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Action-Reaction Force for external forces

  1. Mar 3, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-3-4_1-25-52.png

    1) Are F and F' as shown in figure, the action-reaction forces between the two boxes?

    2) What is the action and reaction force between the two boxes?

    2. Relevant equations


    3. The attempt at a solution
    I think F = 28N and F' = - 28 N Simply adding the action- reaction forces for each case of 4 N and 24 N external forces.
     
    Last edited: Mar 3, 2017
  2. jcsd
  3. Mar 3, 2017 #2

    kuruman

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    What do you think?
    What kind of motion do you think these masses are undergoing? Are they moving, accelerating or what? Is there friction? Showing the full statement of the problem would be very helpful.
     
  4. Mar 3, 2017 #3
    The floor is friction less. and the boxes were at rest prior to force application. That's all.
     
  5. Mar 3, 2017 #4

    kuruman

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    OK, that's good. You left out the "Relevant equations" in the template. What equation(s) do you think are relevant here?
     
  6. Mar 3, 2017 #5
    Well I don't find a particular equation. It's just that I found out all the forces that can be acting using free-body diagram. And I made sum of forces
     
  7. Mar 3, 2017 #6

    kuruman

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    Have you studied Newton's Laws?
     
  8. Mar 3, 2017 #7
  9. Mar 3, 2017 #8
  10. Mar 3, 2017 #9

    kuruman

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    Which ones of Newton's laws involve equations? Can you write them down? Do you know what they mean? They are very relevant to this problem.
     
  11. Mar 3, 2017 #10
    For 1st law, it mathematically states,

    if Fnet = 0 on any object, then it continues in its state of motion or rest forever


    For 2nd law,

    F is proportional to rate of change of momentum (not only proportional, but also are equal. i.e. F =ma)

    For 3rd law,

    If A applies force F on B, B applies force F on A, (direction is opposite
     
  12. Mar 3, 2017 #11

    kuruman

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    Good. More correctly the second law is ##\vec{F}_{net}=m \vec{a}##. The subscript "net" means that you have to add up all the forces and set the result equal to mass times acceleration. Also note that the net force is always in the same direction as the acceleration.

    Knowing all this, can you find the acceleration of the blocks? Note that they move together as one which means that they have the same acceleration and velocity at all times.
     
  13. Mar 3, 2017 #12
    This way, it becomes that a = 2m.s^-2
     
  14. Mar 3, 2017 #13
    That is what I get
     
  15. Mar 3, 2017 #14

    kuruman

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    OK. Now let's take one mass at a time. First the 3 kg mass. Can you write an expression showing the sum of the two forces acting on it? It must have two terms.
     
  16. Mar 5, 2017 #15
    Yes!

    Let, 7kg mass be m1. 3kg mass be m2

    For m2, we can write

    $$ Fnet = Faction-reaction + Fapplied from right $$

    $$ Or, 6 = Faction-reaction + (-4) $$

    $$ Therefore, Faction-reaction = 10 N $$

    Similarly, we get -10 N for the m1 using the strategy above !!!

    Am I correct now?

    THANK YOU SO SO MUCH!!!! You really helped me a lot in this discussion.
     
  17. Mar 5, 2017 #16

    kuruman

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    You are correct. Good job. :smile:
     
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