1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Buoyant Force

  1. Apr 30, 2017 #1
    1. The problem statement, all variables and given/known data
    A 9 cm square wooden post (density=420kg/m3) is 1.3m long and floats in sweater (density=1029 kg/m3. How deep is the post submerged in the water? The picture shows a 9cm x 9cm cross section of the post.

    2. Relevant equations
    Fb=density of liquid x volume of liquid submerged x gravity
    Fmg=density of post x volume of post x gravity
    P=density x gravity x height
    P=Force/area

    3. The attempt at a solution
    I first found the force mg.
    420 x (.09x.09x1.3) x 9.8=.103194 N
    Then I found pressure
    P=.103194/ (.09x.09)=12.74

    I don't think this is right, and I am unsure where to next. Can anyone help me understand how I should be doing this?
     
  2. jcsd
  3. Apr 30, 2017 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Use Archimedes's principle.
     
  4. Apr 30, 2017 #3
    Ok, Archimedes's principle states that the magnitude of the buoyant force=weight of the fluid displaced by object

    Fb=density of fluid x volume of object submerged x g
    =1029 x (.09 x .09 x 1.3) x 9.8
    =106.187 N

    I don't see where to go next.
     
  5. Apr 30, 2017 #4

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Is the entire volume of the wood under water? The answer is probably not because the problem is asking you to find how deep into the water the wood is when it floats. What you calculated is the maximum buoyant force that you can have as in if you pushed the wood entirely under water.

    I also noticed this
    I am sure you meant to say that the post floats in water not sweater. :smile:
     
  6. Apr 30, 2017 #5
    ha! oh yes-I did mean water. Sorry.

    Ok, no the wood is not entirely submerged. But the problem does not given us the percent it is submerged-how do we find that?
     
  7. Apr 30, 2017 #6

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Can you get a number for the buoyant force? If yes, then that is equal to the weight of the displaced water. Hint: The post floats.
     
  8. Apr 30, 2017 #7
    If the post is floating, wouldn't the buoyant force have to be greater? Would we use the density of the wood on the water to find it instead of the density of the water?
     
  9. Apr 30, 2017 #8

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Greater than what? If you place the post on a table, how big is the table force that supports the post? How different from the table force is the buoyant force that supports the post?
     
  10. Apr 30, 2017 #9
    Greater than the gravitational force or m x g. But on the table. the force acting on the post is equal to that of the post acting on the table. In other words, the gravitational force is equal to the normal force. Am I wrong to say that the buoyant force must be greater than the gravitational force? Is it equal?
     
  11. Apr 30, 2017 #10

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    You can answer that yourself. What would happen to the net force on the post if the buoyant force were greater than the gravitational force? How big is the gravitational force anyway?
     
  12. Apr 30, 2017 #11
    I believe the buoyant force is equal to the gravitational force, as the object is at rest.
     
  13. Apr 30, 2017 #12

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Very good. Start with ##BF=mg##. Re-express each side of the equation in terms of volumes and densities. Remember Archimedes's principle.
     
  14. Apr 30, 2017 #13
    Ok, it can either be

    densitywaterx Volumepost x g=masspostxg

    or

    densitywaterx Volumepost x g=densitypost x Volumewater x g

    does this look right?
     
  15. Apr 30, 2017 #14

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    No. Left side (BF) first. Archimedes says the buoyant force is the weight of the displaced water.
    Wdw = mdw g. What is the mass of the displaced water mdw? Answer: It is the density of water ρw, times the volume of the displaced water. You don't know what the volume of the displaced water is, so call it a name and write an expression for the buoyant force. Then we will work on the left side.
     
  16. Apr 30, 2017 #15
    Fb=densityw x Volumedw x g
    =1029 x Vdw x 9.8

    would we set this equal to the density of the post x Volume of post x g?
     
  17. May 1, 2017 #16

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
     
  18. May 1, 2017 #17
    So setting it equal...

    1029 x Vdw x 9.8=420 x (.09 x .09 x 1.3) x 9.8
    Vdw=.0043 m3

    How do we find the height from this? Do we subtract it from the volume of the post?
     
  19. May 1, 2017 #18

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    What does does the submerged volume look like? It has a square 0.09 cm x 0.09 cm base and unknown height h that you are looking for.
     
  20. May 1, 2017 #19
    I know Volume would equal .09x.09xh, but wouldn't this height be referring to be side of the 1.3m? aren't we looking for the part of the .09 that is submerged? the post seems to be laying horizontally in the water, not straight up.

    Also, when I reduce and isolate the height, I obtain .531m or 5.31 cm. The correct answer is 3.67cm
     
  21. May 1, 2017 #20

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    OK, then. You have a rectangle 0.09 m x 1.3 m base and unknown height h that you are looking for. If you don't get the correct answer, you need to show what you did in detail to reduce and isolate the height.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Buoyant Force
  1. Buoyant force (Replies: 7)

  2. Buoyant Forces (Replies: 5)

  3. Buoyant force (Replies: 1)

  4. Buoyant force (Replies: 4)

  5. Buoyant Force (Replies: 36)

Loading...