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

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Homework Help Overview

The discussion revolves around a problem involving a 5.0-cm-diameter cylinder floating in water and the work required to push it 11 cm deeper into the water. The problem is situated within the context of fluid mechanics and buoyancy.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between force and work, questioning how to calculate the work done based on the displacement of water and the forces involved. There are inquiries about the variables in the equation provided and the correct interpretation of the cross-sectional area of the cylinder.

Discussion Status

Several participants have offered guidance on identifying variables and clarifying concepts related to buoyancy and work. There is an ongoing exploration of the calculations needed, with some participants expressing confusion about the units and the application of formulas. No consensus has been reached yet, as participants continue to clarify their understanding and address assumptions.

Contextual Notes

Participants note that the problem is a homework assignment and express the need for clarity on the definitions of terms and the correct application of formulas. There are discussions about the assumptions made regarding the body of water and the implications for the calculations.

BlackPhysics
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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.
 
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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.
 
Baluncore said:
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
 
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 ?
 
Welcome!
Could you identify the variables shown in the posted equation?
Could you post any work on the solution?
 
BlackPhysics said:
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?
 
Lnewqban said:
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
 
BlackPhysics said:
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?
 
BlackPhysics said:
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.
 
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  • #10
BlackPhysics said:
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?
 
  • #11
erobz said:
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
 
  • #12
BlackPhysics said:
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?
 
  • #13
erobz said:
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
 
  • #14
BlackPhysics said:
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
 
  • #15
BlackPhysics said:
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?
 
  • #16
BlackPhysics said:
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?
 
  • #17
erobz said:
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.
 
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  • #18
BlackPhysics said:
A 5.0- cm -diameter cylinder floats in water.
Correct. So, what is its cross sectional area?
 
  • #19
Lnewqban said:
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
 
  • #20
erobz said:
Correct. So, what is its cross sectional area?
78.54cm^2
 
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  • #21
BlackPhysics said:
78.54cm^2
What formula have you used to get that?
 
  • #22
BlackPhysics said:
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)
 
  • #23
erobz said:
What formula have you used to get that?
(pi * r^2) = area of a circle
 
  • #24
BlackPhysics said:
(pi * r^2) = area of a circle
You haven't used it correctly. What does ##r## stand for in that formula?
 
  • #25
erobz said:
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.
 
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  • #26
Ok, fix that then we will move on.
 
  • #27
erobz said:
Ok, fix that then we will move on.
A=πr^2
78.54 = π * 2.5^2
 
  • #28
BlackPhysics said:
f = force
a = area
p = rho/density
g = gravity
x= displacement
ok, but what force exactly?
BlackPhysics said:
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?
 
  • #29
BlackPhysics said:
A=πr^2
78.54 = π * 2.5^2
No. It doesn't.
 
  • #30
BlackPhysics said:
A=πr^2
78.54 = π * 2.5^2
##\pi## is not equal to 10.
 
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