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Sliding Ladder

  1. May 12, 2012 #1
    2zeff2e.png
    The slender ladder has a mass of 10 kg. It is released from rest in the position shown. Friction at the two contact surfaces are negligible. Determine the angular acceleration of the ladder.

    I summed forces in the x,y direction, summed moments about the ladder's center of mass, wrote a kinematics equation relating the angular acceleration to the acceleration of its center of mass for a total of 5 unknowns (Force at bottom, ax, ay, angular acceleration, force at the vertical wall) and 5 equations. I have angular acceleration = 3.75 rad/s^2. This was on a multiple choice test, but there was no answer of 3.75 rad/s^2. I've been at this problem for days, making the same "mistake" over and over. I have found that the solutions manual of a similar problem is also doing the same thing I'm doing. Did my professor make a mistake?
     
  2. jcsd
  3. May 12, 2012 #2

    tiny-tim

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    hi eurekameh! :smile:

    (pleeeease don't post such wide images :redface:)

    (it should be possible with just one τ = Iα equation, and one constraint equation … anyway:)

    show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
     
  4. May 12, 2012 #3
    3.75 rad/s^2 is correct.
     
  5. May 12, 2012 #4
    tiny-tim:
    I don't think it's possible with just one tau = (I)(alpha) equation and one constraint equation because then you'd have too many unknowns with not enough equations.
    Denoting Fx as the force from wall, and Fy as the force from floor:
    For forces in x;
    Fx = max
    For forces in y;
    Fy - mg = may
    And for moments about center of mass;
    (Fy)(cos60) - (Fx)(sin60) = (1/12)(m)(l^2)(alpha)
    The two constraint equations gave me:
    ay = -(alpha)(cos60)
    ax = (alpha)(sin60)
    5 unknowns in 5 equations gave me 3.75 rad/s^2.
    Thanks for the confirmation, Quinzio. :)
     
  6. May 13, 2012 #5

    tiny-tim

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    hi eurekameh! :smile:
    yes you can if you do it τ = dL/dt about P, the point where the normals meet …

    τ = dL/dt = d/dt {Ic.o.mω + mrc.o.m x vc.o.m} = Ic.o.mα + mrc.o.m x ac.o.m

    give it a try! :wink:

    (i don't know why i said Iα before :redface:)
     
  7. May 13, 2012 #6

    sk9

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    YES, so it by instantaneous axis of rotation. search it up on google if you dont know about it. it becomes very easy
     
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