How Is Hydraulic Lift Calculated in Cheerleader and Football Players Scenario?

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The discussion centers on calculating the diameter of the piston used by football players in a hydraulic lift scenario involving a 55 kg cheerleader and four 110 kg football players. The initial calculations indicate that the diameter of the football players' piston is 45 cm, derived from the area and radius formulas. There is confusion regarding the appropriate formulas to use, specifically between those for static scenarios versus those involving work done during lifting. The first formula is deemed suitable for holding the players in position, while the second applies when lifting is involved. Clarification on these formulas is sought to ensure accurate application in similar problems.
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Homework Statement


A 55kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1 m. If her piston is 16 cm in diameter, what is the diameter of the football players' piston?


Homework Equations





The Attempt at a Solution



(55kg)(9.80)/(.02 m^2) = (440kg)(9.80)/A2
A2 = .16 m^2
.16 m^2 = pi * r^2
r=.2257 m
D= .45 m = 45 cm

Is this correct? I am a little bit confused as to the difference between these formulas:

p0 + F1/A1 = p0 + F2/A2 + ro*gh This is the one I used because she is just holding them in position and no work is being done.

for the formula DeltaF = ro*g (A1+A2)d2 You should use this one when the problem involves lifting something (doing work)
 
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