Question

A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.

March 17, 2024

Answer

Radius of smaller piston (r₁) = 5/2 = 2.5 cm

Area of smaller piston (A₁) = π (2.5)² = 6.25 π cm²

Radius of larger piston (r₂) = 25/2 = 12.5 cm

Area of larger piston (A₂) = π (12.5)² = 156.25π cm²

Force applied on the smaller piston (F₁) = 50 kgf

{50\over 6.25 π} = {F_2\over 156.25π}

⇒ F₂ = ${50\over 6.25 π}×156.25π$ = 1250 kgf

Hence, force exerted on the larger piston = 1250 kgf.

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