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 area of the piston are 5 cm2 and 25 cm2.ans step by step

Answer 1

Answer:

in a hydraulic machine

132 *

force on larger piston/force on smaller piston=area of larger piston/

area of smaller piston respectively F=5*5*50=25*50=1250kgf

let force on larger and smaller piston be F and f respectively

diameter of larger and smaller piston are 25cm and 5cm

F/f=R^2/r^2=R^2/r^2 F/50=25*25/5*5

2092000 ·

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

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