In a car lift used in a service station, compressed air exerts a force on a smal
ID: 1471376 • Letter: I
Question
In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radius of r1 = 5.05 cm. This pressure is transmitted by an incompressible liquid to a second piston of radius 14.7 cm.
(a) What force must the compressed air exert on the small piston in order to lift a car weighing 13,300 N? Neglect the weights of the pistons. N
(b) What air pressure will produce a force of that magnitude? Pa
(c) Show that the work done by the input and output pistons is the same. (Do this on paper. Your instructor may ask you to turn in this work.)
A hydraulic lift has pistons with diameters 7.90 cm and 35.3 cm, respectively. If a force of 825 N is exerted at the input (smaller) piston,
what maximum mass can be lifted at the output piston? m = kg
Explanation / Answer
r1= 0.0505 m, r2= 0.147m
a) F2= 13000N
Use Pascal’s principle,
F1/A1 = F2/A2 --------------(1)
F1/r12 = F2/ r22
F1/r12 = F2/r22
F1 = F2*( r12/ r22) = 13300*(0.05052/0.1472) = 1569.6 N
b) P1=P2 = F2/A2 = F2/ r22 = 13300/ (3.14*0.1472) = 196014 = 1.96*10^5 Pa
c) If we move small piston downward by distance d1 large piston will move upward by distance d2 , such that same volume of liquid is displaced by both piston.
Thus V= A1d1=A2d2
d1 = d2*(A2/A1)------------(2)
From (1) and (2)
W = F1d1= [F2*(A1/A2)]*[ d2*(A2/A1)] = F2d2
d)
r1=0.079m, r2= 0.353m
From (1)
F2 = F1*(A2/A1)
mg = F1*(A2/A1) = F1*(r22/ r12) = F1*(r22/r12)
m = [F1*(A2/A1) = 9.8*825*(0.3532/0.0792)]/9.8 = 1680.8 kg