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Buoyancy Problems, Slight Shortcut May 8, 2009

Posted by Chris Sullivan in Training.
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A typical buoyancy question on a diving exam goes something like, A 440 pound object that displaces 4.5 cubic feet of water is on the bottom in 56 feet of salt water. How much water must be displaced by a lift bag to bring it to the surface?

The usual way of solving this problem is to:

  1. Calculate the weight of water displaced by the object, which is 4.5 cubic feet x 64 pounds/cubic foot which equals 288 pounds. Note that if you write out the units (cubic feet and pound/cubic foot) you can see that the volume units, cubic feet, cancel each other out leaving the answer in pounds. So by including the units in your calculation you can make sure your answer makes sense.
  2. Subtract the positive buoyancy from the weight of the object, so 440 pounds less 288 pounds = 152 pounds net. Underwater the object feels like it weighs 152 pounds.
  3. Divide the 152 pounds by the density of sea water, 152 pounds / 64 pounds/cubic foot, so the answer is 2.375 cubic feet. Note that once again the units cancel out, pounds divided by pounds cancels out, dividing twice by cubic feet is the same as multiplying (i.e. y/y/x = x) so the answer is in cubic feet.

This is simple enough, but can be made even simpler. All you have to do is figure out the total displacement of water equivalent to the weight of the object, then subtract the volume.

  1. Calculate weight of the object in cubic feet of sea water: 440 pounds divided by 64 pounds/cubic feet is 440/64 = 6.875 cubic feet.
  2. Subtract the volume of the object: 6.875 cubic feet – 4.5 cubic feet = 2.375 cubic feet.

Same answer, one less step! By the way, giving the depth of 56 feet is a red herring, it makes no difference. Sometimes these questions are phrased as “how much air would you have to pump from the surface?” and in this case the depth is important, and you would have to multiply your answer by (depth+33)/33, in our case that would make the answer 6.41 cubic feet.

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