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Drawbridge Equilibrium

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose the weight of the drawbridge in Figure is supported entirely by its hinges and the opposite shore, so that its cables are slack. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances? The mass of the bridge is 2500 kg.
    Here's a link to the figure
    https://cnx.org/resources/8186f1166e382ac9ec92644a6613ce07451f29c1/Figure_10_03_10a.jpg
    Figure_10_03_10a.jpg

    2. Relevant equations

    I use net fore = 0 and net torque=0
    T=0 because it is slack, N = normal force
    Fsinθ + N = w
    1.5Fsinθ = 7.5N

    3. The attempt at a solution
    I find N by solving the simultaneous equation to get Fsinθ= 20416 and N=4083 so the fraction for a is 1/6
    I am stuck on question b. I know Fsinθ=20416 but I don't know how to find the value of F and θ.
     
  2. jcsd
  3. Feb 25, 2017 #2

    TSny

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    The condition "net force = 0" is actually two conditions. One condition for the vertical components of the forces and the other condition for the horizontal components of the forces.
     
  4. Feb 25, 2017 #3
    so I have to include the equation Fcosθ - Nx=0
    Nx = Normal force in the x direction, since depending on the value of Fcosθ, the bridge will press against the wall

    Now I have 3 variables and only two equations (i.e Fsinθ=20416). How do I solve for F and θ?
     
  5. Feb 25, 2017 #4

    TSny

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    You can assume that the right end of the drawbridge just sits on the ledge without pressing on the wall. So, the normal force is vertical.
     
  6. Feb 25, 2017 #5

    TSny

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    It might help to note that the picture is drawn for the general case where there is tension in the cable. For the special case where there is no tension in the cable, the force F might not be in the direction as shown in the picture. You will need to decide what the direction of F must be in order to satisfy the condition that the total horizontal force on the bridge is zero.
     
  7. Feb 25, 2017 #6
    thank you
     
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