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Brain blocks pt II

  1. Sep 5, 2004 #1
    I have a box when suspended from a cable in a vaccuum is 4450N and is .608m wide (it's also a cube).

    W = 4450N = T1 (tension on the cable)
    V = 0.2248m^3

    When I suspend it into a liquid, L/2 m from the surface, of density 944 kg/m^3, I want to know the buoyant force, the new tension of the cable, the force pushing down from the top and the force pushing up from the bottom from the liquid the box is suspended in.

    L/2 = 0.608m / 2 = 0.304m

    I have the buoyant force as being T1 - T2 = Fb = D( V )g (of the liquid)

    944kg/m^3 * 0.2248m^3 * 9.8m/s^2 = 2080N = Fb

    T2 = T1 - Fb = 4450N - 2080N = 2370N = T2

    But how do I find the forces from the liquid above and below the box?

    Thanks for any help!
     
  2. jcsd
  3. Sep 5, 2004 #2
    D*g*h = the pressure acting on the top or bottom of the box at a given height "h"

    i assume L/2 means half the box's length.

    you can go D*g*(L/2) = pressure on top

    you also know that P = F/A

    solve for F = P * L^2

    That should be the force on top. Do the same for the bottom, except use (3L/2) as the height and those should be the answers.
     
    Last edited: Sep 5, 2004
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