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 ?
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.
Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Pressure is defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
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
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.