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Leaning tower of pisa and rotational dynamics

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    The leaning Tower of Pisa (Fig. 9-47) is 55 m high and 7.0 m in diameter. The top of the tower is displaced 4.5 m from the vertical. Treat the tower as a uniform, circular cylinder.


    (a) What additional displacement, measured at the top, will bring the tower to the verge of toppling?

    (b) What angle with the vertical will the tower make at that moment?

    2. Relevant equations

    Net force in x-direction = 0
    Net force in y-direction = 0
    Net torque = 0

    3. The attempt at a solution

    So this problem is confusing me a little. I'm a little confused as to what forces are being applied to this tower and where are the forces being applied.

    I know that there is a normal force, perpendicular to the ground from the base of the structure, there if a friction force in the x direction to the base of the structure, and there is gravity, which is being applied the the tower itself. However, where is gravity applying the object? At its center of mass or at the top of the tower?
  2. jcsd
  3. Oct 26, 2009 #2


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    The weight of the structure (gravity force) should be taken as acting at its center of mass. Assume friction at the base is high enough to prevent any lateral sliding at the base. Where do the normal and weight forces act when the tower is on the verge of tipping over?
  4. Oct 26, 2009 #3
    Well, when the tower is about to fall over, the normal force acts in the positive y-direction at the base of the tower and the weight acts in the negative y-direction at the tower's center of mass.
  5. Oct 26, 2009 #4


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    yes, but where at the base?
    yes, but where is the center of mass when the tower is leaning at its point of tipover? Hint: the tower starts to topple when the center of mass goes beyond the edge of the base.
  6. Oct 26, 2009 #5
    Well the normal force is at the edge of the base.

    The center of mass, at the point where the it aligns with the base, is when it starts to tip, based on what you said.

    But what is the proof for that? The sum of all forces in the y-direction at this point would be past 0 because the gravity force of the object is greater than the normal force right?
  7. Oct 26, 2009 #6


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    yes, the right edge
    yes, the weight downward at the c.m. also aligns with the right edge of the base.
    when the c.m goes beyond the right edge, there is a net torque about that edge, and it will tip over. Before that point, the torque from the resultant normal force, still left of the edge balances the torque fom the weight force
    No. The tower at the verge of tipover is still in equilibrium, rather unstable for sure, but still standing still. So the normal force must still equal the weight of the tower. This becomes a geometry/trig problem to solve for the angle and additional displacement at this point.
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