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Pisa tower

  1. Aug 23, 2005 #1
    ok the question goes sth like this,an equation of p=c1+c2*r^0.625cos@,where p is the pressure exerted on the soil,c1 & c2 are constant to be determined,r& @ are polar coordinates.(@ measured fr the x-axis which is of direction south)
    now the quesiton states that pisa lean at an angle of 5.6degree towards the south from the vertical.other values such as weight and measurements are given.however i have no idea how to approach this question.
    i tried setting @ as 90 degree,so that the pisa is vertical.hence pressure is given by weight over its area,thus c1 can be determined.however this is flawed and do not concure with the results given by my tutor.Hence this approach is not viable.
    The tutor appoach was complex and i simply do not understand.Owing to the fact that when the pisa tower leans,the pressure when it exerts on the soil is varying across its base.hence some forms of intergral is used.The tutor makes use of delta-a=delta-r * delta-@,where a in this case i reckoned is the area swept out when the tower leans onto her sides.Then there are double intergral ivolved,which i had no idea.He too,take moment about some point.(This quesiton came out from a mechanics book,so we assume that the pisa tower is at equilibrium)
    can anyone here suggest a simpler approach to tackle such problems??any imput in appreciated,thanks!
     
  2. jcsd
  3. Aug 23, 2005 #2
    I'd also start by making @=90, so that you get p=c1 + c2, with p being known from F/a.

    We'd need one more equation for c1 and c2 to actually solve them, so I'd assume that when the Pisa tower is at 5.6 degrees, it's at equilibrium, and I'd set the torque from the pressure of the floor equal to the torque from gravity.
    Torque from gravity is simple, you just need the center of mass.
    On the other hand, the pressure torque would be harder. You'd have to integrate your pressure function with respect to r, and multiply that by the total area to get your net torque from pressure.

    Then set the two equal to get a second equation for c1 and c2, combine that with yoru first equation for c1 and c2, and you should get them.

    Of course, without the numbers, I'm not sure if this method works.
    If you posted the numbers, I could try it out.
    Also, I'll ask a friend who teaches physics to see if he can give a definite method of solving this problem.
     
  4. Aug 23, 2005 #3
    hmm,why the torque from the pressure is equal to the intergral of pressure wrt to r?
    in my tutor solution,he intergrate pressure wrt to area,and the right hand side he intergrate wrt to both the angle and the term r.Thats a double intergral at the right hand side.
     
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