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Rigid Body on Frictionless Surface

  1. Jul 12, 2012 #1
    Hey all,

    Say you have a slab sitting vertically(like a door) at rest on a frictionless surface. If you push with a horizontal force somewhere near the top(as long as it is above the center of mass), would the slab rotate(tip over)?

    My friend and I were originally thinking of how much horizontal force you would have to push a block with to tip it over on a surface with friction. This turned into a question of whether or not you actually could tip a block if it was on a frictionless surface. We both said no, but then we thought about if you replaced the block with a door and it made us think twice.

    Right now we are leaning towards the object rotating, or tipping over, but we aren't completely sure and would like some more input on the problem. Does it depend on the weight of the object? Does it depend on how much force you apply? Does it depend on where you apply the force?

    Also, we are saying the body is rigid. We have tried to break it down into a simpler model, with no avail. Hopefully you guys can help.

    Thanks
     
  2. jcsd
  3. Jul 12, 2012 #2

    haruspex

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    It's a question of moments. Suppose a force F is applied at height h to a rectangular block of mass m, width 2w (in the direction of the applied force) . Taking moments about the potential tipping point, it will tip if Fh > mgw. The reason it can be counterintuitive is that pushing with a steady force when there's no friction is hard to do. The block will accelerate rapidly, and it's hard for the force to keep up. Best way may be to think of applying the force by attaching a horizontal thread that runs over a pulley to another mass suspended vertically.
     
  4. Jul 12, 2012 #3
    Ah that makes a lot of sense. Also the thread idea is great. I think we just put too much thought into the problem.

    Thanks
     
  5. Jul 13, 2012 #4

    jbriggs444

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    Suppose that the block has width w, height h, mass m and is in gravitational field g.

    You apply force f at the top. Your applied torque is f * h.

    Let's say you're pushing to the left. The block starts tipping around the lower left corner. The force of gravity is m * g and its torque is exerted over a moment arm of w/2.

    As long as the force you apply is less than m*g*w/2 / h then you will fail to accomplish anything -- the block will resist tipping.

    If the force you apply is greater than this, the block will begin tipping. Keep it up until the block is at the balance point with the top right corner directly above the bottom left corner and it will continue on its own.

    Note that you can get away with less force using a technique known as "rocking". This is easier if there is friction, but is also possible without it. The resonant frequency is increased [by a factor of 2?] in the frictionless case since you are effectively rotating around the center of mass rather than around the frictionally anchored base.

    Edit

    In the frictionless case, the correct choice of axis of rotation is at the center of mass.
    That means that your moment arm is only f*h/2 and the required force is f = m*g*w.

    [It is possible to do the analysis with the axis of rotation chosen at the lower left hand corner, but then the linear motion of the block contributes to angular momentum and must be accounted for in the comparison of torques]
     
    Last edited: Jul 13, 2012
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