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Finding tension in a string from a suspended object

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.02 m x 0.01 x 0.03 block of copper (density = 8.8 g/cm3) is suspended submerged in milk (density = 1.03 g.cm3) by a string. How much tension is in the string?

    2. Relevant equations


    3. The attempt at a solution
    My initial thought for an equation was:

    T+Fb(buoyancy)-mg=0
    T+ ρgVdisp-mg=0
    T+60.564-58.8=0
    T=-1.76N

    This is definitely not the right answer, and I'm not 100% sure this is even the right equation. Can anyone please help me figure out what the tension in the string should be?
     
  2. jcsd
  3. Apr 11, 2016 #2

    TSny

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    Gold Member

    Hello and welcome to PF!

    Your set-up looks good. But you haven't used the right numbers. Watch the units.
    If you post the details of your substitution of numbers, we can pinpoint the problem.
     
  4. Apr 11, 2016 #3

    Merlin3189

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    Gold Member

    I would say your method is fine, but you got a bit mixed up with units.
    Your dimensions are given in m, the density in g/cm3 and you appear to have used g=9.8 N/kg
    You don't mention units in your calculation: I'd check.
     
  5. Apr 11, 2016 #4
    This is what I did:

    Fb (buoyancy) = ρgVdisp
    Fb=1.03g/cm^3(9.8)(6cm)
    Fb=60.564

    mg=
    ρ=m/v ⇒ m=ρv
    m= 8.8(6cm)
    m=52.8

    Then 52.8(9.8) = 517.44

    This doesn't look correct either though, so I'm not sure where to go...


    I understand what y'all are saying about my units, but I'm just not sure where to fix it. Physics is DEFINITELY not my forte.
     
    Last edited: Apr 11, 2016
  6. Apr 11, 2016 #5


    This is what I did:

    Fb (buoyancy) = ρgVdisp
    Fb=1.03g/cm^3(9.8)(6cm)
    Fb=60.564

    mg=
    ρ=m/v ⇒ m=ρv
    m= 8.8(6cm)
    m=52.8

    Then 52.8(9.8) = 517.44

    This doesn't look correct either though, so I'm not sure where to go...

    I understand what y'all are saying about my units, but I'm just not sure where to fix it. Physics is DEFINITELY not my forte.
     
    Last edited: Apr 11, 2016
  7. Apr 11, 2016 #6

    Merlin3189

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    Gold Member

    The question gave you mixed units, so it's not like you caused the problem! But you do have to be aware and recognise it when someone lese throws it at you.
    I would say, the first step is to understand that all measurements and quantities have units, then to make sure you always decide what they are and write them down. (You've put units in some places, but not everywhere they're needed.)

    You correctly converted 0.02m x 0.01m x 0.03m into 2 cm x 1cm x 3cm and got 6 cm3 for the volume. I would say that - at least the answer, "volume = 6 cm3"
    Since the density is in g/cm3 you can use that to get the mass in grams. (And write it, "mass = 73.1g" or whatever it comes to.)
    Similarly you can get the mass of water displaced, in grams. (And write, "mass of water = 5.7g" or whatever it comes to.)
    And at that stage, personally I'd just work out the difference in grams. That's what the string has to lift.

    Now, if you feel the need to work out the tension in Newtons, you need to think carefully, because the constant g is usually given as 9.8 N/kg not as 0.0098 N/g. So maybe you should *now* convert your grams into kg, before calculating the Newtons.

    The 9.8 is the error and you used it twice in your calculations. Saving a difficult step like that to the end and doing it just once, might be better.

    I expect the experts on the forum would not agree with me and would approve your "big formula" method. But for people like me who are not very good, I think it is better to work in little steps and make sure I know exactly what I'm doing at each one. To me, each little step is understandable, but a big formula is just gobbledegook I'd have to copy from somewhere else.

    Here, you have the dimensions, so you can work out the volume of the block.
    The block volume and density gets you the mass of the block.
    The block volume and the density of water gets you the mass of water displaced, which causes the buoyant force.
    The difference in these two masses is the amount the string needs to support.
    Whatever mass that is, is in grams, so I convert to kg, then find the Newtons needed to lift it.
     
  8. Apr 11, 2016 #7
    Okay, so I've started from scratch and this is what I've got. Hopefully, it looks correct!

    Volume of cube = .000006m or .0006cm
    Density ρ of cube = 8.8 g/cmcm3 or 8800kg/m^3
    Density ρ of milk = 1.03 g/cm^3 or 1030 kg/m^3
    T=?

    scale +Fb-mg=0
    T + ρgVdisp - mg=0
    T+ 0.060564-.51744=0
    T=.457N



    Fb= ρgV
    1030kg/m^3(9.8)(.000006m)
    .060564=Fb


    mg: m=vρ
    8800(.000006)=.0528 kg

    mg= .0528(9.8)⇒.51744
     
  9. Apr 11, 2016 #8

    SteamKing

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    What are the units of volume?

    Remember, 0.01 m = 1 cm. 0.01 m3 ≠ 0.01 cm3

    If your block of copper measures 0.02 m x 0.01m x 0.03m, what are its equivalent dimensions in centimeters?

    Do you wish to change your answer as to the volume of the block in cm3?
    These density conversions look OK.
     
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