# A cylinder with cross-section area A floats with its long axis vertical

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f = force
a = area
p = rho/density
g = gravity
x= displacement
Water acts like a compressing spring on your cylinder: the lower it is pushed, the higher is the resistive vertical force.
As area, density and gravity acceleration are all constant, we could make their product a constant k.
Then, we have a problem similar to the work done by, or on, a spring.

https://courses.lumenlearning.com/suny-physics/chapter/7-4-conservative-forces-and-potential-energy/

erobz
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I'd just like to clear up the issue of "average force". @BlackPhysics is quite right that what we have calculated here, half the maximum force, is not the average force. Average force is defined as ##\frac{\Delta p}{\Delta t}=\frac{\int F(t).dt}{\int .dt}##. That has to be the definition to be consistent with average acceleration, so that ##F_{avg}=ma_{avg}##. Also, the formula still makes sense in vector form. So to calculate it we need to know the force as a function of time.
In the special case of a constant force, and with the displacement and force being parallel, this is equivalent to ##\frac{\Delta W}{\Delta x}=\frac{\int F(x).dx}{\int .dx}##, which we can correctly call the "average force with respect to displacement". Note that this makes sense in vectors only if we write it in the usual form, ##\Delta W=\int \vec F(x).\vec{dx}##, because one cannot divide by a vector.
Far too many educators who should know better set questions asking the student to find the "average force" based only on work and displacement. It is an interesting exercise to compute the average by both definitions in the case of a half cycle of SHM.

In the current thread, we don’t care what the average force is; rather, we want the average force wrt displacement. And since F(x) is linear, this equals ##\frac 12(F_{min}+F_{max})##. Hence ##\frac{\Delta W}{\Delta x}=\frac 12(F_{min}+F_{max})##, or ##\Delta W=\frac 12(F_{min}+F_{max})\Delta x##.

erobz