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 diameter of the piston are 5 cm and 25 cm.

Answer 1

Answer:

1250kgf.

Explanation:

F₁ = force applied on smaller piston.

F₂ = force applied on larger piston.

According to PASCAL'S LAW, Applied force will be equal to resultant force, which gets distributed in several directions.

So,

    P₁ (pressure at smaller one) = P₂...................(1)

Pressure = Force/area

P₁ = F₁/A₁             and           P₂ = F₂/A₂

                     50/π(25) = F₂/π(625)    [From (1)]

                       $\frac{50}{25} = \frac{F2}{625}\\\\\frac{50}{F2} = \frac{25}{625}\\\\\frac{50}{F2} = \frac{1}{25}\\\\F2 = 50 . 25\\ F2 = 1250.$

So, the force is 1250kgf.

Answer 2

Ratio of diameter of smaller piston to bigger piston = 5:25

.•. Ratio of area of smaller piston to bigger piston = 25:625

Force applied on smaller piston, F1 = 50kgf

Let F2 be the force on the bigger piston.

By the principle of hydraulic machine,

Pressure on narrow piston = Pressure on wider piston

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies \: \frac{F1}{F2} = \frac{A1}{A2} \\ \\ \implies \: \frac{50}{F2} = \frac{25}{625} \\ \\ \implies \: F2 \: = \: \cancel{50} \: \: ²\times \frac{625}{ \cancel{25} } \\ \\ \ = 1250 \: kgf \: \: \: \: \: (ans)$