Question

What force is to be applied on a piston of area of cross section 2 m2

to obtain a force of 150N

on a piston of area of cross section 12 m2

in a hydraulic machine?​

Answer 1

Hydraulic Lift :

It is based on the principle of law of conservative of pressure, pressure at one end of the lift is equal to the pressure on another end of the lift. So, from here we can conclude that,

⇒ P1 = P2

⇒ F1/A1 = F2/A2

⇒150/2 = F2/12

⇒ 75 = F2/12

⇒ F2 = 75 * 12

⇒ F2 = 900 N

Answer 2

By the principle of hydraulic machine,

Pressure on smaller piston = pressure on wider piston.

.・. P1 = P2

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies\frac{F1}{2 \times {10}^{ - 4} } = \frac{150}{12 \times {10}^{ - 4} } \\ \\ \implies \: F1 \: = \frac{150}{12 \times {10}^{ - 4} } \times 2 \times {10}^{ - 4} \\ \\ \implies \: F1 \: = \frac{ \cancel{150} \: \: ²⁵ }{ \cancel{12 }\times \cancel{ {10}^{ - 4} }} \times \cancel{2 }\times \cancel{{10}^{ - 4} } \\ \\ \ = 25 \: N....... \: \: \: (ans)$