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I am reading about applications of Pascal's Principle and hydraulic systems. I was able to derive a relationship between the forces in this hydraulic jack by applying Pascal's principle. The pressure due to $F_1$ acting on area $S_1$ is $p_1=\dfrac{F_1}{S_1}$. According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure $p_2=\dfrac{F_2}{S_2}$ is felt at the other piston that is equal to $p_1$. We see that $\dfrac{F_1}{S_1}=\dfrac{F_2}{S_2}$ and $F_2=\dfrac{S_2}{S_1}F_1$. Now I am asking myself if the work done by $F_1$ must be equal to the work done by $F_2$? In other words, is $A_1=F_1h_1$ equal to $A_2=F_2h_2$? If this is true, can you tell me why? I was searching on the Internet for more relationships, but I didn't find a useful website. enter image description here

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You got it all right. Just what you were forgetting is the fact that the fluids are (almost) incompressible. Here is how it goes:

You can write the fact that pressure is transferred undiminished as $$\frac {F_1}{S_1}=\frac {F_2}{S_2} \tag{1}$$

But you also know that volume does not change. So $$h_1S_1=h_2S_2 \tag{2}$$

Now multiply both of these equations ($(1)$ and $(2)$) and cancel out common terms (if any) and you get:

$$F_1h_1=F_2h_2$$

And hence the same work is done.

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Definitely yes:

$$S_1h_1 = S_2h_2$$

because these are volumes of of water, and the full volume of water keeps unchanged. Consequently

$$F_1S_1h_1 = F_1S_2h_2$$ $$F_1h_1 = F_1{S_2 \over S_1}h_2$$ $$F_1h_1 = F_2h_2$$

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Work is defined as force times displacement, where only the component of the force that is parallel to the displacement is involved. Obviously $F_1$ is in the direction of displacement of piston 1 and $F_2$ is in the direction of the displacement of piston 2. Also note, that the volume of the displaced fluid is constant, so the distance traveled by piston 1 is far greater than the distance traveled by piston 2, due to the much smaller area of piston 1 compared to piston 2. This means that the work input to piston 1 is equal to the work output by piston 2, assuming that the small increase in gravitational potential energy of the hydraulic fluid at piston 2 can be ignored.

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