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Buoyancy and Tension

  1. Dec 8, 2006 #1
    I'm not very good at fluids.. and I was wondering if someone could check my work if they had time ^^;;
    What I did makes a lot of sense to me though.. so usually that's a sure sign that I did it wrong.
    Thanks for your time ^^;

    1. The problem statement, all variables and given/known data

    A 10 kg piece of lead is completely submerged in water and suspended from a cable. This takes place at sea level at 4*C. What is the tension in the cable?

    2. Relevant equations

    F_b = W_fluid
    That is, the force of bouyancy equals the weight of the fluid that the lead is submerged in.
    At 4*C water has a density of 1000 kg/m^3.

    3. The attempt at a solution

    Well.. I know tension is acting on the cable (upward) and the force of the bouyancy is (upward) as well. The weight of the lead is (downward). There should be no acceleration. Therefore:

    F = ma = T + F_b - mg
    F = 0 = T + F_b - mg
    T = -F_b + mg
    T = -1000 kg/m^3 + (10 kg)(9.8 m/s^2)
    T = -1000 kg/m^3 + 98 kg-m/s^2
    T = -902 N
    :redface:
     
  2. jcsd
  3. Dec 8, 2006 #2

    Doc Al

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    This is correct: The upward buoyant force equals the weight of the displaced fluid. But how much fluid does the lead object displace? Hint: What's it's volume?

    Good.
    So far, so good.
    What you put for F_b is density of water, but what you need is the weight of the displaced water.
    Sanity check: How can the string tension be greater than the block's weight? (And negative, to boot!)
     
  4. Dec 8, 2006 #3
    The volume of water at 4*C should be 22.4 L.
    Therefore, T = -22.4 L + (10 kg)(9.8m/s^2)
    T = 75.6 N

    Now.. is the volume of water necessarily its weight, or how much water it will hold?
     
  5. Dec 9, 2006 #4

    OlderDan

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    You need the volume of the displaced water, which is the same as the volume of lead. To find the volume of lead you need the density of lead and its mass. You have the mass; look up the density. The weight of displaced water is found from its density and volume. 22.4L is the volume of 1 mole of an ideal gas. That has nothing to do with this problem.
     
  6. Dec 9, 2006 #5
    Oh okay.. I had it in my notes and for some reason I thought it was the volume of water X_X;

    So the density of lead is 11300 kg/m^3
    In order to cancel out units, I can do (11300 kg/m^3)(1/10 kg) = 1300 m^3 of lead. I think that is its volume..
    So this volume must be the weight of the displaced water? Or should I further multiply by the density of water?
     
  7. Dec 9, 2006 #6

    vanesch

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    It seems to me that you are randomly putting together the data you have in order to obtain other numbers... Can you figure ? 1300 m^3 for a block of lead of 10 kg ? 1300 m^3, that's the volume of a BIG HOUSE ! And if it is made of lead, it would then weight 10 kg :bugeye:

    Somewhere else you write:
    First of all, why should the volume of water be 22.4 L ?? You know what 22.4 L is ? It is the molar volume of a perfect gas in normal situations. Water is not a gas, and we don't have one mole.

    But worse, after that, you ADD that VOLUME to A WEIGHT to produce a force :grumpy:

    Honestly, putting randomly some data in some mathematical operations doesn't usually produce anything sensible...

    The force of boyancy, as given by Archimedes, is the WEIGHT of the displaced water. To know the WEIGHT, you need:
    - its volume (and the volume is that of the block of lead)
    - its density
    ==> those two together give you its MASS
    - the gravitational acceleration
    ==> with the mass, it gives you the weight, which you are after.
     
  8. Dec 9, 2006 #7
    The density of lead is 11300 kg/m^3
    The volume of lead should be then 1130 m^3 because you are given that it weighs 10kg... and units cancel out. But this is wrong though.... I dont understand how I am supposed to find the volume of the lead with no other data given other than the lead is 10kg and it's 4*C. Is it just a certain quantity that is supposed to be known? I'm assuming that once you get the volume of lead, if you multiply it by the density then you get the mass, like you said, and then multiplying it by the gravitational acceleration, 9.8 m/s^2, to get the weight of the fluid (water) which is the force of bouyancy
     
  9. Dec 9, 2006 #8
    I had also tried
    (10 kg)/(11300kg/m^3) to get the m^3 on top, but I got a really small number (8.85 x 10^-4 m^3) and assumed it was way too small..
     
