Question

What force is applied on a piston of area of cross section 2 cm 2 to obtain a force 150 N on the piston of area of cross section 12 cm 2 in a hydraulic machine ?

March 17, 2024

Answer

A₁ = 2 cm²

A₂ = 12 cm²

F₂ = 150 N

{F_1\over A_1 } = {F_2\over A_2}

⇒ ${F_1\over 2 } = {150\over 12}$

⇒ F₁ = ${150\over 12}×2$ = 25 N

Hence, force applied = 25 N

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