# Why doesn't a hydraulic lever violate conservation of energy?

Suppose I apply some force on one side of Hydraulic lift where area is less, and the fluid in the lift raises some heavier object on the other side where area is more, Now work done is $$Force\times displacement$$ and displacement on both side is same (incompressible liquid) but force on one side is less, so we get more energy on other side. Then why doesn't the law of Conservation of energy fail here.

• Your statement “displacement on both sides is same” is incorrect. – Farcher Apr 12 '19 at 12:22
• do you think the levers also violate energy conservation? – user8718165 Apr 12 '19 at 12:41
• displacement means "volume", right? – JEB Apr 12 '19 at 13:58
• @JEB hits the point. Displacement here means a distance moved and not the volume displaced. – JimmyB Apr 12 '19 at 14:44

The displacement produced is not the same. That is why, energy is conserved.

When you apply force on one side of the opening (with smaller $$A$$, i.e. $$A_1$$), the displacement in the piston that does the work on the water is say, $$x$$. The displacement on the other side of the lift with $$A_2$$ where $$A_2>A_1$$, has a displacement smaller than $$x$$, which we'll call $$y$$.

What happens here is that the water absorbs energy from the piston and sends it straight to the lift on the other end with area $$A_2$$. The volume of water remains the same. But the displacements need not be the same.

Consider the work done $$W=P\Delta V$$ where $$\Delta V$$ is the change in volume. Since the first and the second openings are subjected to the same pressure (from the piston to the water, and from something that lifts the object in the larger opening), $$\Delta V=A_1x = A_2y$$.

$$y=\frac{A_1x}{A_2}$$

Since, $$A_2 >A_1$$, clearly, $$y. Displacement in both sides is not same. If on one side of lift the area is $$A_1$$, and on other side it is $$A_2$$, and we apply a force $$F_1$$ on one side to distance $$d_1$$ then volume decreased in one side is $$=A_1 \times d_1$$

Equal amount of volume will raise in the other side.

So $$A_1 \times d_1=A_2 \times d_2$$

$$A_1 \not= A_2$$, so $$d_1 \not=d_2$$.

Actually, we need to apply the little force $$F_1$$ for a greater distance $$d_1$$.