# Question about a hydraulic jack

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. 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.

Definitely yes:

$$S_1h_1 = S_2h_2$$

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

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

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.