Question

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?

Answer 1

it is given that ,
Maximum mass of a car that can be lifted, m = 3000 kg

Area of cross section of piston carrying the load , A= 452 cm²
= 0.0452 m²

Maximum force exerted by the car due to its weight (F) = mg
= 3000kg × 9.8m/s²
= 29400 kgm/s²

now maximum pressure = maximum force/area
= 29400/0.0452
= 6.504 × 10^5 N/m²

Therefore the maximum pressure on the smaller piston is 6.504 × 105Pa.

Answer 2

Hii dear,

◆ Answer-
Pmax = 6.92×10^5 Pa

◆ Explanation-
# Given-
Mmax = 3000 kg
A = 425 cm^2 = 0.0425 m^2
g = 9.8 m^2

# Solution-
Maximum pressure the piston can bear is given by-
Pmax = Fmax / A
Here, Fmax = Mmax × g
Pmax = Mmax × g / A
Pmax = 3000 × 9.8 / 0.0425
Pmax = 6.92×10^5 Pa

Maximum pressure the piston have to bear is 6.92×10^5 Pa.

Hope this is useful...