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

[BlackPhysics]

Summary: A 5.0- cm -diameter cylinder floats in water. How much work must be done to push the cylinder 11 cm deeper into the water?

F =Aρgx

A 5.0- cm -diameter cylinder floats in water. How much work must be done to push the cylinder 11 cm deeper into the water?

F =Aρgx

x being the distance being pushed down.

 

[Baluncore]

Welcome to PF.
Is this a homework problem?
If it sinks a further 11 cm, how much additional displacement will there be?
Force is not work done.

 

[BlackPhysics]

Welcome to PF.
Is this a homework problem?
If it sinks a further 11 cm, how much additional displacement will there be?
Force is not work done.

Im confused with that as well. This is all the information I am given
This is homework

 

[Baluncore]

Work done is force times distance moved.

In this case, you will apply a force to displace a volume of water. That force will be zero at the start, then will increase to a maximum when the rod is 11 cm deeper.

What volume of water will then be displaced ?
What will the final force be ?

 

[Lnewqban]

Welcome!
Could you identify the variables shown in the posted equation?
Could you post any work on the solution?

 

[haruspex]

A 5.0- cm -diameter cylinder floats in water. How much work must be done to push the cylinder 11 cm deeper into the water?

It depends how broad the surface of the water is. If the cylinder fits snugly in a tank barely any wider then very little work is needed.
So assume an infinite expanse of water. That way, the level at the top of the water does not change.
Given that, what volume of water is displaced, and how much is its mass centre raised?

 

[BlackPhysics]

Welcome!
Could you identify the variables shown in the posted equation?
Could you post any work on the solution?

A cylinder with cross-section area AA floats with its long axis vertical in a liquid of density ρρ.

F=Aρgx

A 5.0-cmcm-diameter cylinder floats in water. How much work must be done to push the cylinder 11 cmcm deeper into the water?this is all that is given

 

[erobz]

A cylinder with cross-section area AA floats with its long axis vertical in a liquid of density ρρ.

F=Aρgx

A 5.0-cmcm-diameter cylinder floats in water. How much work must be done to push the cylinder 11 cmcm deeper into the water?this is all that is given

If you make the assumption of a very large body ( as @haruspex has pointed out ) of water (look up waters density), you have the information necessary to solve the problem if you know the definition of Work?

 

[Steve4Physics]

A cylinder with cross-section area AA floats with its long axis vertical in a liquid of density ρρ.

F=Aρgx

A 5.0-cmcm-diameter cylinder floats in water. How much work must be done to push the cylinder 11 cmcm deeper into the water?

Hi @BlackPhysics. Welcome to PF. The rules here require you to show evidence of your own effort before we offer help. It's worth taking a quick look at the guidelines:
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/
- especially Rule 4!

The general approach here is that we guide you through a problem so you can work it out for yourself.

You need to check what has been said in previous posts. For example @Lnewqban (in Post #5) asked you to give the meaning of each variable in your equation. But you haven't answered yet.

 

[Lnewqban]

A cylinder with cross-section area AA floats with its long axis vertical in a liquid of density ρρ.

F=Aρgx

A 5.0-cmcm-diameter cylinder floats in water. How much work must be done to push the cylinder 11 cmcm deeper into the water?this is all that is given

All that information I have read it in your first post.
Again, help us to help you:
Could you identify the variables shown in the posted equation?
Could you post any work that you have done about trying to find the solution?

 

[BlackPhysics]

If you make the assumption of a very large body ( as @haruspex has pointed out ) of water (look up waters density), you have the information necessary to solve the problem if you know the definition of Work?

what i have done is take the area of the cross sectional area which is

pi * 2cm^2 * 1000 kg.m^3 * 9.8 * 11cm

 

[erobz]

what i have done is take the area of the cross sectional area which is

pi * 2cm^2 * 1000 kg.m^3 * 9.8 * 11cm

That appears to be the change in the Buoyant Force of water you have calculated, not the cross-sectional area of the cylinder?

