Lifting an Elephant with Hydraulics: A Physics Puzzle

[kikko]

Homework Statement

The 70kg student in the figure balances a 1200kg elephant on a hydraulic lift. A second 70kg student joins the first student. How high do they lift the elephant?
diameter of the piston elephant is on = 2.0m
diameter of the piston student is on = 0.48m
The liquid inside the lift is oil, therefore rho = 900 kg/m3
Out would be elephant, In would be the students

Homework Equations

Fout/Fin = Aout/Ain
Fout=Aout/Ain x Fin - \rhoghAout

The Attempt at a Solution

(1200)(9.8) = ((pi*(2/2)^2)/(pi*(.48/2)^2))*(2)(70)(9.8) - (900)(9.8)(h)(pi)

h = .435m = 43.5 cm

This answer comes out wrong. I noticed the .435^-1 = 2.3m, and the answer is 2.3 cm, but I'm not sure if there is a relation there.

Anyways, any tips in the right direction?

 

[zgozvrm]

Initially, the problem looks to be invalid.
Are you sure the elephant is on the larger piston?

 

[kikko]

Yes. Here is a picture, I solved .48 for the diameter of the first piston and got it correct.

http://session.masteringphysics.com/problemAsset/1070646/3/15.P41.jpg

 

[zgozvrm]

Yes. Here is a picture, I solved .48 for the diameter of the first piston and got it correct.

http://session.masteringphysics.com/problemAsset/1070646/3/15.P41.jpg

Yes, of course. After thinking about it a bit more that makes sense.

But, I don't see how they can be in balance.
There would have to be equal pressure on each side of the system.

On the student's side, you have:

$$\frac{70 kg}{(0.24)^2 \pi ~m^2} = \frac{70 kg}{(0.576) \pi ~m^2} \approx \frac{70 kg}{0.181 ~m^2} \approx 386.83 ~kg/m^2$$


On the elephant's side, you have:

$$\frac{1200 kg}{(1)^2 \pi ~m^2} = \frac{1200 kg}{\pi ~m^2} \approx \frac{1200 kg}{3.142 ~m^2} \approx 381.97 ~kg/m^2$$



So, since there is more pressure being exerted on the oil on the student's side than on the elephant's side, the elephant should be rising (at least to the extent that shortest piston's travel allows, which means either the student's piston "bottomed out," or the elephant's piston "topped out.")



Now, if the student only weight 69.12 kg, you'd have:

$$\frac{69.12 kg}{(0.24)^2 \pi ~m^2} = \frac{69.12 kg}{(0.576) \pi ~m^2} \approx \frac{69.12 kg}{0.181 ~m^2} \approx 381.97 ~kg/m^2$$

which is the same pressure that's being exerted on the elephant's side.




... I think the weight of the pistons would affect the system though.