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Finding Weight Of A Bird?

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A man is flying a kite. The string of the kite makes an angle of 50 degrees with respect to the horizontal and exerts a pull of 15 Newtons. A small bird decides to sit on the kite and causes the kite to reach a new equilibrium position with the sting at 30 degrees with respect to the horizontal. Assume that the force exerted by the wind on the kite alone is exactly the same as the force exerted by the wind on the kite and bird together. Calculate the weight of the bird.




    3. The attempt at a solution
    For this all i did was find the forces in the x direction for both cases and found the delta force so:
    Fx1 = 15 N cos50 = 9.64
    Fx2 = 15 N cos30 = 12.99

    So delta Fx = 12.99-9.64 = 3.35N
    So than 3.35N/9.81 = 0.3415kg

    It just seems to simple and i think i did something wrong, any insights would be appreciated.
     
  2. jcsd
  3. Mar 26, 2012 #2

    PeterO

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    In the first case, the weight of the kite + the tension in the string + the force of the wind = 0 since the kite is stationary - from that we can work out the size of the weight force and the wind force. Take particular notice of the direction of the sum of the Weight + wind forces.

    Once the bird is added, the wind force stays the same, but the weight increases. This gives a change in angle of the sum of weight and wind. The tension in the string once the bird is added may well be different to 15N
     
  4. Mar 27, 2012 #3
    Ok so heres what i did for that one. For the scenario with know bird on it i did
    mg - T1sin theta = 0
    soo... m = (15Nsin50 degrees)/9.81 = 1.17kg = the mass of the kite at 50 degrees
    Than for scenario 2 with the bird i did:
    mg - T2sin theta = 0
    soo... m = (15Nsin30)/9.81 = 0.76kg

    And than i took difference of the two masses: 1.17 - 0.76 = 0.41 kg = 4.02 Newtons

    Is this Correct??
     
  5. Mar 27, 2012 #4

    PeterO

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    Notice that you have a smaller mass for the kite+ bird, than for the kite alone; something is wrong.

    The error is assuming that when the kite sinks to 30 degrees the Tension will still be 15N - it is a reaction force and can change. It is the force from the wind that remains the same.
     
  6. Mar 27, 2012 #5
    So basically, the tension in sequence number 2 is not 15 Newtons? Ya but in the question it states that the force exerted by the wind is the same both times doesnt it?
     
    Last edited: Mar 27, 2012
  7. Mar 27, 2012 #6

    PeterO

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    That's right - the force exerted by the wind is the same. However, the weight involved changes, so the Tension may well change, not only in direction [they tell you that] but possibly magnitude as well.

    Case 1.

    Kite - weight down
    Wind - Force in some direction
    String - Tension at 50o , so partially down and partially left.

    The wind force have an up and a right component.

    The right component balances the left component of the tension.
    The up component balances the combined weight plus down component of the tension.

    Case 2.

    The wind force doesn't change. - so its up and right components are presumably unaltered.
    The weight force increases.
    So the tension has the same left component, but a smaller down component [since (increased) weight plus tension component are the same; as the wind force is the same]

    Not surprisingly, the angle on the tension alters - and also its magnitude; Pythagoras will tell us that.

    Presumably there is just enough information to get the actual weight of the bird. I am not convinced yet.

    EDIT: Just re-read the question and there is enough information. You need to calculate the components of the Tension [left and down] the consider what I wrote above.
     
  8. Mar 27, 2012 #7
    So for the left direction in sequence 1 the left direction: T1 cos 50 = 15cos50 = 9.64 N and than for the y direction: T1sin50 = 15sin50 = 11.49N. So using pythogorean: hypotenuse = square root of 9.64^2 + 11.49^2 = 14.99832324N

    The left direction in sequence 2 it is T2 cos 30 = 15cos30 = 12.99 N and than in the y direction for sequence 1: T2 sin 30 = 15sin30 = 7.5N
    So using pythagorean u get the hypotenuse to equal square root of 12.99^2 + 7.5^2 = 14.99967N

    so than delta = 14.99967 - 14.99832324 = 1.34 x10^-3N

    Is that how i would do it?
     
