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

[BlackPhysics]

Ok, but the units are $\rm{m}^2$ on area. you have $\rm{m}$, otherwise the values are ok.

Now we can move on to the definition of Work. What is it?

Work = force * distance

 

[erobz]

Work = force * distance

Only if the Force is constant over the applied displacement (and in the same direction). Is the Buoyant Force going to be constant over the $11 \, \rm{cm}$ displacment?

 

[haruspex]

Work = force * distance

Not necessarily. A formula means nothing unless you can state exactly what the variables mean in relation to each other. For this equation, what must be the relationship between the distance and the force?

 

[BlackPhysics]

Only if the Force is constant over the applied displacement. Is the Buoyant Force going to be constant over the ##11 \, \rm{cm} displacment?

it is not constant...

 

[BlackPhysics]

it is not constant...

so knowing its not constant now what? how can i calculate the Force.

 

[erobz]

And actually $F$ is the applied force here...It's not the Buoyant force. My bad. How $F$ varies over the displacement $x$ relates to the Buoyant Force.

 

[erobz]

so knowing its not constant now what? how can i calculate the Force.

Have you had any Calculus? If you are given this problem, you should probably be acquainted with the subject.

EDIT:
Or because of the linearity in this specific problem there is an equivalent route that doesn't use Calculus.

 

[BlackPhysics]

Have you had any Calculus? If you are given this problem, you should be acquainted with the subject.

yes i know calculus. Do i just take the integral of the formula?

 

[haruspex]

yes i know calculus. Do i just take the integral of the formula?

You could, but there's an easier way. Think of the displaced mass of water. Where was its centre of mass at first, and where is it at the end?
Edit: on second thoughts, that way is no simpler. Stick with the integration or use the linearity of the force as a function of displacement.

 

[BlackPhysics]

im guess it would be in the center to start and after the water is displaced it has to move

 

[erobz]

yes i know calculus. Do i just take the integral of the formula?

Yes, $W = \int F \cdot dx$ or you could consider that since the relationship is linear $\int F \cdot dx = W = F_{avg} \Delta x$ or you could do whatever @haruspex has planned.

 

[BlackPhysics]

Yes, $W = \int F \cdot dx$ or you could consider that since the relationship is linear $\int F \cdot dx = W = F_{avg} \Delta x$ or you could do whatever @haruspex has planned.

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

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

 

[haruspex]

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?

$\pi 2.5^2cm^2$ is not 0.0196… m². Check your conversion from $cm^2$ to $m^2$.
You still have not said what that F represents. What force is it?

 

[BlackPhysics]

$\pi 2.5^2$ is not 0.0196… Check your conversion from $cm^2$ to $m^2$.
You still have not said what that F represents. What force is it?

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

 

[haruspex]

F is the buoyancy force

No it isn't. The buoyancy force depends on the weight of the cylinder, which we do not know.

 

[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!