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Forces revisted

  1. Oct 23, 2007 #1
    I am trying to help a friend with this webassign problem:

    Jean, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from Dulles Airport . She notices that the string makes an angle of 23° with respect to the vertical while the aircraft accelerates for takeoff, which takes about 18 seconds. Estimate the takeoff speed of the aircraft.

    I just need a kick in the rear to get me started! I know that there is Tension and weight acting on the watch. I know that average a=v/t so average v=at and I know that F=ma
    But I am having trouble putting it all together since I do not have any mass or other numbers besides 18s.

    Anyone got a hint for me?

  2. jcsd
  3. Oct 23, 2007 #2


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    The vertical component of the tension supports the weight of the watch while the horizontal component provides the acceleration force.
  4. Oct 23, 2007 #3
    Right, I am just not sure what to do without T or mg.....hmmmm
  5. Oct 23, 2007 #4
    Try breaking the Tension up into it's components and apply Newton's 2nd law.
  6. Oct 23, 2007 #5
    Obviously I am missing something big here. I do not have T; if I break it into components, I just get T_x=Tsin23 and T_y=cos23.... so by Newton's 2nd get [tex]\sum F_x=ma[/tex] so [tex]T\sin23=ma[/tex]

    ...I am missing how to use 18 seconds....
    ....and I don't see how I could cancel anything..arrrgggggghhh:mad:

    What am I overlooking?
  7. Oct 23, 2007 #6
    Try applying Newton's 2nd law to [tex]\sum F_{y}[/tex] too.

    This is actually a neat and useful solution, and simple enough that once you solve it you'll be able to remember it always :)
    Last edited: Oct 23, 2007
  8. Oct 23, 2007 #7
    So...[tex]\sum F_y=0[/tex] -->[tex]T_y-mg=0[/tex] --->[tex]T_y=mg=T\cos23[/tex] so...hey....[tex]m=\frac{T\cos23}{g}[/tex]....looks like T will cancel if I plug that in.

    I can't believe that they would give this to a high school student as the first of 15 problems....way to discourage the crap out of a bunch of kids who can't stand physics/math anyway.

    I really think they should ease into the assignment a little more. Get these kids brains warmed up.

    Christ...I would have given up if it were assigned to me.

    thanks for the help,
  9. Oct 24, 2007 #8


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    the horizontal component of the tension will be

    [tex]T_H = T\sin(\theta)[/tex]

    and the vertical

    [tex]T_V = T\cos(\theta)[/tex]

    but the horizontal component need to accelerate the watch

    [tex]T_H = ma[/tex]

    and the vertical need to cancel the weight

    [tex]T_V = mg[/tex]


    [tex]\frac{T_H}{T_V} = \frac{ma}{mg} = \frac{T\sin(\theta)}{T\cos(\theta)} = \tan(\theta)[/tex]


    [tex]a = g \ tan(\theta)[/tex]
  10. Oct 24, 2007 #9
    This can also be solved geometrically:
  11. Oct 24, 2007 #10
    Thanks guys, but I already solved this in post #7. I was just babbling towards the end there :)

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