  10. Dec 9, 2006 #9

    OlderDan

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    It is not enough to canel the kg units. You must have the correct units left after the kg cancel

    (kg/m³)/kg = 1/m³

    1/m³ is NOT a unit of volume; it is a unit of 1/volume

    Try starting from a definition

    D = m/V
    V = m/D = (10 kg)/(11300kg/m^3)

    So what if it's a small number? Think how big 1m³ really is. What is that number 11300kg/m^3 telling you about the mass of a lump of lead that has a volume of 1m³?
     
    Last edited: Dec 9, 2006
  11. Dec 9, 2006 #10
    What is that number 11300kg/m^3 telling you about the mass of a lump of lead that has a volume of 1m³?

    Oh.! It's telling you that all of the 11300 kg can be held in only that small amount of space.. then getting 8.85 x 10^-4 m^3 as an answer seems to fit in!

    According to what vane said:
    The force of boyancy, as given by Archimedes, is the WEIGHT of the displaced water. To know the WEIGHT, you need:
    - its volume (and the volume is that of the block of lead)
    - its density
    ==> those two together give you its MASS
    - the gravitational acceleration
    ==> with the mass, it gives you the weight, which you are after.

    In order to get the mass, I do (8.85 x 10^-4 m^3)(11300kg/m^3)(9.8 m/s^2) = 98 N for the weight
    But then -98N + 98N = 0.. could there possibly be 0 N of tension?
    Thinking again.. the string could not affect the lead because the force of bouyancy is keeping it up just enough for it to stay in its place but for the string to not affect it..

    Or maybe I should have added gravity instead of multiplied it to get 88 N, as in the original equation
     
    Last edited: Dec 9, 2006
  12. Dec 10, 2006 #11

    vanesch

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    This joins my remark earlier: it is not because applying a formula gives you a number which you don't like for a good or bad reason, that you have the right to invert the formula ! Try that to your bank account :tongue2:
     
  13. Dec 10, 2006 #12
    Okay ^_^;;
    I will keep that in mind.. Thank you.

    But using that number, I could multiply it with the density to get the weight, which would be 10 kg and what was given. Is this the force of bouyancy, or do I need to also multiply the 9.8 for the gravity to gte the weight in N. Then, it would be -98 N and adding that to the weight of the lead would give me 0 N for the tension.. is that possible?
     
  14. Dec 10, 2006 #13

    vanesch

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    That number was the volume of your block of lead, right ?
    Now, the force of boyancy is not given by the weight of the displaced lead, but by the weight of the displaced water. So the second time you need a density, it is the one of the water (which is 1000 kg/m^3), because you need the mass of the displaced WATER.

    That is possible if the block were immerged in liquid lead: then the force of boyancy would be equal to its weight (that's in fact the calculation you did: you calculated the boyancy using the density of lead, so it is the weight of the displaced lead (in a container of liquid lead) that you then have to use. But we're putting the block of lead in WATER, so now we have to use the density of WATER to find the force of boyancy.
    (but not of course, to find the volume of the block of lead: that was correctly done with the density of lead).
     
  15. Dec 10, 2006 #14
    Okay.. the volume of lead inside the density of water...

    (8.85 x 10^-4 m^3)(1000 kg/m^3) = .885 kg. that is the mass of the lead in the displaced water. Therefore, it's "weight" has to be 98 times that, or 86.73 N, which is the force of bouyancy
    -86.73 N + 98 N (the weight of the block in the water) = 11.27 N! That looks like it would work ^_^! Because I found the mass of the lead inside the displaced water, and then added the mg which was derived in the free body diagram..

    ^^ Thanks for your help
     
  16. Dec 10, 2006 #15

    vanesch

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    9 POINT 8 times ...
     
  17. Dec 10, 2006 #16
    Oh ^_^;; Okay 9.8 times, I just must have not hit the period key hard enough.. Thanks for warning me ^^;;
     
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