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Mass of water displaced by a block

  1. Jul 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A block is suspended from a scale and then lowered into a bucket of water. The density of the water is 1 gm/cm3. The initial reading on the scale is 19N and one complete revolution of the scale is a change of 10N. (The block is completely submerged after one complete revolution)
    What is the mass of the the water displaced by the block? What is the volume of the block? What is the density of the block?

    2. Relevant equations

    Magnitude of buoyant force = weight of displaced fluid.
    The mass density is the mas of the substance divided by its volume.

    3. The attempt at a solution

    I found the mass of the block when completely submerged in water to be .92kg by [(Initial reading-final reading)/gravity]=[(19N-10N)/9.8]
    I'm not sure how to use this information to go about finding the mass of the water displaced by the block.
     
  2. jcsd
  3. Jul 20, 2011 #2

    tiny-tim

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    Welcome to PF!

    Hi Jgoshorn1! Welcome to PF! :smile:

    You know that the weight of the block is 10 N.

    So call the volume of the block V cm3.

    What is the weight of the water displaced? :wink:
     
  4. Jul 20, 2011 #3
    Hi tiny-tim.
    After staring at this problem for a little while long I decided that the mass I had previously calculated for the block when it is completely submerged would be 19N/9.8=1.95 rather than the (19N-10N)/9.8 that I previously thought.
    So the apparent weight when submerged is 10N and volume=mass/density. I'm not sure how to find the volume of the block with the information I've been given. I'm super confused.
     
  5. Jul 21, 2011 #4

    tiny-tim

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    Hi Jgoshorn1! :wink:

    I'm not really understanding your terminology :confused:

    the mass of the block is fixed, it's the same whether it's in air water or outer space.

    Call the volume V (so the density is m/V) …

    then how much would the block weigh (as shown on the scale) when submerged? :smile:
     
  6. Jul 21, 2011 #5
    Hi!
    So I worked on this problem for a very long time last night and I still wasn't getting an answer provided we can chose from [1.1 kg, 1.6 kg, 2.4 kg, 3.5 kg, 4.8 kg]

    The way I tried it was:
    Fnet=0 because it's in equilibrium
    I made a FBD with Force(gravity) pointing down, Force(buoyancy) and Force(Tension from the scale) pointing up.
    (Fb+Ft) - (Fg) = 0
    Fb = m(block)g - Ft
    Fb = (1.94 kg)( 9.8m/s^2) - 9N
    Fb = 10.012
    Fb = m(water)g
    10.012 = m(water)(9.81)
    10.012/9.81 = m(water) = 1.02kg

    Does this make sense? I can't figure out any other way to really do it... It is part of an online lab (and we can discuss labs on our discussion posts but nobody seems to be getting an answer provided) and there is a really simple animation that goes with it where the block is lowered into the water attached to the scale.. the scale goes backwards through one complete revolution (started at 19N ending at 9N).. the water level rises in the animation as to be expected and I even tried to calculate the volume of the water displaced from taking measurements on the x-y coordinate system but it doesn't make any sense because if you look at it, when you put the block (aprx 4x1) into the water the water level raises by aprx 10x1. Also there are no given units so that would all just be an assumption that each grid block on the animation plane is 1cm x 1 cm x 1 cm.

    Super frustrating.
     
  7. Jul 21, 2011 #6

    I did that:
    Magnitude of buoyant force = weight of displace fluid
    Fb = m(water)g
    10.012 = m(water)(9.81)
    10.012/9.81 = m(water) = 1.02kg

    The problem is that 1.02kg wasn't a given option and I wanted to know if the way I was doing it was wrong or not.. Or if there is any other way of doing it.
     
  8. Jul 21, 2011 #7
    Sorry... somehow I completely missed your post #5 when I was responding.

    Yes, that looks correct for the mass of the fluid. As for why it doesn't match up with one of the answers provided, that usually indicates a rounding problem somewhere: either you are rounding somewhere, or the creators of the lab are rounding somewhere that you aren't.

    Edit: I caught my error and attempted to delete the offending post, but managed in doing so to cross my deletion with Jgoshorn1's reply. :( Yikes. The full text of the missing post is as he quoted. Sorry for making a mess of this thread.
     
    Last edited: Jul 21, 2011
  9. Jul 21, 2011 #8
    It's okay, I wasn't offended. I emailed my prof to figure out whats going on with this problem.. I don't care if I am getting it wrong or miss the credit for the problem.. I just want to know how to do it correctly!

    Until then I'm struggling with the second/third question: What is the volume/density of the block.
     
  10. Jul 21, 2011 #9
    As you stated in post #1:

    "The mass density is the mass of the substance divided by its volume."

    What is the mass of the water? What is its density?

    Given the volume of the displaced water, what is the volume of the block?
     
  11. Jul 21, 2011 #10
    Mass of water = 1.02kg = 1020 g
    Density of water = 1gm/cm^3
    v(water)=m/d = 1020g/(1g/cm^3) = 1020 cm^3 = volume of water = volume of block?

    The options to chose from are [750 cm^3, 850 cm^3, 1100 cm^3, 1425 cm^3,2400 cm^3]

    The volume of the displaced water should = the volume of the block, right? Am I not converting something correctly? Also, on the lab the prof writes "Density of water = 1gm/cm^3" I'm not sure what the unit gm/cm^3 is.. I figured he just meant to write g/cm^3.
     
  12. Jul 21, 2011 #11
    The density of water is one gram per cubic centimeter, so "gm" should just mean gram.

    Your work looks correct.

    Last step. Given the volume of the block, how do you get its density?
     
  13. Jul 21, 2011 #12
    My work looks correct and I don't know any other way to do it but I'm not getting it to match (or even come really close to) any of the other answers.

    Mass of block = 1.94kg = 1940g
    For density of block:
    d=m/v = 1940g/1020cm^3 = 1.90g/cm^3

    Answers to chose from: [1.2 gm/cm^3, 1.8 gm/cm^3, 2.4 gm/cm^3, 3.5 gm/cm^3, 4.8 gm/cm^3]

    So, given my previous posts; for mass of water displaced would you chose 1.1kg, for volume of block would you chose 1100cm^3, for density of block would you chose 1.8gm/cm^3???
     
  14. Jul 21, 2011 #13
    "Coming close" is relative. Like I said above, my suspicion is that you have either a rounding mistake or a measurement mistake somewhere early in your work, which is throwing things off down the line. Given your initial measurements, your math looks correct.
     
  15. Jul 21, 2011 #14
    Thanks!

    I went back and re-did the volume and density of the block using 1.1kg as the weight of the water displaced and I got exact answers.

    I appreciate your encouragement!
     
  16. Jul 21, 2011 #15
    Sure thing.

    As always, the general trick involved in problems like this is writing down all of the various relationships you know about involving the quantities you are given and the quantities that are being asked for and finding a chain of relationships which gets you from point A to point B. Often times, the specific numbers given by the problem get in the way of finding the chain of relationships you want; that seems to be the source of your frustration here.

    So, as general advice, forget the numbers. Think about the relationships you need, and plug in the numbers as a last step. Thinking about things this way helped me when I took undergrad physics (and often, in the higher level classes, you aren't even given the numbers, so this is the way you are required to think).
     
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