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 and variance for the following frequency distribution.
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
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.