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Homework Help: Fluid Statics: Find vertical and horizontal forces in tank

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data

    The tank shown in the figure is filled with water and fastened to the ground. There is a small opening on top of the semi-spherical part of the tank.
    a) What is the magnitude of the horizontal force on the screws? Explain using a free body diagram.
    b) Consider the proper area element (dA) and determine the vertical force in the screws. Assume the weight of the container can be neglected.


    2. Relevant equations

    p = ρ*g*h
    FH = pc dA = ∫Ap dA
    FV = ρg∀ or FV = dFy
    (∀ = Volume)

    3. The attempt at a solution
    Note that you are only given what is seen in the image, and the tank is completely filled.

    a) FH = pc * A = ∫p dA, p = ρgr, A = 1/2 π r2, dA = 2πr dr
    0R 2ρgr2 dr = (ρgπR2)/4
    I then incorrectly defined h in my free body diagram and used that to calculate FH for the lower part of the tank, which is apparently not needed.
    The solution provided for this problem is that all FH will cancel out, so FH = 0 (this is the entirety of the solution provided, no other information was given). Why is this?

    b) FV = ρg∀, ∀=(4/3)πR3 * (1/2) = 2/3πR3, FV = (2/3) ρgπR3. I came up with this solution after once again incorrectly defining h for the height of the cylinder part of the tank, but why is h not needed here? Why is there no vertical force in the screws from the cylinder?
    The solution provided is seen below, with the correct answer being FV = (1/3) ρgπR3 (warning: large image). Why is the integration from 0 to π/2, and not 2*∫0π/2?

    Any help is very much appreciated.
    Last edited: Dec 4, 2014
  2. jcsd
  3. Dec 5, 2014 #2
    the cylinder is hollow and filled with water which applies a perpendicular horizontal force on it only . there is no vertical force due to the cylindrical water . net Horizontal force due to cylindrical part is zero as every small force dF applied on one side (say towards the right of the figure) is cancelled out by the diametrically opposite side's dF of the same magnitude(that is the force towards the left ) .
    hence force on screws will be determined by the spherical part only as it is given in the solution .

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