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Leaning Tower of Pisa

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is as attached.


    3. The attempt at a solution

    I tried doing double integration of r from 0 to 9.8 and θ from 0 to 2pi. But here im assuming all forces from the soil act along the tower (no friction) and managed to solve C1.

    But im not sure how to solve C2?
     

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    Last edited: Aug 29, 2012
  2. jcsd
  3. Aug 29, 2012 #2

    Simon Bridge

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    It helps if you show us how you have been trying.
    Can you not find the max and min locations without knowing C2?
     
  4. Aug 29, 2012 #3
    What are the equilibrium conditions?
     
  5. Aug 29, 2012 #4
    1) Vector Sum of forces about any direction = 0

    2)no resultant moment about any point
     
  6. Aug 29, 2012 #5
    That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.
     
  7. Aug 29, 2012 #6
    Yes, but the cos θ becomes sin θ and then from 0 to 2∏ it cancels out to be zero..
     
  8. Aug 29, 2012 #7
    Here's what I got:

    Let infinitesimal moment be dM.

    ∫ dM = ∫ r dF = ∫ rP dA = ∫∫ r2P dr dθ...

    Then when ∫cosθ becomes sin θ the limits going from 0 to 2∏ it becomes 0, so C2 cannot be found..
     
  9. Aug 29, 2012 #8
    Observe how the second term of pressure behaves. Its moment about O must be non-zero.
     
  10. Aug 29, 2012 #9
    what about..
    you know that Pmax is right underneeth the center of gravity which is r=2.67 and θ=0
    and also Pmax/min is the second derivative of function p..which i suppose is
    [itex]\partial[/itex]^2p/[itex]\partial[/itex]r^2 * dr+ [itex]\partial[/itex]^p/[itex]\partial[/itex]θ^2 * dθ
    and your only variable is C2
     
  11. Aug 29, 2012 #10
    And with that, you only get r = 0... when u equate each = 0 ..
     
  12. Aug 29, 2012 #11
    Like I said, this cannot be zero. The specific mistake you are making is that of ignoring the vector nature of r and F. What you denote as dM is a vector product of r and dF. Work it out.
     
  13. Aug 29, 2012 #12
    yes, dM = r x F

    but I assumed all F acts upwards, so vectors r and F are perpendicular so it simply becomes r*F...
     
  14. Aug 30, 2012 #13
    Draw the resultant moment in a few different places and you will see different directions. You can't add them together as scalars.
     
  15. Aug 30, 2012 #14
    That is true...I realized that only for the same angle they are pointing in the same direction

    So how do we find the net moment? It's like trying to sum up all the vectors in different directions..
     
    Last edited: Aug 30, 2012
  16. Aug 30, 2012 #15
    Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?
     
  17. Aug 30, 2012 #16

    d[itex]\overline{M}[/itex]
    = [itex]\overline{r}[/itex] x d[itex]\overline{F}[/itex]
    = ([itex]\overline{r}[/itex] x [itex]\overline{p}[/itex]) dA
    = (r cosθ [itex]\widehat{i}[/itex] + r sinθ [itex]\widehat{j}[/itex]) x (C1 + C2 r0.625 cosθ)[itex]\widehat{k}[/itex]
    = -( C1 cosθ + C2 r1.625 cos2θ) [itex]\widehat{j}[/itex] + (C1 r sinθ + C2 r1.625 sinθcosθ ) [itex]\widehat{i}[/itex]


    Along j

    [itex]\int[/itex] d[itex]\overline{M}[/itex]
    = - [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [itex]\int[/itex][itex]^{2∏}_{0}[/itex] dθ [C1 r cosθ + C2 r1.625 cos2θ]
    = -478.9 C2

    Along j
    Everything cancels out to be zero!
     
    Last edited: Aug 30, 2012
  18. Aug 30, 2012 #17
    Do the vector product :)
     
  19. Aug 30, 2012 #18
    I am sure you can integrate that now.
     
  20. Aug 30, 2012 #19
    Already did! (Page 1 please) Sorry for the long edit....was trying to figure out how latex works
     
  21. Aug 30, 2012 #20
    So, have you solved the entire problem?
     
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