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Help with calcuating force constant ?

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data

    A 320 kg wooden raft floats on a lake in equilibrium. When a 75 kg man steps carefully onto the raft, it floats 3.5 cm deeper into the water. When he steps off, the raft oscillates for a while.

    (a) The buoyant force of the water on the raft is a restoring force that is directly proportional to how deep the raft sits in the water. In this way, the water acts like a mechanical spring. Calculate the force constant of this spring.

    2. Relevant equations

    3. The attempt at a solution

    I am not sure how to get the force constant out of this, as if it were a spring.

    k = f/displacement ?
  2. jcsd
  3. Jan 10, 2012 #2


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    Homework Helper

    So what is the force the man exerts on the raft (and hence the water) by stepping onto it? This is exactly balanced by the buoyant force. Express this in N (Newton).

    And what is the displacement of the raft in the water caused by that force? Express this in m (metre).

    After you work out what those two are, just divide as per the equation you quoted. The units are [itex]Nm^{-1}[/itex]
  4. Jan 10, 2012 #3
    so, 75kg * 9.8 = 735 J


    735j = (x)(0.035) solve for x gives me 21,000 NM

    is that correct?
  5. Jan 10, 2012 #4


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    Homework Helper

    Why J (Joule) as the unit for force? Remember, weight = mg, and this is a force measured in Newton. Potential energy due to Earth's gravitation = mgh, and this is energy measured in Joule. Big difference.

    So the weight of the man is mg = 75*9.8 = 735N (Newton).

    The displacement x is 3.5cm = 3.5/100m = 0.035m.

    Hence the force constant k = F/x = 735/0.035 = 21000N/m (Newton-per-metre, not the slash signifying division between Newton and metre. Instead of the slash one can also write [itex]Nm^{-1}[/itex] (and many modern texts prefer this notation). This is NOT the same as Nm (Newton-metre, obtained by multiplying Newton and metre, and which is a unit of energy equivalent to Joule).

    Please be careful with your units. Wrong unit = wrong answer. Other than that, your figures are fine.
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