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Forces Acting on a Moving Hinged Bar

  1. Nov 24, 2008 #1
    Hi all,

    Okay, so the problem. Say we have a bar fixed to a horizontal surface by a hinge, at the end of the bar is a weight, which is fixed so that it cannot move. The bar is lifted to some angle [tex]\theta[/tex] and let go of. I'm ignoring air resistance and the weight of the bar.

    I'm trying to calculate the velocity of the weight at the end of the bar by finding the resultant force on it and then using a = F/m, then v = u + at.
    This is all happening in something I've programmed, so at set intervals I'll be recalculating the forces acting, then the acceleration and hence the velocity. The weight is then moved by the velocity (x and y components) and I draw a line from the weight to the hinge to indicate the bar.

    I've said that the weight has weight W. Parallel to the slope the component of the weight is W*sin([tex]\theta[/tex]) will cancel with the compression of the bar pushing back up (as these will be equal, assuming the bar does not change length).
    This leaves W*cos([tex]\theta[/tex]), the perpendicular component of W, as the force moving the system.

    I then say that, in x and y components, this comes out to be:
    Fx = W*cos([tex]\theta[/tex])*sin([tex]\theta[/tex])
    Fy = W*cos2([tex]\theta[/tex])

    This almost works: the motion is in the correct direction and looks great for some situations. However, the bar ends up changing length - and quite substantially in some situations.

    So, what have I done wrong? What are the actual x and y components of the force moving the system?

    Thanks in advance for any help!
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 24, 2008 #2


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    Staff: Mentor

    An easier way to do this would be to use the fact that KE + PE = Total Energy is a constant for this system. So if the weight started out at some height with no velocity before being dropped, that would give it a starting PE (do you know how to calculate that?). Then as it falls through its arc, it gives up PE and gains KE. The velocity is always along the arc, so that KE translates into horizontal and vertical velocity vector components.

    Maybe try it from this perspective, and see if you get a more natural-looking fall.

    BTW, this looks to be a work or personal project, or maybe upper division programming/simulation work. If it's simple homework, I should move this to the Homework Help forums. Can you say what this is for?
  4. Nov 24, 2008 #3
    Thanks for the reply, berkeman.

    This is for a personal project. I'm actually trying to get some fundamentals down before I attempt to create some frameworks; I played World of Goo recently and wondered whether I could produce some similar, although simpler, structures.

    Thinking about it, would your energy method also work for larger structures? I imagine so, at a glance.

    The potential energy of the weight at its starting height would be PE = mgh. And KE = 1/2mv2. So Total energy, E = 1/2mv2 + mgh.

    How would vx and vy be calculated, though, I'm not sure?
  5. Nov 24, 2008 #4


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    Staff: Mentor

    Sounds like a fun project. Vx and Vy are constrained by the hinge and rigid rod. The velocity vector will always be along the path of the weight, so around in a circle centered on the hinge (if I understood your explanation correctly). Just use the triangle formed by the hinge, the weight, and the point directly below the weight on the horizontal surface to give you the angles you need to use some trig to calulate Vx and Vy from Vtangential.
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