Question

$❤️QUESTION❤️$
What force is applied on a piston of area of cross section 2cm²to obtain a force 150N on the piston of area of cross section 12cm²in a hydraulic machine?


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Answer 1

Answer:

We know that in a hydraulic machine, Pressure in equal on both sides.

Pressure on narrow piston = Pressure on wider piston

∴ P1 = P2

∴ A1F1 = A2F2

∴ 2×10−4F1 = 12×10−4150

∴ F1 = 25 Newton

Answer 2

Answer:

We know that in a hydraulic machine, Pressure in equal on both sides.

Pressure on narrow piston = Pressure on wider piston

∴ P1 = P2

$\frac{F1 }{A1} = \frac{F2}{A2}$

$\frac{F1}{2 \times {10 }^{ - 4} } = \frac{150}{12 \times {10}^{ - 4} }$

F1 = 25 NEWTON