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enter image description here

The above pic is describing the system how the car will be lifted if car is placed on the larger piston. But if the small piston is moved through 1m (say) ,then the piston is merely moved by millimetres. Then how, in practical life,small piston can actually be used to lift a car in a carwash station. If helpful, please suggest books, E-books or a few good articles.

The following portion is edited area after the question was declared answered

My question was very nicely answered and now my confusions have been eradicated and I'm grateful for that. I have hit upon a conclusion in the light of answer given and want the experts to confirm my following statement: If small multiple displacements "h" are not considered(see the answer with figure) and only stroke of the smaller piston is considered,

THEN more fluid enters the large piston than fluid lost by by smaller piston and vice versa (on pushing small piston) i.e more fluid will come out of the larger piston than fluid gained by smaller piston upon pulling the smaller piston upwards.

Note: I also wrote similar text in comments section but now I edited question thinking I followed wrong approach

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  • $\begingroup$ Try applying continuity of mass : what leaves (large or small piston) = what arrives in (small or large piston) or m1 = a1 . V1 and m2 = a2 . V2, and m1 = m2 $\endgroup$ – Solar Mike May 16 '17 at 17:01
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The main idea of the hydraulic press is that we use a piston for which we have to apply a small force $F_1$ and small area $A_1$ to displace a specific volume of the incompressible fluid. Imagine we want to displace a volume $V=A_{\text{base}}h.$ For the piston with a small area, we have large height displacement $h$.

But the displaced volume in both pistons has to be approximately the same because water is in very good approximation incompressible, as long as velocities are small. So, $$V_1=V_2 \implies A_{1}h_1=A_{2}h_2 \implies \frac{h_1}{h_2}=\frac{A_2}{A_1} \implies h_1=\frac{A_2}{A_1}h_2.$$

Depending on your choice of $A_2$ and $A_1$ you can create different transmission ratios. Using an additional valve which is connected to the reservoir we suck fluid back into the circuit in the backstroke. You can use repetitive motion with small $h$ displacements per cycle and accumulate them for a large total displacement.

The following picture shows the working principle. Note the both valves. The left one opens in the work stroke and closes in the backstroke. The right valve is operating in an inverted fashion.enter image description here

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  • $\begingroup$ In the light of above description, I want you to confirm whether my following statement is true or not. If we concern only 1 stroke of piston(not multiple small displacements h), if the small piston is pushed downwards,comparitivly more volume will enter the large piston section and Same would happen i.e if piston is pulled, more fluid will come out of larger piston as compared to fluid gained by the small piston. $\endgroup$ – Muhammad Hashim May 13 '17 at 4:04
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To put the answer simply : the volume delivered by the small piston is delivered many times - this then moves the large piston accordingly. That is what the maths above shows.

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The 2 keys to a hydraulic jack (which is what a hydraulic car lift is) are: the often quoted Pascals's law.. AND valves, a non return one to "ratchet" the big piston up, and a controllable one to let it down. Brilliant simplicity.

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