he small piston of a hydraulic lift has a cross-sectional area of 3.50 cm2, and its...
he small piston of a hydraulic lift has a cross-sectional area of 3.50 cm2, and its large piston has a cross-sectional area of 200.0 cm2 (please see the figure below). What force must be applied to the small piston for the lift to raise a load of 14.8 kN? (In service stations, this force is usually exerted by compressed air.)
Homework Answers
We know that pressure applied at one location will extend through the fluid and will have the same value everywhere.
Therefore,
F1 / A1 = F2 / A2
F1 / 3.50 = 14.8 / 200
F1 = 0.259 kN
OR
F1 = 259 N
Add Answer to:
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