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Homework Help: 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?
     

    Attached Files:

    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?
     
  22. Aug 30, 2012 #21
    I found C1 and C2;

    C1 = 477 300
    C2 = 799 832

    which gives:

    pmin = -2853243
    pmax = 3807843

    How can there be negative pressure? It doesn't make any sense... (Given the initial assumption that all forces act upwards)
     
    Last edited: Aug 30, 2012
  23. Aug 30, 2012 #22
    Your answers are obviously incorrect, pressure cannot be negative. I guess the equation of forces or torques is incorrect.
     
  24. Aug 30, 2012 #23
    I can't see what's wrong with my equations for torque?

    But here's what I used for my force equations:

    dF = p dA

    dA = r dr dθ

    [itex]\int[/itex] dF
    = [itex]\int[/itex][itex]\int[/itex] pr dr dθ
    = [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [itex]\int[/itex][itex]^{2∏}_{0}[/itex] dθ [C1r + C2r1.625 cosθ ]
    = 2∏ [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [C1r]
    = (9.8)2 ∏ C1
    = 301.7 C1


    Sum of forces in z-direction must be zero:

    301.7 C1 = 144*106
    C1 = 477 300


    Not sure what's wrong here...

    My assumptions are:
    1) All forces ground acting on base are upwards (z-direction)
     
  25. Aug 30, 2012 #24
    First of all, you are ignoring that the base slab is inclined. So the total force of pressure is different from the total weight.

    Secondly, in the torque integral I think you did not express the area element correctly.

    Thirdly, since you have not shown the torque equation, I can't say much else :)
     
  26. Aug 30, 2012 #25
    Ok, given that the slab is inclined at 5.6 degrees, we can simply take the answer * cos 5.6 but it would roughly give the same answer..

    I'm not sure where I went wrong for my torque equation:

    = ([itex]\bar{r}[/itex] x [itex]\bar{F}[/itex])
    = ([itex]\bar{r}[/itex] x [itex]\bar{p}[/itex]) dA
     
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