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Free Body diagram & reaction force

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    This is just a little part to a bigger problem I am having trouble with. I have simplified it to get to the point.

    In the free body diagram shown, I would like to find the value of the horizontal reaction force (RH) applied to the wheel from the surface, assuming there is no slip.

    freebodydiagram-2.jpg

    (just in case my drawing isn't clear, the yellow thing is like a wheel, and the line with the mass on it is rigid and attached to the wheel thing).

    2. Relevant equations

    Possibly equilibrium equations, but I'm not sure because the system isn't necessarily in equilibrium.

    3. The attempt at a solution

    My best attempt so far is based on my guess that since there is no slip between the surface and the wheel, the reaction force is equal in magnitude to the force that that part of the wheel exerts to the surface.

    If this is true, then:

    RH = 1/rr*(torque from mg - torque from B - torque from C) + A + B + C

    RH = 1/rr*(mg sin(theta) - B*rb - C*rc) + A + B + C

    I'm just not sure about the validity of my claim that R will oppose both the moments AND the translational forces.

    Thanks guys!

    Andrew
     
  2. jcsd
  3. Sep 28, 2009 #2

    lanedance

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    what are A,B,C - horizontal forces?

    no slip condition means the velcoity at that point is zero, ie the frictional reaction force avaiable is sufficient to prevent any slippage

    either way, i would keep in mind that it is likely there to be a horizontal & rotation accerleation that should be included in your force balance ie for each body... that is unless you know they are zero.

    [tex] \sum_i F_x_i = ma_x [/tex]

    and similarly for vertical & rototions
     
  4. Sep 29, 2009 #3
    Yes, A, B and C are horizontal forces.

    If the frictional reaction force is sufficient to prevent slippage, does that mean it is equal to (the sum of the forces at that point on the wheel resulting from the torques) + (the sum of the horizontal forces)?

    Or is it only the sum of the forces at that point resulting from the torques?

    Or is it only the sum of the horizontal forces?

    My physical aptitude is letting me down here a little in that I am having trouble understanding the magnitude of the horizontal force that the wheel applies to the surface at that point.

    By the way, thanks for your help lanedance.
     
  5. Sep 29, 2009 #4

    lanedance

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    no worries, but as you need to include the acceleration, i think the answer is not as easy as that... all the following must be satistfied consistenty:

    sum of external forces acting on free body balanced with acceleration
    [tex] \sum_i F_x_i = ma_x [/tex]
    [tex] \sum_i F_y_i = ma_y [/tex]
    sums of external torques with angular acceleration
    [tex] \sum_i T_i = L\alpha[/tex]

    now without trying it in detail (disclaimer) I would think you will have to:

    1) if the whole struture starts to roll what is the link between angular acceleration & horizontal acceleration? (around centre of rotation, I think is probably easiest where A, is ...)
    [tex] a_x = (???) \alpha[/tex]

    2) now sum the horizontal forces
    [tex] \sum_i F_x_i = ma_x [/tex]

    3) now sum torques (around centre of rotation) and equate to angular acceleration, also what is moment of inertia L?
    [tex] \sum_i T_i = L\alpha[/tex]

    now use 1) to relate 2) & 3) and solve for the horizontal force
     
    Last edited: Sep 29, 2009
  6. Sep 29, 2009 #5

    lanedance

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    or you could look at the torque around the point of no slip to get angular acceleration without RH
     
  7. Sep 29, 2009 #6
    Thanks that helps a lot!!
     
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