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Forces and Speed

  1. Apr 8, 2007 #1
    The question reads: Two blocks are in contact on a frictionless table. A horizontal force is applied to the first block. If the m1 = 2.0kg, m2= 1.0kg and Fapp = 3.0N, find the force of contact between the two blocks.

    What does the question mean by the force of contact?
  2. jcsd
  3. Apr 8, 2007 #2
    The force by one block on the other block.
  4. Apr 8, 2007 #3
    That's what I thought, but how do I go about finding this force, don't tell me exactly, but if you could point me on the right path.
  5. Apr 8, 2007 #4
    The two blocks are in contact, right. What else does that imply about their motions?
  6. Apr 8, 2007 #5
    They are stationary or moving at the same speed/acceleration.
    So, is this right? -
    Fnet = Fapp
    2a = 3
    a = 1.5m/s^2

    block 2
    Fnet = Fapp
    1a = Fapp
    Fapp = 1.5N

    The force of contact is 1.5N??
  7. Apr 8, 2007 #6
    Looks okay except for notation. Let me be picky. Fapp is on the first block not the second block. The contact force acts on the second block so it is the net force in the horizontal direction on the second block. Make sense?
  8. Apr 8, 2007 #7
    Oh thats more typing - I have Fcontact written down, I just can't copy! I also have the calculations in columns headed with horizontal - but the math is correct though?
  9. Apr 8, 2007 #8
    Yes but this is physics class right. :biggrin:
  10. Apr 8, 2007 #9
    I am working on Sir Isaac Newto Contest Questions, which are a little tougher. Would you point me in the right direction with this next one too?

    A 2kg chicken rests at point C on a slack clothes line ACB. CA and CB slope up from the horizontal at 30degrees and 45degrees. What minimum breaking strength must the line have to ensure the continuing support of the bird?

    Normally I would be fine with this type of question, but I don't have any lengths.

    ~~~ I just had I thought, do I just substitute in letters and work from there??
  11. Apr 8, 2007 #10
    The lengths are irrelevant here. Draw a free body diagram.
  12. Apr 8, 2007 #11
    I'm having some trouble, this is what I have done so far.
    Vertical Direction
    Fnet = Fg + Fn + T1v + T2v
    0 = 19.6 [D] + Fn + Tsin30 + Tsin45
    Fn = 19.6 - 1.21T

    Where do I go from here??
  13. Apr 8, 2007 #12
    You should always specify what system/object you are writing Fnet down for. It looks like you have written it for the bird, what is providing Fn?
  14. Apr 8, 2007 #13
    Okay, I shall make sure I do that - I was doing it for the bird. The rope clothesline is providing the Fn I suppose, is that right?
  15. Apr 8, 2007 #14
    If you state that, then I must ask what is providing the tension forces you state?
  16. Apr 8, 2007 #15
    The force of gravity on the bird.
  17. Apr 8, 2007 #16
    Now wait, you have Fg in your equation, the force on the bird due to gravity, then Fn from the string, then two more forces that look like tension. I think you have "double counted."
  18. Apr 8, 2007 #17
    I think I have double counted. I put the component of tension (vertical) for both triangles - the 30 degree and 45 degree
    So I only need one, right?
  19. Apr 8, 2007 #18
    Now if you are trying to decide which string to use, how could you decide? That is not what I was getting at.
    You need to consider the tensions here, but what is providing that normal force you have??
  20. Apr 8, 2007 #19
    To be honest, I am not sure. My teacher has not gone into things like this, he just tells us. This is why I am struggling a little now I think. How should I go about thinking about this, because I don't really want you to just tell me the answer.
  21. Apr 8, 2007 #20
    Free body diagrams area great tool but you must know what forces act on the object. To do that, just run through the list of forces you studied. Look for things pushing or pulling on the object. Remember that there are strange forces such as gravity that act as "invisible pushers."
    A normal force is usually reserved for contact forces between objects. The string is in contact with the bird, strictly speaking, but you have already included it in your force equation as two tensions.
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