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How do we get the formula for the contact force ?

  1. Nov 12, 2011 #1
    My teacher said there isn't a special kind of formula, you just derive it from f=ma.


    This is the diagram:

    Ss8D2.gif
     
  2. jcsd
  3. Nov 12, 2011 #2

    Doc Al

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    I assume you want the contact force between those two masses. Your teacher is right. Just apply Newton's 2nd law.

    Hint: First find the acceleration of the system.
     
  4. Nov 12, 2011 #3

    Yes. I found the acceleration.
     
  5. Nov 12, 2011 #4

    Doc Al

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    Good. Now apply Newton's 2nd law to M1 alone.
     
  6. Nov 12, 2011 #5
    To find the the answer in N ?

    So-> F= m1 * acceleration that I found for the whole system ?

    Ok, I found the answer. But scientifically, I don't understand how.
    My teacher said that the contact force for both blocks is the same. How is that possible ? So if a bee flying at 10 km/h hits a car moving at 30 km/h the forces between both bodies is the same. How is that possible ?
     
  7. Nov 12, 2011 #6

    Doc Al

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    Right.
    That's Newton's 3rd law. Whenever two objects interact, they exert equal and opposite forces on each other.
    Definitely.
     
  8. Nov 12, 2011 #7
    Man I love physics. lol.

    Ok but why did we use m1 rather then m2 when we applied newtons second law on m1 ?
     
  9. Nov 12, 2011 #8

    Doc Al

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    If you're applying Newton's 2nd law to m1, why in the world would you use m2?
     
  10. Nov 12, 2011 #9

    SammyS

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    Draw two Free Body Diagrams, one for each block
     
  11. Nov 12, 2011 #10
    Yes but here is what we did:

    We found acceleration for the whole system.
    Then the questions asked:

    Find the magnitude of the contact force between the two blocks.

    Why would you use the mass of m1 and not m2 ?
     
  12. Nov 12, 2011 #11

    Doc Al

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    When applying Newton's 2nd law to m1, use the mass of m1.

    You can also apply Newton's 2nd law to m2, in which case you'll use the mass of m2.

    You'll get the same answer for the contact force either way. (Analyzing m1 is a bit easier, since only one force acts.)
     
  13. Nov 12, 2011 #12
    Here's the problem to make it clearer:

    SkTQo.png

    My acceleration is: 1.32 m/s^2

    And then I just multiplied 1.32 m/s^2 by 1.39 m/s^2.
    If I do 1.32 m/s^2 *3.46 m/s^2 I don't get the same answer...
     
  14. Nov 12, 2011 #13

    Doc Al

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    Right.
    That's correct.
    The contact force is not the only force acting on m2. To apply Newton's 2nd law to m2, you must use the net force: ƩF = m2*a.
     
  15. Nov 12, 2011 #14
    I get it. Thanks.

    Can you help me to draw a FBD for this problem ?
     
  16. Nov 12, 2011 #15

    Doc Al

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    Sure. First identify the forces acting on each block. What forces act on m2? On m1?
     
  17. Nov 12, 2011 #16

    Not sure but

    m1-> force coming from m2 and force going forward.

    m2->force from F (6.40N) and force of contact.
     
  18. Nov 12, 2011 #17

    Doc Al

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    What do you mean by 'force going forward'? The only thing in contact with m1 is m2.
    Right. They act in different directions, of course.
     
  19. Nov 12, 2011 #18
    <-m1

    <-m2->

    or

    <-m1->

    <-m2->
     
  20. Nov 12, 2011 #19

    Doc Al

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    Only one force acts on m1. In what direction does it act?
     
  21. Nov 12, 2011 #20
    Well there is the force of contact... Does this count ?
     
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