Question

a hydraulic lift has been constructed to lift vehicles of maximum 3000 kg weight.the area of cross-section of the piston platform on which the vehicles are lifted is 580 sq cm.calculate the maximum pressure experienced on the small piston

Answer 1

Pressure on the piston P=

A

F

Force F=m×a

=3000×9.8

=29400 N

Area of cross section A=425×10

−4

sqm

Therefore the pressure P=

425×10

−4

3000×9.8

=6.92×10

5

Pa.

Answer 2

Answer:

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 = F/A

29400 / (425 × 10-4)

= 6.917 × 105 Pa

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 105 Pa

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