Hydraulic lift inquiry big problem

[mrjoe2]

Hi all,

I have quite a perfect understanding of hydraulic lift, yet there is one huge problem that is contradicting my understanding. for a hydraulic lift where you push on on side of diameter d1 piston, and the object on a piston of diameter d2, you get the equation:

F1/A1 = F2/A2 just to support the object (car) without lifting it. but when you want to lift the car a certain height, you get F1/A1 = F2/A2 + pgh where p is the density of the fluid. this all makes sense.

but why when you do calculations to find the change in pressure force to lift the car a certain height h, it doesn't work. what I am doing is:

F1' = A1(F2/A2 + pgh) minus F1 = A1(F2/A2) to get the change in pressure force to life the car a certain height. so Delta F = A1pgh right?
But subbing in values does not get the correct change in pressure. they use some question DeltaF=pg(A1 + A2)h which is something i have never seen before. I don't get the flaw in my logic, please someone help I am about to cry.

 

[alphysicist]

Hi mrjoe2,

Hi all,

I have quite a perfect understanding of hydraulic lift, yet there is one huge problem that is contradicting my understanding. for a hydraulic lift where you push on on side of diameter d1 piston, and the object on a piston of diameter d2, you get the equation:

F1/A1 = F2/A2 just to support the object (car) without lifting it. but when you want to lift the car a certain height, you get F1/A1 = F2/A2 + pgh where p is the density of the fluid. this all makes sense.

but why when you do calculations to find the change in pressure force to lift the car a certain height h, it doesn't work. what I am doing is:

F1' = A1(F2/A2 + pgh) minus F1 = A1(F2/A2) to get the change in pressure force to life the car a certain height. so Delta F = A1pgh right?
But subbing in values does not get the correct change in pressure. they use some question DeltaF=pg(A1 + A2)h which is something i have never seen before. I don't get the flaw in my logic, please someone help I am about to cry.

It's not so much a flaw in your logic; it appears that you are just not considering the correct height to use in your equation.

For example, to start, let the car be initially only supported by the hydraulic lift, with the pistons at the same vertical level. The car is sitting on the large piston with area A, and the applied force is at the small piston with area a.

Now you want to raise the car up a height H. But here to make the large piston go up a height H, the small piston is going to go down a height h. (And so the height in your original formula is the sum of these heights.) In other words the force difference is given by:

$$\Delta F = a \rho g (H+h)$$

You next can get eliminate h in terms of H. When the large piston rises, it is because a certain amount of fluid is moving to that side of the lift; that same volume of fluid is leaving the side of the lift where the small piston moves down. By setting those volumes equal you can eliminate h and get the equation you have in your post (which in the variables I am using is:

$$\Delta F = \rho g (a+A) H$$

Do you get that equation?

 

[thomate1]

I can understand your frustration and confusion with the calculations for a hydraulic lift. It is important to note that the equations used in hydraulics are based on certain assumptions and ideal conditions, and may not always accurately represent real-world scenarios.

One possible explanation for the discrepancy in your calculations could be the assumption of a perfect, incompressible fluid. In reality, all fluids have some level of compressibility, which can affect the pressure and force calculations in a hydraulic system. Additionally, the equations used for hydraulic lifts often do not take into account factors such as friction, leaks, and other losses, which can also impact the overall accuracy of the calculations.

It is also important to consider the design and construction of the hydraulic lift itself. Any imperfections or variations in the size and shape of the pistons or cylinders can affect the force and pressure distribution within the system.

In order to accurately calculate the change in pressure force needed to lift a car to a certain height, it may be necessary to consider these factors and make adjustments to the equations accordingly. It may also be helpful to consult with other experts in the field and compare your calculations with those used in real-world hydraulic lift systems. Remember, science is all about questioning and refining our understanding, so keep exploring and seeking answers to improve your understanding of hydraulic lifts.