Hydraulic Jack Question: Calculating Piston Movement for 2000 kg Car

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The discussion revolves around a hydraulic jack problem involving a 2000 kg car being raised 1.5 meters. The setup includes a large piston with a diameter of 30 cm and a small piston with a diameter of 5 cm. Participants are exploring the relationship between the movements of the two pistons.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are considering whether the movement of the small piston can be directly calculated from the movement of the large piston or if a more complex relationship exists. There are discussions about applying the concept of work and volume transfer between the pistons.

Discussion Status

Some participants have suggested using the work principle to relate the movements of the pistons, while others are exploring the volume relationship. A specific value for the small piston’s movement has been proposed, but it is based on assumptions that are still being examined.

Contextual Notes

There is an ongoing examination of the assumptions related to pressure and work done on both sides of the hydraulic system. Participants are also considering the implications of the diameters of the pistons in their calculations.

frigid
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A 2000 kg car is raised 1.5 meters high on a hydraulic jack, which has a large piston with a diameter of 30cm. The small piston has a diameter of 5cm (btw the reservoir is filled with oil)

d) When the car is raised by 1.5 m, what distance will the small piston move?

Is it simply that you have to take off 1.5 m on the side of the small piston in order for the big piston to raise by 1.5 m? Or is there something more complicated about this?
 
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You need to apply the idea of work. The work on one side will be the same as the other piston. Calculate the work for the large side and then use that amount for what the small piston must do.
 
Alternatively, you can remove Volume fom one side,
to put it on the other side.
(you get the same answer as using Work ...
W = F dh = P A dh = P dV)
 
is d = 0.25 m ?

I got this from the assumption 1/6th the work done at the small piston (from the pressure relationship).
 

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