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Totally lost, buoyant force

  1. Dec 18, 2011 #1
    1. A juniper wood plank measuring 7 ft by 0.5 ft by 0.1 ft is totally immersed in water. What is the net force acting on the plank? ()



    2. a) F top = P top * A top(l x w) ; b) P bottom = P top + (Dw * h); c) F bottom = (F top) + (Dw*l*w*h); d) Fb = F bottom - F top e) Weight = Dw * l * w * h; f) F net = Fb – weight;



    3. I believe the above equations are all I need to solve this problem and in that order. I believe my stumbling blocks are the units.

    a. I converted the area from feet to inches and formulated F top to be 14.7 lb/in^2 * 504 in^2 leaving 7408.8 lbs of force (the in^2 cancels out, right?)

    b. This is where I get lost. The Dw of Juniper wood is 35 lb/ft^3. Since the height is .1 feet, I don't have to convert, so I just multiply 35 lb/ft^3 * .1 ft leaving 3.5 lb/ft^2. This is where I am lost. I have 7408.8 lbs + 3.5 lb/ft^2. So I am left with 7412.3 lb/ft^2 which cant be right. In step a I had in^2 and canceled them out. Now what?
     
  2. jcsd
  3. Dec 18, 2011 #2
    Yeah, never mind. I was totally off on a tangent here with some crazy formula in my text. It is as simple as calculating the weight = Dw of the wood * V = 35 * .35 = 12.25lb. Then buoyant force = Dw of the water * V = 62.4 * .35 = 21.8lb. Fnet = Fb - W = 21.8 - 12.25 = 9.55 lb. Easy once it's laid out that way. Hope this helps someone else out there some day.
     
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