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Buoyant force of children

  1. Nov 13, 2006 #1
    Three children, each of weight 356 N, make a log raft by lashing together logs of diameter 0.30 m and length 1.80 m. How many logs will be needed to keep them afloat in fresh water? Take the density of the logs to be 800 kg/m^3.

    I'm interpreting this problem as a buoyancy problem for starters.
    I have:
    B=1068 N
    V(obj)= 0.127
    V(liq)= ? is this the same as the obj?
    p(obj)= 800kg/m^3
    p(liq)= 1.0e3 kg/m^3

    I'm thinking I need to find the density of one log based on this information and divide the given density of all the logs by this. However I'm having a hard time finding a way to get to this using the equations for buoyancy. For a floating object the equation I was originally going to use was
    p(liq)gV(liq)=p(obj)gV(obj) but I'm not sure what if the Volume of the liquid is the same as for the object since V is the volume of the fluid displaced from the object. Any guidance on this problem is greatly appreciated.
  2. jcsd
  3. Nov 13, 2006 #2


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    If the object is completely submerged, then the volume displaced is equal to the volume of the object.

    You're given the density of the logs in the question statement.
  4. Nov 13, 2006 #3


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    I think you will do better thinking of the children and all logs as the floating object, so its weight is the combined weight of children and logs. Assuming the children wish to stay dry, the volume of displaced water will be no more than the volume of the logs.
  5. Nov 13, 2006 #4
    Okay so I took Dan's advice and added up children and logs to get a total force. What I did was I used D=M/V and solved for M to get the mass of the logs being 101.6kg. Then used F=mg to get the force which is 995.7N and added this to (356)(3)=1068N to get 2063.7N. I used the equation F= p(obj)gV(obj) and solved for p(obj) is this right?
  6. Nov 13, 2006 #5


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    You are on the right track finding the weight of a log. The children can use as many logs as they need. You can calculate the fractional number of logs needed and then round up to get the answer. Think in terms of n logs instead of just one. You want the weight of n logs plus the weight of the children to equal the weight of water having the same volume as n logs. That way, only the logs will get wet.

    You can think of the log/children combination as having an "effective density" by assuming the children have weight but no volume; their weight gets added to the weight of the logs, but the volume is just the volume of the logs. In order to float, the usual condition needs to be satisfied, which is: The _________ of the object must be ___________ the ___________ of water.
  7. Nov 15, 2006 #6
    The equation in my book is V(liq)/V(obj)=p(obj)/p(liq) and by the way I'm reading your explanation I feel like it's a proportion problem. However I'm not sure how to fit weight into a similar equation, unless there's something I'm forgetting. So the weight of n logs plus children is 995.7n+1068N which has to equal the weight of the water in order to stay afloat. The volume of water is equal to 0.127n (from pie (r)^2*h). How do I use all of this to solve for the number of logs? It makes sense step by step but I just can't seem to put it all together.
  8. Nov 15, 2006 #7


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    You can find the weight of the water from its volume and density. n has to be big enough so that the weight of displaced water is at least as much as the weight of the n logs plus children.
  9. Feb 6, 2010 #8
    how do you know this: V(obj)= 0.127 ? and is this the volume of the logs or of the children?
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