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Homework Help: Pressure, Tension, Torque

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    A symmetric building has a roof sloping upward at [tex]36.0^\circ[/tex] above the horizontal on each side.

    (a) If each side of the uniform roof weighs [tex]1.10\times10^4[/tex]N, find the horizontal force that this roof exerts at the top of the wall, which tends to push out the walls.

    (b) Which type of building would be more in danger of collapsing: one with tall walls or one with short walls? Explain.

    (c) As you saw in part (a), tall walls are in danger of collapsing from the weight of the roof. This problem plagued the ancient builders of large structures. A solution used in the great Gothic cathedrals during the 1200s was the flying buttress, a stone support running between the walls and the ground that helped to hold in the walls. A Gothic church has a uniform roof weighing a total of [tex]20000[/tex]N and rising at [tex]40.0^\circ[/tex] above the horizontal at each wall. The walls are [tex]40[/tex]m tall, and a flying buttress meets each wall [tex]10[/tex]m below the base of the roof. What horizontal force must this flying buttress apply to the wall?

    2. Relevant equations



    3. The attempt at a solution

    For part (a), I can't see how any force is being applied in the x-direction, rather than just in the y-direction, caused by the force due to gravity. Once this force is recognized, I can solve for the torque about the base of the walls but I don't see how the force applied isn't parallel to the lever arm.

    For part (b), it's given and intuitive that tall walls are more subject to falling due to a longer lever arm and, consequently, a greater net torque, driving the system out of equilibrium.

    For part (c), I'd choose the axis of rotation to be about one of the walls with a torque in the positive direction caused by the Flying Buttress and a torque in the negative direction applied by the roof. From there, however, I can't seem to see this system in its true, physical light.

    Enlighten me?
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: Consider the force that one side exerts on the other where they meet at the top. What direction must that force be in?
  4. Nov 12, 2008 #3
    I don't understand this either. I know that the two parts of the roof connect and impose some kind of frictional force due to gravity but I don't know why they don't just fall. I don't see why the axis of rotation isn't the center of mass with no external torques acting upon the roof component to drive it to push outward on the wall or against the other side of the roof.
  5. Nov 12, 2008 #4


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    Homework Helper

    Where the peak of the roof meets the weight of the roof will press one side against the other. Since that total weight translates down to the top of the wall there is a force vector that has both an x and y component at the top support on the wall. The y component is carried vertically to ground and hence there is only the compressional forces to deal with bearing the weight.

    However there are horizontal forces acting at the top of the wall, that comes from bearing the roof. These component forces will be determined by the angle given by the roof.

    So which wall - shorter or taller will be able to withstand this outward force?
    (Hint: Ever seen pictures of Notre Dame in Paris?)
  6. Nov 12, 2008 #5
    The shorter one can withstand the outward force better; but, I only understand this conceptually. The middle-ground seems obscure and abstract to me.
  7. Nov 12, 2008 #6


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    Homework Helper

    The purpose of the flying buttress was not to apply torque to the roof. It was to apply counterbalancing torque to the top of the wall.

    The force of the roof is a vector.

    Look at the x,y components of the force at 36 degrees to determine the outward force.
  8. Nov 12, 2008 #7
    I see the vector but WHY does it have x- and y-components? Why isn't it just the weight acting upon the center of mass, like I've been taught?
  9. Nov 12, 2008 #8
    Looking at it, I solved for the weight acting along the roof's axis, giving [tex]\vec{w}\sin(36\pi/180)[/tex]. Then, because this weight force acts non-orthogonally against the wall, we have to solve for the horizontal component of this weight, giving [tex]\vec{w}\sin(36\pi/180)\cos(36pi/180)[/tex]. This doesn't work. My trig.'s right for what I'm trying to set-up, but obviously my set-up is incorrect.
  10. Nov 12, 2008 #9
    What's wrong with my work? :/
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