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Equilibrium problem: sliding door

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    There is a sliding door lifted at a constant rate by two cables, A and B. The sides of the door fit loosely into grooves in the walls.

    Dimensions of door: 9 ft tall by 10 ft wide
    Weight of the door: 200 lbs
    Cable A is attached at the top of the door, 2 ft from the left side
    Cable B is attached at the top of the door, 2 ft from the right side
    The coefficient of kinetic friction between the door and grooves is 0.3

    Suppose cable B breaks. What force must cable A exert to continue raising the door at a constant rate?

    2. Relevant equations

    Equilibrium equations:

    [tex]\sum[/tex]Fx = 0, [tex]\sum[/tex]Fy = 0, [tex]\sum[/tex]M = 0

    Friction equation (N is force normal to surface):

    f=[tex]\mu[/tex] * N

    3. The attempt at a solution

    It seems as if I don't have enough information to solve the problem.

    To highlight the difficulty I have chosen to sum the moments about the center of mass of the door.

    [tex]\sum[/tex]Fx = Nleft - Nright = 0

    [tex]\sum[/tex]Fy = TA - 200 - (.3)Nleft - (.3)Nright = 0

    [tex]\sum[/tex]M = 5*0.3*Nleft - 5*0.3*Nright - 3*TA = 0

    What I immediately discover is that the two normal forces are equal, and therefore from the last equation TA = 0. Obviously this is not the correct answer!

    I thought that perhaps I could add a couple to the final equation to reflect the fact that the unbalanced door is trying to rotate and can't due to the door frame, but then I end up with four unknowns :-/

    What am I missing about this problem?

    Thanks,
    -Zac
     
  2. jcsd
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