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

[erobz]

F = (0.01963m^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 2.1166W = F * (Delta X)
0.2328 = 2.1166 * 0.11m

Does it look like this?
The average force is just the mass * gravity?

No, the average Force is not $mg$.

F is the buoyancy force
F = (0.1963m^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 192.42251

$F$ is not the Buoyant Force here, it is the force which you push with to counter the Buoyant Force.

And re -re check the area...it has gotten worse!

 

[haruspex]

F is not the Buoyant Force here, it is the force which you push.

Since the force of push is variable, it's not exactly that either.

 

[erobz]

Since the force of push is variable, it's not exactly that either.

$$dF = \gamma d {V\llap{-}}_D$$

Is that not the force required to overcome the buoyancy of the cylinder in a quasistatic equilibrium?

 

[BlackPhysics]

F = (0.001963M^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 2.1166W = F * (Delta X)
0.2328J = 2.1166 * 0.11m

ok so i converted cm^2 to m^2

and the force is not the buoyant force.
So I am guessing its the force that is pushing the cylinder in the water?

 

[erobz]

F = (0.001963M^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 2.1166W = F * (Delta X)
0.2328J = 2.1166 * 0.11m

ok so i converted cm^2 to m^2

and the force is not the buoyant force.
So I am guessing its the force that is pushing the cylinder in the water?

But the work is the average force times the displacment. You have not calculated the average force, you have calculated the final force at depth $x= 11 \, \rm{cm}$

Also, the units of force are missing.

 

[BlackPhysics]

But the work is the average force times the displacment. You have not calculated the average force, you have calculated the final force at depth $x= 11 \, \rm{cm}$

Also, the units of force are missing.

How do you calculate that without time? average force = f = m(VI-VF)/T

 

[erobz]

How do you calculate that without time? average force = f = m(VI-VF)/T

The force is a function of $x$, not time.

Or, just do the integral if you are familiar with calculus.

 

[haruspex]

you have calculated the final force at depth 11cm

Yes, that was my point.

How do you calculate that without time?

That's a fair question. You are right that average force is change in momentum/elapsed time. But here we are dealing with work, not momentum, so you want the force at the average displacement.
When just starting to push the cylinder down, the slightest force will do. The force needed increases linearly with depth up to the maximum calculated by your equation. So, if you integrate the force wrt depth, what do you get?

 

[BlackPhysics]

Yes, that was my point.

That's a fair question. You are right that average force is change in momentum/elapsed time. But here we are dealing with work, not momentum, so you want the force at the average displacement.
When just starting to push the cylinder down, the slightest force will do. The force needed increases linearly with depth up to the maximum calculated by your equation. So, if you integrate the force wrt depth, what do you get?

2.327? is that right?

 

[BlackPhysics]

2.327? is that right?

W = 2.327 * x?

 

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