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What force is applied on a piston of area of cross section 2 cm 2 to obtain a force 150 N on the piston of area of cross section 12 cm 2 in a hydraulic machine ?

Fluids Pressure

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Answer

As we know, by the principle of hydraulic machine

Pressure on piston A = pressure on piston B

Hence,

F1A1=F2A2\dfrac{F{1}}{A{1}} = \dfrac{F{2}}{A{2}} \\[0.5em]

Given,

A1 = 2 cm 2

Converting cm 2 into m 2

100 cm = 1 m

So, 100 cm x 100 cm = 1 m 2

Hence, 2 cm 2 = 1×210000\dfrac{1 \times 2}{10000}

Therefore, A1 = 2 x 10-4 m 2

A2 = 12 cm 2

Hence, 12 cm 2 = 1×1210000\dfrac{1 \times 12}{10000}

Therefore, A2 = 12 x 10-4 m 2

F2 = 150 N

Substituting the values in the formula above we get,

F12×10−4=15012×10−4⇒F1=150×2×10−412×10−4⇒F1=25N\dfrac{F1}{2 \times 10^{-4}} = \dfrac{150}{12 \times 10^{-4}} \\[0.5em] \Rightarrow F1 = \dfrac{150 \times 2 \times 10^{-4}}{12 \times 10^{-4}} \\[0.5em] \Rightarrow F_1 = 25 N \\[0.5em]

Hence, force applied = 25 N

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