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 ?

Updated On: Nov 3, 2023
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Solution and Explanation

The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load - carrying piston, A = 425 cm2 = 425 × 10 - 4 m2
The maximum force exerted by the load, F = mg
= 3000 × 9.8 

= 29400 N
The maximum pressure exerted on the load - carrying piston, \(P =\frac{ F }{A} \)

\(= \frac{29400 }{ 425 × 10 ^{- 4}} \)

= 6.917 × 10 5 Pa
Pressure is transmitted equally in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 10 5 Pa.

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Concepts Used:

Pressure

Pressure is defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

Everyday examples of pressure are:

  • The working of the vacuum cleaner is an example of pressure. The fan inside the vacuum creates a low-pressure region which makes it easy to suck the dust particles inside the vacuum.
  • Using a knife for cutting is another example of pressure. The area exposed from the knife is small but the pressure is high enough to cut the vegetables and fruits.

Formula:

When a force of ‘F’ Newton is applied perpendicularly to a surface area ‘A’, then the pressure exerted on the surface by the force is equal to the ratio of F to A. The formula for pressure (P) is:

P = F / A

Units of Pressure:

The SI unit of pressure is the pascal (Pa)

A pascal can be defined as a force of one newton applied over a surface area of a one-meter square.