EDIT:
Well...its not even really that. Where is the $2 \, \rm{cm}^2$ coming from?

 

[BlackPhysics]

That appears to be the change in the Buoyant Force of water you have calculated, not the cross-sectional area of the cylinder?

EDIT:
Well...its not even really that. Where is the $2 \, \rm{cm}^2$ coming from?

that was me using the a*p*g*x the A is the cross sectional area is what i am thinking

 

[BlackPhysics]

that was me using the a*p*g*x the A is the cross sectional area is what i am thinking

i guess it's supposed to be 2.5cm^2

the area of a circle is pi * r^2

 

[erobz]

i guess it's supposed to be 2.5cm^2

the area of a circle is pi * r^2

Thats not the area in terms of the diameter. And where is that decimal point coming from?

 

[erobz]

that was me using the a*p*g*x the A is the cross sectional area is what i am thinking

Sure, if the buoyant force were constant that would be the Work?

 

[BlackPhysics]

Thats not the area in terms of the diameter. And where is that decimal point coming from?

A 5.0- cm -diameter cylinder floats in water.

 

[erobz]

A 5.0- cm -diameter cylinder floats in water.

Correct. So, what is its cross sectional area?

 

[BlackPhysics]

Welcome!
Could you identify the variables shown in the posted equation?
Could you post any work on the solution?

f = force
a = area
p = rho/density
g = gravity
x= displacement

 

[BlackPhysics]

Correct. So, what is its cross sectional area?

78.54cm^2

 

[erobz]

78.54cm^2

What formula have you used to get that?

 

[BlackPhysics]

78.54cm^2

As of right now i am getting 8.46 J and it is wrong

f= (78.54cm) * (1000 kg/m^3) * (9.8 m/s^2) * (11cm)

 

[BlackPhysics]

What formula have you used to get that?

(pi * r^2) = area of a circle

 

[erobz]

(pi * r^2) = area of a circle

You haven't used it correctly. What does $r$ stand for in that formula?

 

[BlackPhysics]

You haven't used it correctly. What does $r$ stand for in that formula?

radius. the diameter is 5.0 cm so half of that is the radius.

 

[erobz]

Ok, fix that then we will move on.

 

[BlackPhysics]

Ok, fix that then we will move on.

A=πr^2
78.54 = π * 2.5^2

 

[haruspex]

f = force
a = area
p = rho/density
g = gravity
x= displacement

ok, but what force exactly?

As of right now i am getting 8.46 J and it is wrong

f= (78.54cm) * (1000 kg/m^3) * (9.8 m/s^2) * (11cm)

f is a force, not work, as was pointed out in post #4. Its units may be Newtons, not Joules.
The 78.54 is presumably supposed to be the horizontal area, but I don’t know how you got that from $\pi 2.5^2$, and the units should be $cm^2$.
When you have corrected that equation, what force does f represent?

 

[erobz]

A=πr^2
78.54 = π * 2.5^2

No. It doesn't.

 

[haruspex]

A=πr^2
78.54 = π * 2.5^2

$\pi$ is not equal to 10.

 

[BlackPhysics]

$\pi$ is not equal to 10.

im being dumb sorry its 19.63

 

[BlackPhysics]

im being dumb sorry its 19.63

19.63= π * 2.5^2

 

[erobz]

19.63= π * 2.5^2

Ok, now that you have that. Is it a good idea to multiply units of $\rm{cm}^2$ and $\rm{m}$ etc...as you have done in post #22? Firstly, its best to work with all variables until the calculation step, but if you insist, I think you should convert all figures to standard units.

 

[BlackPhysics]

Ok, now that you have that. Is it a good idea to multiply units of $\rm{cm}^2$ and $\rm{m}$ etc...as you have done in post #22? Firstly, its best to work with all variables until the calculation step, but if you insist, I think you should convert all figures to standard units.

(0.01963m) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
Area Density G X

 

[erobz]

(0.01963m) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
Area Density G X

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?

 

[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.