Physics

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.

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(a) As we know, by the principle of hydraulic machine

Pressure on the smaller piston = pressure on the larger piston

Hence,

$\dfrac{F$

Given,

Diameter of the smaller piston (d₁) = 5 cm

Hence, A₁ = 𝜋 $(\dfrac{5}{2})^{2}$ = 6.25 𝜋

Diameter of the larger piston (d₂) = 25 cm

Hence, A₂ = 𝜋 $(\dfrac{25}{2})^{2}$ = 156.25 𝜋

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

Substituting the values in the formula above we get,

$\dfrac{50}{6.25 \pi} = \dfrac{F$

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

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