University of Wisconsin Green Bay

Two children (with masses of 32 kg and 35 kg) are pretending to be pirates on a wooden raft in a freshwater lake. The raft is 1.8 m x 2.0 m x 0.18 m and it is made of pine which has a density of about 560 kg/m3. If the children hang two 25 kg “treasure chests” underneath the raft, will they get their feet wet? Each chest has a volume of 0.021 m3.

  • In this problem, you are asked to determine whether or not the children will get their feet wet. This depends, of course, on whether or not the raft will sink—in other words, on whether or not it is buoyant enough to float while carrying the children and the chests. So with a little thought, you can see that you are asked to relate motion (the floating or sinking of the raft and its occupants and cargo) to force (buoyant force).

    Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem.

  • Objects in a Horizontal Circle Objects in a Horizontal Circle

    Note that there is a great deal of detail in this problem. You don’t need to worry about any of the numbers (masses, densities, volumes, etc.) until you actually fill into the algebraic equation that comes out of Newton’s 2nd Law. At this stage, you only want to identify the best system to consider and the types and directions of forces on that system.

    Fb Fb Fb gravity Fb Fb Fb gravity
  • ΣF=ma

    The key equation for any problem that relates forces and motion is Newton’s Second Law. Regardless of what quantity you are asked to find, begin with the Second Law. If additional information is needed, it will become apparent as you proceed.

  • At this point, we can say that if the raft has a volume of at least 0.44 m3, the raft and chests will displace enough water for the entire system to float. The total volume of the raft is (1.8 m x 2.0 m x 0.18 m) = 0.65 m3, so the raft will float and the children will not get their feet wet.

    Although no further information is requested in this problem, some similar problems will ask you to go a step further and determine how much of the raft is above water, which just requires one additional step to divide out the area.

    Fb Fb Fb 0 0 MtotalAy Fb Fb Fb 360 kg/m^3 kg/m^3 g g g g .021m^3 .021m^3 m^3 m^3 ay Fy .44m^3

  • ΣF=ma
    FB1 + FB2 + FB –mtotalg = 0

    In this problem, the upward buoyant force of the water acts against the downward weight of the raft, the children, and the chests. Buoyant force depends on the density of the fluid and the volume of fluid that is displaced by the object in it. Part of the raft and all of the chests are under water, and so the buoyant force depends on the total volume of the chests as well as the submerged volume of the raft.

    We were not told if the raft is able to float. By assuming a floating raft (a = 0) we could solve for the volume of the raft that would be required to stay afloat. That volume shows up in buoyant force on the raft (FB raft = ρfluidVsubmerged part of the raft g) and was found to be less than the total volume of the raft and so it is able to stay afloat.