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

[erobz]

W = 2.327 * x?

Based on your numbers, it doesn't look right to me.

The average Force is nothing more than how you would average two numbers for this problem because the function is linear.

$$ F_{avg} = \frac{F(0)+F(x)}{2} $$

 

[BlackPhysics]

Based on your numbers, it doesn't look right to me.

0.52889375 is that right?

 

[erobz]

0.52889375 is that right?

No, I don't get that either.

 

[haruspex]

W = 2.327 * x?

You calculated the final (i.e. maximum) force to be 2.116 N (but you keep leaving out the units!).
How did you get W = 2.327 * x from W=F*x? Looks like you multiplied F by 1.1 first.
And you want the force at the average displacement, not at the final displacement. Or to put that another way, you want to integrate the force from its initial 0 to its final value wrt displacement: $W=\int F(x).dx$.

 

[BlackPhysics]

Based on your numbers, it doesn't look right to me.

The average Force is nothing more than how you would average two numbers for this problem because the function is linear.

$$ F_{1.05778} = \frac{F(0)+F(2.115575)}{2} $$

1.0577875 is what i got as the avg force

 

[erobz]

1.0577875 is what i got as the avg force

Units!

 

[BlackPhysics]

Units!

1.0577875 N

 

[erobz]

Now finish it out, and don't forget the units. Do you understand why this works for calculating the area under a line?

 

[BlackPhysics]

Now finish it out, and don't forget the units. Do you understand why this works for calculating the area under a line?

Yes i do.

Avg F = 1.0577875 N
Work = 1.0577875 N * .11M
0.11635 J = 1.0577875 N * .11M

Is this right?

 

[erobz]

Yes i do.

Avg F = 1.0577875 N
Work = 1.0577875 N * .11M
0.11635 J = 1.0577875 N * .11M

Is this right?

Now it's time for the Physicist's to blast you about sig figs! Good Luck!

 

[Lnewqban]

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.

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

 

[haruspex]

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 $\Delta W=\frac 12(F_{min}+F_{max})\Delta x$.