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Buoyancy problem from NEXT STEP

Discussion in 'MCAT Study Question Q&A' started by nai54, Aug 2, 2015.

  1. nai54

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    The question asks:

    What is the tension on the rope in the diagram assuming the density of water is 1g/cm^3, the density of the object is 2g/cm^3 and its volume is 10cm^3. The picture is a ball completely submerged in water with a string attached to holding it in the water.

    The answer is 0.19N but I cannot get to the answer :(

    I started off by doing Tension=mg- Fb--> T=density(o)V(o)g-density(f)V(o)g and plugged in the numbers. Am I doing this wrong?
     
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  3. theonlytycrane

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    The object is more dense than water so it would sink if it wasn't attached to the string holding it in the water.

    The volume of the object is 10 cm^3. 10 cm^3 * 2g / cm^3 = 20g = .02kg. The Force of gravity on the object (down) is .02kg * 10 m/s^2 = .2N.

    The buoyant force of water = mass of water displaced * gravity. Water is incompressible so the mass of water displaced = volume of the object * density of water = 10 cm^3 * 1g / cm^3 = 10g = .01kg. The buoyant force (up) is .01kg * 10 m/s^2 = .1N.

    We have .2N down and .1N up, so the tension in the string should be .1N (up) to keep the object from sinking.

    I'm not sure how the key got 0.19N :confused:
     
    #2 theonlytycrane, Aug 2, 2015
    Last edited: Aug 2, 2015
  4. nai54

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    hmm yes that makes sense. That answer seemed a little off it might be a mistake. Does it matter that the units are g/cm^3? Should it be converted to kg/m^3 since g is m/s^2?
     
  5. theonlytycrane

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    The units for volume of the object cancel with density but you're right- Newton = kg * m / s^2. I need to convert g to kg. Editing my post above. Nice catch :)
     
    #4 theonlytycrane, Aug 2, 2015
    Last edited: Aug 2, 2015
    nai54 likes this.
  6. PharmHoppe

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    Just to add to this:

    For a submerged object: Weight = buoyant force + Normal force. So tension on rope= N

    Here, 0.2 = 0.1 + N. and N = 0.1
     

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