Calculating Piston Heights in a Car Lift: A Fluid Pressure Problem

[tod88]

Homework Statement

A car lift has two pistons. The one that lifts the car has an area of 2 meters squared, the one that is pushed down has an area of 1 meter squared. If a car weighing 2000kg is to be lifted 2m, how far must the smaller piston be pushed down? The pistons are of negligible mass.

Homework Equations

F1/A1 = F2/A2

The Attempt at a Solution

So:

(2000)(9.81)(2) / 2 = mgh or (9.81)mh

I'm getting stuck where I find the height on the other side. Since the pistons were of negligible mass I thought that meant that it would just be equal to 9.81h and then divided by 9.81 to solve for h. Is this the correct approach or am I missing something?

 

[dynamicsolo]

Something missing in the problem statement is that the car is being lifted at constant velocity (which is the typical situation in such a problem). That is what let's you write that the fluid pressure on that platform must be mgh/A_1 (the net force on the car must be zero).

The other side of the equation would also have to be a force over an area, so it can't just be mgh. Since it has to be F_2/A_2 = (mgh / 2 m^2) , what is F_2?

Now, what is the work done by the fluid force to lift the car 2 meters? That work is ultimately supplied by whatever is pushing down on the other piston and it has to be the same amount of work. (The hydraulic fluid only provides a medium for energy transfer, so to say -- it is not a source of energy.) To do that much work, through what distance does force F_2 have to act?

(There's another approach that is entirely separate from the work/energy issue. For the pressure to be uniform all through the hydraulic fluid, it has to be incompressible, which means the volume of fluid doesn't change. So if the first piston has an area of 2 m^2 and rises by 2 m, how much fluid must enter that side of the hydraulic system? Therefore, how much fluid must be displaced from the side with the second piston? How far does the second piston have to descend in order to force out that volume?)

 

[tod88]

Thank you SO much ... That was exactly what I needed.