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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?
Solution 1
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/(425xx10^(-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.
Solution 2
Pressure on the piston due to car = `"Weight of car"/"Area of piston"`
`P = (3000xx9.8)/(425xx10^(-4)) Nm^(-2) = 6.92 xx 10^5 "Pa"`
This is also the maximum pressure that the smaller piston would have to bear.
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