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

[bcjochim07]

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)

 

[bcjochim07]

any thoughts?

 

[Moetasim]

with the hydraulic lift.


Your solution is correct. The formula you used, p0 + F1/A1 = p0 + F2/A2 + ro*gh, is the correct one to use in this situation because the cheerleader is simply holding the football players at a constant height, not lifting them. This formula takes into account the balance of forces at equilibrium.

The other formula, DeltaF = ro*g (A1+A2)d2, is used when the hydraulic lift is actively lifting something. In this case, the lift would be exerting a force on the football players to raise them to a higher height. This formula takes into account the work done by the lift in raising the object.

Overall, your understanding and use of the correct formula is correct. Keep in mind that in real-world situations, other factors such as friction and fluid viscosity may also need to be taken into account for a more accurate calculation.