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A force of 3600 N is exerted on a piston that has an area of [tex]0.030 \, \text{m}^2[/tex]. What force is exerted on a second piston that has an area of [tex]0.015 \, \text{m}^2[/tex]?

Answer :

Answer:

18,000 N

Explanation:

apex

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Rewritten by : Jeany

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
[tex]p_1 = p_2[/tex]
which becomes
[tex] \frac{F_1}{A_1}= \frac{F_2}{A_2} [/tex]
where
[tex]F_1=3600 N[/tex] is the force on the first piston
[tex]A_1=0.030 m^2[/tex] is the area of the first piston
[tex]F_2[/tex] is the force on the second piston
[tex]A_2=0.015 m^2[/tex] is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
[tex]F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N[/tex]