Calculating Pressure for a Hydraulic Lift: A Scientific Approach

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To calculate the gauge pressure required to lift a car with a hydraulic lift, the correct formula is p = F/A, where F is the force (mass times gravitational acceleration) and A is the area of the piston. The area of the piston should be calculated using A = π(r^2), where r is half the diameter of the piston. The initial confusion stemmed from a miscalculation of the area, particularly regarding the use of the diameter in the formula. Once the area is correctly determined, the pressure can be accurately calculated in pascals. Properly applying these principles will yield the correct answer for the homework problem.
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Hello! I was wondering if you could help me with a homework problem:

The piston of a hydraulic automobile lift is X meters in diameter.

What gauge pressure, in pascals, is required to lift a car with a mass of Y kg?


I know that a piston is a cylindrical shaped apparatus - however I believe that does not affect my calculations.

I also know this about pressure: p = F/A

F = mass x a so in my case that would be: Y kg x 9.8m/s

A (of a circle) = 2 x pi x r^2 r being half my diameter so X/2


From using this formula, the pressure is calculated in Pascals. The answer I get however is wrong. Am i using the wrong formula? should find the volume of the piston? I'm not sure how that would help... hope someone can give me some hints! =)
 
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If you are considering the pressure on just the bottom of the piston (or top) then you only need to consider the area of one circle, not two.
 
that's right, i should only consider the area of one circle. that "2" in front of pi r squared was a typo... =) I've been trying to calculate it assuming it's one circle but I'm doing something wrong. =/
 
Is your final result

\frac{4(Y kg)(9.8 m/s^2)}{\pi x^2}?
 
Why is the equation being multiplied by 4?
 
The 4 comes from the denominator, since you are squaring X/2, you get x^2/4, and instead of having stacked fractions, you can multiply top and bottom by 4 to get what I gave.
 
My final equation looks like this:


(Y kg)*(9.8 m/s^2) / (pi)*(r^2) = p (pascals)
 
regarding the equation that I am using (p = F/A), does that even look like the correct equation I need to solve this type of problem?
 
Express it in terms of x and y, I think that's what you need to do. Is this online homework? Is that the complete problem?
 
  • #10
sghaussi said:
regarding the equation that I am using (p = F/A), does that even look like the correct equation I need to solve this type of problem?

Yes it is the right equation. Your first calculation of area was wrong. If you get the area right, I think you will have it.
 
  • #11
yes, it is online homework and that is the complete problem except "x" and "y" there are real numbers. so instead of multiplying by 4 i just divided x by two before I plugged it into the equation.
 
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  • #12
Did you divide correctly? ;)
Lets see the numbers.
 
  • #13
got it.. i WAS getting the area wrong.. i was imputing it incorrectly into the calc. thanks so much for your time and patience!
 
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