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Hard Physics Problem on Forces

  1. Oct 20, 2012 #1
    Frensley_Forces_Horizontal_006.gif


    1. The problem statement, all variables and given/known data
    An airplane begins its takeoff sequence moving with a constant acceleration a. A passenger holds up a pocketwatch during the takeoff sequence and notices that the watch makes an angle θ = 14° with the vertical, and that 13.3 seconds pass before the plane leaves the runway.

    (a) What is the plane's constant acceleration?


    (b) How far does the plane travel on the runway?


    2. Relevant equations

    F = ma

    3. The attempt at a solution

    There is no given mass, I don't know how to find it without it.
     
  2. jcsd
  3. Oct 20, 2012 #2

    Doc Al

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    Staff: Mentor

    Just call the mass 'm'. You won't need the actual value.
     
  4. Oct 20, 2012 #3
    can you explain how?
     
  5. Oct 20, 2012 #4

    Doc Al

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    Apply Newton's 2nd law. What forces act on the pocketwatch? Analyze horizontal and vertical force components separately.
     
  6. Oct 20, 2012 #5
    I tried but still nothing, the vertical component is mg and the horizontal one is 9.5m(mg*cos14), am I missing something?
     
  7. Oct 20, 2012 #6

    Doc Al

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    First things first. What forces act on the watch? (There are two forces acting: Name them.)
     
  8. Oct 20, 2012 #7
    Force tension and force weight
     
  9. Oct 20, 2012 #8

    Doc Al

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    Excellent. What are the horizontal and vertical components of each force? (The weight is mg; call the tension force "T".)
     
  10. Oct 20, 2012 #9
    Components of force weight:
    horizontal component: 0
    vertical component: mg

    Components of force tension:
    horizontal component: mg*cos14
    vertical component: mg
     
  11. Oct 20, 2012 #10

    Doc Al

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    Good. (The vertical component acts down.)

    No. (The vertical component will end up equal to mg, but you'll get to that in the next step.)

    The tension force is T. It acts parallel to the chain of the watch. What are its components?
     
  12. Oct 20, 2012 #11

    horizontal component: T * sin14
    vertical component: mg
     
  13. Oct 21, 2012 #12

    Doc Al

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    Good.

    But I would have preferred:
    Horizontal: T*sin14
    Vertical: T*cos14

    Nice and simple.

    Now apply ƩF = ma to the vertical and horizontal components.
     
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