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Forces on a rolling body with sliding bodies on it.

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm attaching a picture for the problem.



    2. Relevant equations

    F - N - K(that's my problem) = Ma

    K = m2a2

    N = m1a (the acceleration is the same, since the rolling body pushes the m1 body forward)

    m1g - K = m1a1 (this is the acceleration in the y direction)

    a2 = a + a1 (this is obvious)

    3. The attempt at a solution

    Well I know how to solve it now(since I know the force diagram). I just don't know
    how that K tension force started to act on the rolling body.

    BTW! If you know similiar force problems I would really, really love you for it.
    Imho this is pretty complex, I was wondering about non-calculus based, very complex
    force problems, examples so I can prepare.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Oct 22, 2011 #2

    arildno

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    Ok, I haven't checked your solution, since your main question concerned how "K" radiated, as you call it, through the pulley.

    Remember that the rope is a physical object, and that it is idealized in having zero mass. Agreed?

    But, obviously, the rope being physical can be subject to forces, i.e, it must obey Newton's 2.law of motion!

    And, since the rope cannot have infinite acceleration, the 2.law of motion for a physical, massless object must be: F=0, where F is the sum of external forces acting upon it.

    But, note that the force on the rope from box 1 is horizontally acting , whereas the force on the rope from box 2 is vertically acting.

    If you add those forces vectorially (taking the directions into the account, not just the magnitude we call rope tension), you do NOT get 0 as your vectorial sum.

    In order to get that, the PULLEY must provide a force on the rope, to keep in place actually (remove the pulley, and the rope would flop down on the trolley pretty quickly, indicating a force acting upon it from the removed pulley).

    And THEN, it is simple to understand how the rope "radiates" some force onto the trolley:

    It is simply the result of Newtons 3.law, about action and reaction.
     
  4. Oct 22, 2011 #3
    Why haven't I thought of that?! Thank you very much.

    Just to be clear that I understand it perfectly. If the trolley would be a stationary object,
    Newton's 3rd law would still hold and those red K forces would still apply, we just wouldn't need
    them to calculate anything so it's pointless to draw them. Otherwise the rope would accelerate in a 45degrees(in this picture anyways) path to infinity.
    I hope I'm right. :)

    Thanks again for the help, because this is a huge new knowledge application for me.

    ps.: If the pulley was in let's say 60 degrees from the horizontal, would the red K forces still be the same? I understand the the pulley must provide force on the rope but now it's in a different angle, so the total sqrt(2)*K total force would still "radiate" through the pulley, but the x-y components wouldn't be the same. That is my thought process, please correct me if I'm mistaken.
     
    Last edited: Oct 22, 2011
  5. Oct 23, 2011 #4

    arildno

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    Correct.
    Or, alternatively, you'd have to include in your relevant free body diagram the forces action on the trolley from the ground, in order to keep it stationary.

    (It is fairly conventional, though, that we do not include "statics" forces in the FBD, and only take into consideration those forces relevant to the motion we want to describe.)

    You're welcome!
    The pulley must still provide the necessary force to keep the rope "wrapped" around it.
    Modelling the rope as massless, and with no friction being present necessitates constant tension throughout the rope.
    Thus, if one end of the rope is modelled as strictly horizontal, the other modelled as strictly vertical, then the force from the pulley must be a 45 degree force with "square root of two" times tension in magnitude, irrespective of the pulley's own angle.
     
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