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Relative Acceleration - Particle and Wedge

  1. Jan 28, 2017 #1
    I would be very grateful for help with deriving the following equation...

    "A smooth fixed plane is inclined at 30 degrees to the horizontal. A wedge of mass M and angle 30 degrees is held on the surface so that its upper face is horizontal, and a particle of mass m rests on this face. The system is released from rest. Show that the resultant acceleration of the particle is:

    (M + m)g / 4M + m
    I do not understand how to derive this equation. Right now, I can only envisage the particle adding to the wedge's overall weight, and thereby increasing the normal reaction of the surface and consequently the wedge's acceleration down the slope. It seems to me that the acceleration of the wedge and particle should be the same.

    Any solution or assistance would be most gratefully received
  2. jcsd
  3. Jan 28, 2017 #2


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    Staff: Mentor

    Welcome to the PF.

    In the future, please organize your schoolwork post by using the Homework Help Template that you are provided when posting here. I'll paste in a copy of the Template below so you can see what it looks like.

    Also, can you please Upload a diagram that shows the situation? Please also show us your free body diagram (FBD) for the problem. That will make it a lot easier for us to help you.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  4. Jan 28, 2017 #3
    Thank you very much for your response and my apologies for not adhering to the template. Based on the problem description (which was not accompanied by an image) I came up with the free body diagram attached.

    Does this look correct? As you can probably see, the main problem is that I'm uncertain as to the acceleration of the particle relative to the wedge. I can see the wedge being acted on by the surface and particle, but no net forces seems to act on the particle. I clearly don't understand the system and would really appreciate some assistance.
  5. Jan 28, 2017 #4
  6. Jan 28, 2017 #5


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    Staff: Mentor

    Ah, the diagram helps a lot, thanks.

    Are they assuming that all surfaces are frictionless? They say the inclined surface is "smooth", but that doesn't necessarily imply frictionless.

    If there is no friction between the wedge and the inclined plane, and there is friction between the wedge and the top box/particle, then you are right that the masses would just add. But if there is no friction between the wedge and the top box/particle, I can see that the situation would be changed. The top box would have no horizontal motion as the wedge slipped down and to the left. Can you see how that would change the acceleration of the wedge and particle?
  7. Jan 28, 2017 #6
    Thank you very much for your reply. As all the earlier exercises in the textbook indicated frictionless surfaces, I assume the same applies here and that there's no friction between the wedge and the top particle. Honestly, though, I just don't understand how the system produces an acceleration of '(M + m)g / 4M + m' in the particle.
  8. Jan 29, 2017 #7


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    Staff: Mentor

    Could you use parenthesis to clarify what you mean by this equation? If it's (M + m)g / 4(M + m) I think I know how they got it.
  9. Jan 29, 2017 #8
    Thank you very much indeed for your response. The equation is definitely printed in the book as being:

    (M+m)g / (4M + m),

    I.e. M only is multiplied by four in the denominator. The system description preceding the equation is faithfully copied in this post. The book is "Introducing Mechanics" by Jefferson (OUP). Nevertheless, I would certainly be very interested to see the derivation of (M + m)g / 4(M + m).
  10. Jan 29, 2017 #9


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    Staff: Mentor

    Hmm, I will need to think more about it then...
    The reasoning was that since the top mass cannot be moved sideways (because there is no friction) by the motion of the wedge, it just translates straight downward as the wedge accelerates down and left along the inclined plane.

    So the acceleration of the wedge is (M+m) * cos(60) = (M+m)/2.

    And since the top mass only translates downward, I thought that the acceleration downward was half of the wedge's acceleration down and to the left. But there must be more going on here, so back to the drawing board... Have you made any more progress?
  11. Jan 29, 2017 #10
    I haven't solved it yet, but your your reasoning has generated new impetus and I've now gotten closer to the desired result. I'm going to have another try at getting the actual result, but I might have to resume tomorrow evening as I need to be up early tomorrow.
  12. Jan 29, 2017 #11
    The best I've been able to come up with for now is the following updated equations of motion (where ##R_{m}## is the normal reaction of particle on wedge):

    Horizontal Forces on wedge: ##\frac{1}{2}R = \frac{3}{\sqrt{2}}Ma##

    Vertical Forces on wedge: ##Mg + R_{m} - \frac{3}{\sqrt{2}}R = \frac{1}{2}Ma##

    Vertical Forces on Particle: ##mg - R_{m} = \frac{1}{2}ma## (acceleration of wedge downwards = 1/2a)

    These gave me ##a = \frac{2(M+m)}{19M + m}##, which really sucks. Obviously I'm making a big error but I'll have to look into it again tomorrow.

    Once again, many thanks for your most valuable input.
  13. Jan 30, 2017 #12
    Which subject is it? In vector or in Linear?
  14. Jan 30, 2017 #13
    Hi Rafa_El, thanks for joining the thread. I'm not sure that I quite understand your question - please could you clarify the "it" you refer to? The question text in the original post is copied straight from the book and fully defines the problem. It needs solving using Newton's laws and straightforward vector/linear algebra.
  15. Jan 30, 2017 #14
    Oh i'm sorry im not really fluent english. I was misunderstand it. I was referring to your question
  16. Jan 30, 2017 #15
    No problem at all! The subject of the question is Introductory Mechanics, specifically using forces represented as vectors to solve for an acceleration.
  17. Jan 30, 2017 #16
    I ever face something like this. But usually we use common derivative in my school. Like for horizontal FX = Fx1+Fx2... Fxn then FY= Fy1.... for vertical then we use Fr=square root(Fx+Fy)^2 then F=M.a then we substitute it for desired result like accele,mass or even the force
  18. Jan 30, 2017 #17
    And based on your question i think it ask amount of vertical force which mean against the gravity mean Fy= xN.sin○
    correcte if i'm wrong
  19. Jan 30, 2017 #18
    Yes, I think the consensus in the thread is that the particle can only move vertically. We just need to show that it's acceleration is:
    (M + m)g / 4M + m
    I.e. the actual value of the acceleration doesn't matter.
  20. Jan 30, 2017 #19
    well i need to work on it and hope i get the derivative
  21. Jan 30, 2017 #20
    • Member advised that text-speak is not allowed at PF. Proper spelling, grammar and punctuation should be used.
    Btw i want to see your textbook about it. Maybe i can conclude something
    Last edited by a moderator: Jan 30, 2017
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