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 ?
Pressure on the piston P=AF Force F=m×a =3000×9.8 =29400 N Area of cross section A=425×10−4sqm Therefore the pressure P=425×10−43000×9.8=6.92×105Pa.
