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. What force must the compressed air exert on the small piston in order to lift a car weighing 13,300 N?
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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. What force must the compressed air exert on the small piston in order to lift a car weighing 13,300 N?
(a) 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.)
(d) 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,
(e) what maximum mass can be lifted at the output piston? m = kg
Answer:
Answer:
(a) 1,569.63 N
(b) 195,933.99 Pa
(c) As pressure and volume are equal for each piston, workdone must also be equal
(d) 1,647.47 kg
Explanation:
Let the cross-sectional area (CSA) of small piston = A₁
Let the cross-sectional area (CSA) of the bigger piston = A₂
Let the Force applied at the smaller piston = F₁
Let the Force applied at the bigger piston = F₂
The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.
F₁/A₁ =F₂/A₂———————————————————– (1)
(a) F₁= F₂ xA₁ /A₂
F₂ = 13,300 N
A₁ = π r₁²
=π x (0.0505)²
= 0.008011 m²
A₂= π r₂²
=π x (0.147)²
= 0.06788 m²
Substituting into (1)
F₁ = 13,300 x 0.008011/0.06788
= 1,569.6272
≈ 1,569.63 N
(b) Air pressure = Force/Area
= F₁/A₁
= 1,569.6272/ 0.008011
= 195,933.99 Pa
(c) The pressure is constant for both pistons according to Pascal Law.
Workdone = force x distance—————————————– (2)
force = pressure × area
distance = volume/area from
Substituting into (2)
Workdone = pressure × volume.
As pressure and volume are equal for each piston, work must also be equal
(d) F₂ = F₁ x A₂/ A₁—————————————————- (3)
=π x (0.079)²
= 0.01960 m²
=π x (0.353)²
= 0.3914 m²
F₂= 825 x 0.3914/ 0.01960
= 16,474.7448
≈ 16,474.74 N
= 1,647.47 kg
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