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Homework Help: Dynamics - Object Sliding Along A Frictionless Wall

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    The 500lb box is released from rest in the position shown. The wall and ground are smooth. Determine (a) the angular acceleration of the box, and (b) the reactions at corners A and B at the instant after release.

    **G = center of mass
    **L = length of side = 2.5ft
    **phi = 15 degrees (found through geometry)

    2. Relevant equations

    ƩF = mag

    ƩM = Igα

    aa = ab + α x ra/b

    3. The attempt at a solution

    Okay so the two assumptions I'm working off of here are that points A and B have kinematic constraints, the constraints being that they have to travel along their respective walls; A can only travel downward (-j) and B can only travel sideways (+i) the instant they are released. So these are the two equations I started with:

    ƩF = FA(i) + (FB-mg)(j) = maG

    ƩMG = (√2/2L)sin(phi)FA+(√2/2L)sin(phi)FB = 1/6mL2α

    I then solved for α using the sum of the moments and put it into this equation:

    aA = aG + α(k) x rA/G

    Here, if A can only accelerate in the j direction, then that means its i terms sum to zero so I pull out the i terms:

    {the first term comes from my F=ma equation, aG}
    (FB - mg)/m + (6d2cos2(phi)(FA+FB))/(mL2) = 0

    I then use the same relative acceleration equation with point B and the same kinematic constraints of all the j terms in this equation summing to zero since B can only move in the i direction:

    FA/m + (6d2cos2(phi)(FA+FB))/(mL2) = 0

    Now I have two equations and two unknown reaction forces so it should be straight forward, but I get a negative value for FA... I can't help but think that it boils down to a sign error or a faulty cross product somewhere but I've looked over it for 3 hours and I can't find anything mechanically wrong with it. Also, a friend told me that the reaction force at A should be zero, but I'm not sure if he's correct on this. It kind of makes intuitive sense that it would be zero but I can't for the life of me figure out how to mathematically get a positive value for FA, much less get a zero value for it. Any help on this would be immensely appreciated. I'm sorry for the long post.
  2. jcsd
  3. Mar 17, 2012 #2


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    welcome to pf!

    hi yoft! welcome to pf! :smile:

    (i haven't checked your equations, but …)

    it would be a lot easier to use conservation of energy first, to find the speed and acceleration,

    and then to have a go at the reactions! :wink:
  4. Mar 17, 2012 #3
    I actually had tried using energy conservation but I couldn't really get anywhere with it. It's pretty straight forward to use the distance that the center of mass drops to calculate the linear speed right before it hits the floor but I don't know what to do with that information. I suppose the magnitude of acceleration is just gravity but I don't know which direction it's pointing in. I also am unsure as to how I would relate that to alpha. I know a = alpha cross r but isn't r the distance between the instance center and the center of mass? And I don't know how I would find the instance center.
  5. Mar 17, 2012 #4


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    hi yoft! :smile:

    for conservation of energy, you only ned to know the (linear) velocity of the centre of mass, and the angular velocity

    for the angular velocity, just use |vA - vB|/(AB) for any two points A and B

    (it works because |vA - vB| = |ω x (rA - rB)| = ωAB)

    (in this case, i'd use the two obvious corners!)

    try again! :smile:
  6. Mar 17, 2012 #5
    I guess I still don't really understand what you're suggesting :( it seems to me that doing that adds in 2 additional unknowns (linear velocity at points A and B) while not really moving towards getting angular acceleration? And thanks so much for your help and time!

    Also: is your "AB" term the magnitude of the distance between points A and B?
  7. Mar 17, 2012 #6


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    hi yoft! :smile:
    but they're not additional unknowns, they're "instead-of" unknowns!! :rolleyes:

    the coordinates of A and B completely determine the position of the block, don't they? :wink:

    use them to determine everything else! :smile:

    (and yes, AB is the distance)
  8. Mar 17, 2012 #7
    Okay so using conservation of energy:

    T1+V1+U1-->2 = T2+V2

    I get:

    mgh = 1/2mvG2 + 1/2IG[(VA-VB)/L]2

    I'm guessing I have to put VA and VB in terms of VG somehow and solve for VG? And even if I get VG I'm still not sure how I would relate that to α and the reaction forces. I guess I don't really understand the direction you're showing me :(
  9. Mar 17, 2012 #8


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    call the coordinates of A and B x and y,

    and express everything in terms of x and y​

    (and i'm going out for the evening now :wink:)
  10. Mar 18, 2012 #9
    I think I got it, I used the equations of relative accelerations first to get linear acceleration in terms of alpha and then put them into the sum of moments and forces to some for alpha. Thanks for your help though!
  11. Mar 19, 2012 #10
    So I'm told, but I couldn't (and still don't) understand how to go from linear velocity and angular velocity to angular acceleration and force. And I don't know how much easier it really would have been, it took around 6 lines the way I did it.
    Last edited by a moderator: May 5, 2017
  12. Mar 19, 2012 #11
    It actually turns out that the reaction force at A is supposed to be negative... And then you're supposed to assume that that means the reaction force at A is actually zero since it can't be negative, and then you redo the calculations under the assumption that A is zero. I don't really know how I feel about being supposed to make that assumption but it's good to know that my answer was right the first time because I couldn't understand why conceptually it was the wrong approach.
  13. Mar 19, 2012 #12


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    it isn't assuming that A is zero so much as assuming that the wall isn't there! :biggrin:

    yes, either way is correct :smile:

    (conservation of energy is usually simpler, but in this case since only the initial forces were asked for, the direct approach was probably quicker)
  14. Mar 19, 2012 #13
    Thanks a lot for your time!!
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