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Finding coefficient

  1. Jan 27, 2007 #1
    Hi there, My physics exam is on monday, and i have reviwed a lot, but i can't get these last five questions, i ll post em one by one, so first i need help with the first one,

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

    A 1.25 kg block is released from rest on a ramp tilted 35 degrees above the horizontal, if the block moves 1.84m down the ramp in 1.18s, find the coefficient of friction between the block, and the ramp.

    Plz help me out.
    Thanks
     
  2. jcsd
  3. Jan 27, 2007 #2

    radou

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    Any thoughts on the question?
     
  4. Jan 27, 2007 #3

    arildno

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    What equations do you think might come in handy here?
     
  5. Jan 27, 2007 #4
    uhh, i m think Ff=uFn, and i m not sure, if i should solve it, like a projectile, question.Yep, i m very confused.
     
  6. Jan 27, 2007 #5

    arildno

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    That is, indeed an important equation! :smile:

    Now, however, there is an even more important equation that is ALWAYS used in classical dynamics/kinematics problems.
    Which one is that?
     
  7. Jan 27, 2007 #6
    net force= ma, at least i think so man. Thanks.
     
  8. Jan 27, 2007 #7

    arildno

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    Correct!

    Remember that the direction of the force of gravity is neither parallell to the direction of motion, nor perpendicular to it.
     
  9. Jan 27, 2007 #8
    Yes, is it coming straight down the ramp, and the angle its one of the component make is 35 degrees, i hope i m clear, Thanks again.
     
  10. Jan 27, 2007 #9

    arildno

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    Note:
    You should first figure out what the normal force is, expressed by the object's mass, the angle, and g.
     
  11. Jan 27, 2007 #10
    normal force, is gonna be the hypotenuse, of its components rite,and is force of gravity equal to the y-component of the normal force, if yes, then normal force would = mg/cos angle.
     
  12. Jan 27, 2007 #11

    arildno

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    Not at all.

    The normal force cancels the normal component of gravity, since the object experience no acceleration INTO the ramp, only ALONG it.

    Therefore, the force of GRAVITY itself, mg, should be regarded as the hypotenuse in a right-angled triangle with normal&tangential components as the cathetes.
     
  13. Jan 27, 2007 #12
    k, so the normal force cancels the Y-component of fg, rite , and is , Net force in the y-direction, zero or not. Thanks.
     
  14. Jan 27, 2007 #13

    arildno

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    What do you mean by Y-component of fg???
     
  15. Jan 27, 2007 #14

    arildno

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    Decompose mg in the tangential and normal directions.
     
  16. Jan 27, 2007 #15
    k, check out the diagram and let me know, if u don;t get something. Thanks
     

    Attached Files:

  17. Jan 27, 2007 #16

    arildno

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    I get it perfectly, but I don't think you do.
    At the very least, what you've posted so far is wrong.

    Now, look at it this way, then:
    If the TANGENTIAL direction makes a 35 degree angle with the horizontal, what is the angle between the NORMAL direction and the vertical?
     
  18. Jan 27, 2007 #17
    k let me c, one sec., and TANGENTIAL direction is the one it is heading ?, the angle makes 55 degrees, i guess.
     
    Last edited: Jan 27, 2007
  19. Jan 27, 2007 #18

    arildno

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    Quite so; tangential means ALONG the ramp, normal direction means the direction head-on the ramp, making a 90 degrees angle with the tangential direction.
     
    Last edited: Jan 27, 2007
  20. Jan 27, 2007 #19
    so what would force of normal equal.
     
  21. Jan 27, 2007 #20

    arildno

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    EDIT: 55 degrees is wrong!
    Please make new posts, rather than fill old posts with new stuff you are uncertain of.
     
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