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Homework Help: Buoyancy and Centre of gravity

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A thin wooden plank 1m long is kept on the protruding stone with some of its part immersed in water.
    Will the centre of gravity of the wooden plank be inside the water or outside?


    2. Relevant equations



    3. The attempt at a solution

    For the wooden plank to be in equilibrium, moments of buoyancy force and weight about the point of contact of stone should be equal.
    Buoyancy force acts at the midpoint of the immersed part.
    But what will be the position of the centre of mass (gravity) of the plank?
     

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  2. jcsd
  3. Sep 28, 2010 #2
    How long is the rod?
     
  4. Sep 28, 2010 #3
    1 metre
     
  5. Sep 28, 2010 #4

    rl.bhat

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    The position of the centre of mass (gravity) of the plank is the mid point of the plank.
     
  6. Sep 29, 2010 #5
    then does it mean the torque on both side is equal?
     
  7. Sep 29, 2010 #6
    What I asked was whether it will should lie inside the water or outside to get balanced?
     
  8. Sep 29, 2010 #7

    rl.bhat

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    Assume that x-axis is oriented along the thin plank and area of cross section of the plank 1 cm^2.
    Density of wood = 0.7 g/cm^3 and that of water = 1 g/cm^3
    Let x cm be the portion of the plank in the water.
    Length of the plank is 100 cm. So the mass of the plank = 70 g.
    Mass of the displaced water = x g. Then buoyancy force is proportional to x.

    If R is the reaction of the stone, R = (70 - x)
    Take the moment about the end of the plank out side the water.

    R*10 + x(100 - x/2) = 70*50

    Solve the quadratic. If x is more than 50, C.G. is inside the water. Otherwise it is outside the water.
     
  9. Sep 30, 2010 #8
    How did you get R = 70-x ?
     
  10. Sep 30, 2010 #9

    rl.bhat

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    I have assumed that the component of the weight of the plank perpendicular to x-axis is

    L*d*g*cosθ. Length of the plank is 100 cm, density 0.7 g/cm^3

    Component of the weight of the displaced liquid = x*1*g*cosθ

    For equilibrium, total downward with respect x-axis must be equal to the total upward force. Weight is in the downward direction and buoyancy and the reaction are in the upward direction.
    Hence R*g*cosθ = 70*g*cosθ - x*g*cosθ
     
    Last edited: Sep 30, 2010
  11. Sep 30, 2010 #10
    Thanks!
     
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