Example 15.1 The Car Lift Problem A car lift is used in a service station, compr

Question

Example 15.1 The Car Lift
Problem A car lift is used in a service station, compressed air exerts a force on a small piston of circular cross-section having a radius of 5.92 cm. This pressure is transmitted by an incompressible liquid to a second piston of radius 13.8 cm.

What if the maximum force that the compressed gas can produce is equivalent to 3.60 atm?
(a) What is the maximum weight of the car it can lift up?
W = N

(b) At this maximum output, if the vehicle needs to be lifted up 1.52 m, through what distance the small piston will be displaced?
y = m

Explanation / Answer

weight of the car=pressure*area of second piston=3.6*10^5 N/m^2 * pi*(13.8*0.01)^2 m^2=21.538 kN b.force on small piston=pressure*area=3.6*10^5*pi*(5.92*0.01)^2=3963.65 N so work done on the vehichle to lift it up 1.52 m=weight*height=32737.76 J so small piston displacement=work done/force applied on it=8.259 m

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