\text{load}} = \frac{mg}{A}$$
$$F = mg = 3000 \times 9.8 = 29{,}400 \, \text{N}$$ $$P = \frac{29{,}400}{4.25 \times 10^{-3}} = 6.92 \times 10^6 \, \text{Pa}$$
$$P = 6.92 \, \text{MPa} \quad (\text{or } 69.2 \, \text{bar approx})$$
\(P_\text{max} = \textbf{6.92 × 10⁶ Pa}\)
Smaller piston bears same pressure regardless of area. Force amplification comes from larger area ratio: \(F_\text{small} = P \times A_\text{small}\).