  9. Mar 27, 2012 #8
    I did the exact same thing as stanc, however, i think that since the force of the wind stays the same for both situations, the force in the x would always 9.64N and the force in the y would always equal 11.49N
    You would than have to say to find hypotenuse in situation 2 it is tan 30 = opp/adj = 11-x/9.64 where x represents the mass of the bird.
    it turned out to be 5.93 newtons
     
  10. Mar 27, 2012 #9
    Before the lazy bird landed on the kite
    [itex]F_y=15Sin50^o[/itex]
    [itex]F_x=15Cos50^o[/itex]

    Assuming the tension of the string remains.
    [itex]F_y-mg=15Sin30^o[/itex]
    [itex]mg=F_y-15Sin30^o[/itex]
    mg = 15x0.2660=4N

    If of unequal tension,
    [itex]mg=F_y-XSin30^o[/itex] where X is the new tension.
     
    Last edited: Mar 27, 2012
  11. Mar 27, 2012 #10
    Where did you get the 15x0.2660?
     
  12. Mar 27, 2012 #11
    [itex]F_x and F_y[/itex] remain constant
    [itex]mg=15Sin50^o-15Sin30^o[/itex]
    mg = 15x0.2660=4N
     
  13. Mar 27, 2012 #12
    But wouldnt that be saying 15sin 30 = 15sin50 -mg which is saying the second situation = the first situation subtract the force of the bird whereas the force of the bird would have to be added to the first situation to equal the second situation and also why would you only consider the y components?
     
    Last edited: Mar 28, 2012
  14. Mar 28, 2012 #13
    The weight of the bird only acting downward. Assuming it a single point.
    It has no horizontal component,mgCos90°=0
     
  15. Mar 28, 2012 #14
    Ok but can you explain to me why the equation mg = 15sin50 - 15sin30 is correct? instead of -mg = 15sin50 - 15sin 30?
     
  16. Mar 28, 2012 #15
    I don't know how you get -mg = 15sin50 - 15sin 30.
    Maybe you can show it.
     
  17. Mar 28, 2012 #16

    PeterO

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    No THAT IS NOT THE WAY TO DO IT. [WHOOPS CAPS LOCK STUCK ON - still; that's better]

    The first part of the calculation is correct,

    the left direction component is indeed 9.64 N - indicating the wind force had a "9.64N to the right" component.

    The vertical component was indeed 11.49N down, indicating the wind force had a [11.49N + Weight of kite] upward component.

    Once the bird has landed, the horizontal component is STILL 9.64N since the wind force didn't change, and the combined downward component of Tension and Weight(kite + bird) must also still be the same, since the wind force didn't change.

    Since we now have a new triangle with the Tension, we can see that tan30o = [new downward component]/[old horizontal component].

    That will give us a new, smaller downward component [it will also give us a new overall tension but who cares]

    lets suppose that the new downward component was 8N, that would mean the component is 3.49N smaller, meaning kite + bird is 3.49N bigger than kite alone; so the bird weighs 3.49 N.

    Now, I just invented that value of 8N. You must use your trig on the 30o triangle to find what it is really equal to.

    EDIT: If you want you can calculate what the new tension is - it is simply 9.64 / cos30o
     
    Last edited: Mar 28, 2012
  18. Mar 28, 2012 #17
    Ok so than new downward component would be: tan 30 = downward component/9.64 so downward component would equal 5.56 Newtons and therefore the difference would be 11.49N - 5.56N = 5.924N. But when i find the distance do i not have to find the same downward component using the angle 50 degrees?? and also why can i not use mg = 15sin50 - 15sin30 to solve for the birds weight?
     
  19. Mar 28, 2012 #18

    PeterO

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    The problem is that highlited in red above. The tension in the string does not remain 15N
     
  20. Mar 28, 2012 #19

    PeterO

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    OK, you have the birds weight at 5.924N

    if mg = 15sin50 - 15sin30 happens to give that same answer - then perhaps you can use it - but I don't expect it will.

    I can understand why you would involve a 15sin50 factor, and/or a 15cos50 factor, because when the angle is 50 degrees, the tension is 15N - you were told that.

    HOWEVER, you were told that the angle changed to 30 degrees, and the wind force remained the same. There was no suggestion that the tension remained at 15N. That is why a 15sin30 or a 15cos30, or even a 15tan30 term has no application to this problem.

    Also, what distance were you trying to find, and why?
     
  21. Mar 28, 2012 #20
    Ok i understand, what i was wondering was since i found the new downward component using pythogorean with tan (angle) = downward component / 9.64N shouldnt I do the same for the first sequence by doing tan(50) = downward component / 9.64N? or is 15sin(50) the same thing?
    EDIT: i didnt realize i wrote distance, i meant downward component, sorry..
     
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