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Forces and Normal Force

  1. Feb 5, 2008 #1
    [SOLVED] Forces and Normal Force

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

    In the figure, a crate of mass m = 75 kg is pushed at a constant speed up a frictionless ramp (θ = 32°) by a horizontal force F. The positive direction of an x axis is up the ramp, and the positive direction of a y axis is perpendicular to the ramp. (a) What is the magnitude of F? (b) What is the magnitude of the normal force on the crate?

    2. Relevant equations
    F=ma


    3. The attempt at a solution
    After drawing a free body diagram...there is a movement to the right. F=mg which should mean F=75x9.8 which is 735N. I would think that the Normal Force would be found by F(g)cosx. Im fairly clueless about part A
     
  2. jcsd
  3. Feb 5, 2008 #2

    rock.freak667

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    When you resolve the weight you get mgsin32 down the plane and mgcos32 perpendicular to it right?

    The only force acting in the +ve y direction is the the Normal reaction,R. So you know that answer.

    Since the motion is up the plane the resultant force is

    F-mgsin32=ma

    What do you think the resultant force should be equal to while moving at a constant speed?
     
  4. Feb 5, 2008 #3
    I am really confused with this
     
  5. Feb 5, 2008 #4
    Wouldnt the Normal force be 632.32N? since the gravitational force is -632.32?
     
  6. Feb 5, 2008 #5

    rock.freak667

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    Yes it would.

    Resultant force,ma=F-mgsin32

    If you are moving with a constant speed up the ramp and your direction is not changing, what does that say about your acceleration?
     
  7. Feb 5, 2008 #6
    there is no acceleration?
     
  8. Feb 5, 2008 #7

    rock.freak667

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    Yes that is correct. If there is no acceleration then what is the resultant force acting on it?
     
  9. Feb 5, 2008 #8
    Im sorry, but im really just not getting this one
     
  10. Feb 5, 2008 #9

    rock.freak667

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    ok then, we start over.

    Do you agree that the force acting down the plane due to the horizontal component of the weight is mgsin(32) ?

    A force,F, is pushing the crate up the plane. What is the resultant force here then?

    F-mgsin(32) right?
     
  11. Feb 5, 2008 #10
    That force would be 632.32 - 389.39?
     
  12. Feb 5, 2008 #11

    rock.freak667

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    Nope.


    389.39 is the value of mgsin(32)

    632.32 is the normal reaction. Those two forces are in different directions.


    But anyhow.

    The resultant force =F-389.39

    If you are moving with a constant speed and your direction is not changing you aren't accelerating,right?
    Then a=0 right?

    Now if the resultant force is ma (Newton's 2nd law)

    then ma=F-389.39
     
  13. Feb 5, 2008 #12
    so it would equal 389.39 if acceleration is 0?
     
  14. Feb 5, 2008 #13

    rock.freak667

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    It would be equal to that because the crate is not accelerating.
     
  15. Feb 5, 2008 #14
    Well i put in 632.32 for the Normal force and its apparently wrong
     
  16. Feb 5, 2008 #15

    rock.freak667

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    What is the answer?
     
  17. Feb 5, 2008 #16
    I have no idea...i thought it was 632.32
     
  18. Feb 5, 2008 #17

    rock.freak667

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    Then how do you know the answer is not 623.95N?
     
  19. Feb 5, 2008 #18
    Theonline program rejected it. Wasnt the other answer 389.39? because if it was, that was rejected as well
     
  20. Jun 4, 2009 #19
    Re: [SOLVED] Forces and Normal Force

    F=mg*tan(theta)
    mg=mass*gravity
    Fn=pithagren theorm
    Fn=sqrtroot[(Fnx^2)+(Fny^2)]
    Fny=mg
    Fnx=F
     
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