Why doesn't the hydraulic lift violate the law of conservation of energy

Question

Suppose a force $F_1$ is applied to the left side of the piston of a hydraulic lift and the displacement is $d_1$. So the force on the right side is $F_2$ and the displacement is $d_2$. So $$F_1 d_1=F_2 d_2$$ But since the liquid is moved up, the potential energy of the liquid is changed. Doesn't this ($F_1 d_1=F_2d_2$) violate the law of concervaton of energy. Shouldn't it be $$F_1d_1 = F_2d_2 + \text{change in potential energy of the liquid}$$ If so how can I prove the pressure on both sides are same?

Answer 1

We can use conservation of energy for an almost complete model of this system (neglecting viscosity and assuming incompressible water).

Suppose you raise the car (of mass $M$) of an amount $d_2$. The energy you spent is $$U_2=Mgd_2 + \rho A_2 d_2g d_2$$ where $\rho$ is the density of water and $A_2 d_2$ is the volume of water you displaced ($A_2$ being the surface). This energy has to be equal to the work you applied, that is $F_1 d_1$. Plus you pushed some water to the bottom and that gives you an extra energy of $ \rho A_1 d_1 g d_1$, so $$U_1=F_1d_1+\rho A_1 d_1 g d_1$$ (signs are chosen so that $U_1$ and $U_2$ are the "magnitude" of the energy).

Because water is incompressible the two volumes displaced must be the same, i.e. $$A_1d_1=A_2d_2$$ so that $$d_1={A_2\over A_1}d_2$$ and substituting $$U_1=F_1{A_2\over A_1}d_2+\rho A_1 g\left({A_2\over A_1}\right)^2d_2^2.$$

Because of conservation of energy $U_2-U_1=0$

$$\left(Mg-F_1{A_2\over A_1}\right)d_2+\rho A_2g \left(1-{A_2\over A_1}\right) d_2^2=0$$

which has, as a solution, either $d_2=0$ (of course if nothing moves energy is conserved) or $$d_2=-{Mg-F_1{A_2\over A_1} \over \rho A_2g\left(1-{A_2\over A_1}\right)}$$

Because in the example you drew the force $F_2$ is the weight of the car, we have

$F_2=Mg$ so that the most general formula we can write for the hydraulic lift is

$$ \rho A_2 g \left(1-{A_2\over A_1}\right) d_2= F_2-F_1{A_2\over A_1}$$

Let's discuss some special cases:

1. if we neglect the displacement of water (we can put $\rho=0$ as if it's weightless) we get $A_1 F_2=A_2 F_1$ (NB: to do this properly, setting $\rho=0$ has to be done before solving the equation above otherwise we are dividing by 0 at some point). This means $F_2/A_2=F_1/A_1$ i.e. the two pressures are the same and the amount of force you need is $$F_1=F_2{A_1\over A_2}$$ and, by using incompressibility of water again that means $F_1=F_2{d_2 \over d_1}$ i.e. the formula of "ideal" the hydraulic lift. The result you quoted. By making $A_2$ bigger we make $d_2$ smaller and therefore we need less force $F_1$. This is also valid if the mass of the car is much bigger than the displaced water, and is in general a valid approximation for real life scenarios.
2. if we include the weight of the water, then $$F_1={A_1\over A_2}F_2+\rho g (A_2-A_1) d_2$$ so we now need more force ($A_2>A_1$) to lift the car, due to our force also having to account for the displaced water. Also notice that now "simply" making $A_2$ bigger, as we did before, is not convenient, as the bigger $A_2$ is, the more water is displaced.
3. the case in which the side number one of the lift is horizontal, meaning that we do not have the energy gain due to water coming down. We can find it by doing everything again without the second term of $U_1$ and that leaves us with $$F_1={A_1\over A_2}F_2+\rho g A_2 d_2$$ so we need an even bigger force, as we do not have any help from the water going down.
4. The case in which also the car is horizontal (in this case, of course, $F_2!=Mg$ and is just the force you need to push the car, whatever that is). Now also the second term in $U_2$ vanishes, so that $U_1-U_2=0$ is simply $$F_2d_2 = F_1d_1$$ which is again the "ideal" hydraulic lift.

Answer 2

But since the liquid is moved up, the potential energy of the liquid is changed.

The potential energy is $mgh$. On the left, a small amount of liquid moved down a large distance, so we have a small $m$ times a large $h$. On the right, a large amount moved up a small distance, so we have a large $m$ times a small $h$. If the lift is perfectly balanced, these will multiply out to the same amount.