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Newtonian motion

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known

    You are playing with your daugther in the snow. She sits on a sled and asks you to slide her across a flat , horizontal field. You have a choice of (a) pushing her from behind by applying a force downward on her shoulders at 30 degrees below the horizontal (b) attaching a rope to the front of the sled and pulling with a force 30 degrees above the horizontal. Which would be easier for you and why

    2. Relevant equations
    F =ma

    3. The attempt at a solution
    I think both situations yield the same net force thus the same acceleration
     
  2. jcsd
  3. Apr 12, 2016 #2

    mfb

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    Consider the case where the sled has a constant velocity. Do you still need a force to keep it moving? What about that effect (it is also relevant while still accelerating)?
     
  4. Apr 13, 2016 #3
    Both options are not in constant velocity. Both options have one force in the horizontal direction
     
  5. Apr 13, 2016 #4

    mfb

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    It does not matter.

    There is an effect you missed so far. The force won't all be used for acceleration. There is something else to consider.
    Hint: it is snow, not ice.
     
  6. Apr 14, 2016 #5
    Friction
     
  7. Apr 14, 2016 #6

    mfb

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    Correct. How do you calculate the friction force? Does pushing/pulling at an angle change it?
     
  8. Apr 14, 2016 #7
    I think for both cases it will be -Fcos@
     
  9. Apr 14, 2016 #8

    mfb

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    What is F, and where does the sign come from, and in which direction?
     
  10. May 7, 2016 #9
    F is the applied force
     
  11. May 7, 2016 #10
    The sign is a mistake
     
  12. May 7, 2016 #11

    nrqed

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    I think that what mfb wants to make you think about is the definition of the friction force. What is the fundamental equation that gives the kinetic friction force on a sliding object?
     
    Last edited: May 7, 2016
  13. May 8, 2016 #12
    Fk = ukN
     
  14. May 8, 2016 #13

    nrqed

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    Right. So what can we say about the normal force in the first case compared to the second case?
     
  15. May 8, 2016 #14
    It has a component in one case?
     
  16. May 8, 2016 #15

    mfb

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    What does "it has a component" mean? All forces can always be split into separate components. But where do those components point to?
     
  17. May 10, 2016 #16
    I mean the force is at an angle so it will result in a horizontal and vertical comp.. but the thing is both forces are at angles. But the components of both forces have different direction
     
  18. May 10, 2016 #17
    Are you trying to point that?
     
  19. May 10, 2016 #18

    mfb

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    Right.
    So what is your conclusion?
     
  20. May 10, 2016 #19
    The y comp of first case will be up while down in second case....so maybe more friction in second case? Is that right
     
  21. May 10, 2016 #20

    mfb

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    The second case is you pulling up. You expect more force on the ground there?
     
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