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The small piston of a hydraulic lift has a crosssectional area of 2.87 cm2 and the large piston 314 cm2 . What force must be applied to the small piston for the lift to raise a load of 2.4 kN

Respuesta :

Answer:

P1 = P2      pressure is uniform

F1 / A1 = F2 / A2

F1 = F2 (A1 / A2) = 2,400 N * (2.87 / 314) = 21.9 N