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)
A₁ = π r₁²
=π x (0.079)²
= 0.01960 m²
A₂= π r₂²
=π x (0.353)²
= 0.3914 m²
Substituting into (2)
F₂= 825 x 0.3914/ 0.01960
= 16,474.7448
≈ 16,474.74 N
= 1,647.47 kg
