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A ball hanging by a string while I'm accelerating.

  1. May 3, 2013 #1
    1. The problem statement, all variables and given/known data

    a ball was hanging by an inextensible string to a small airplane. firstly, it was hanging straight down but when the pilot started the motor the string made an angle of 30 degrees wrt the vertical.

    Assuming that the ball is a point mass, find the acceleration (m/s²) of the airplane (neglect air resistance)


    3. The attempt at a solution

    While hanging at 30 degrees, the ball is constant, this means that the vectorial sum of forces is zero (newton's law), but I don't know what to do next !
     
  2. jcsd
  3. May 3, 2013 #2

    jambaugh

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    What forces are acting on the ball? Why do you say the sum of vector forces is zero?
     
  4. May 3, 2013 #3
    The weight and the force of plane's acceleration, I guess ! !

    it is zero because the ball is constant with respect to the plane, Should I take the earth as my reference so the ball is moving ?
     
  5. May 3, 2013 #4
    Yes. To apply Newton's second law, you need to express it with respect to an inertial reference frame. Also, if you do a free body diagram on the ball, you will see that you have omitted one of the forces acting on the ball. Also, what exactly do you mean by the term "the force of the plane's acceleration?" Relative to an inertial reference frame, this has no meaning. Also, relative to an inertial reference frame, the net force on the ball is not zero.
     
  6. May 3, 2013 #5
    You mean that I should change my inertial reference frame ?
    I think I omitted the tension of the chord.

    Can someone please give me the first steps ! this problem is killing me.
     
  7. May 3, 2013 #6
    I would suggest you draw a free body diagram of the ball. Be sure to include tension, and gravity, those are the only two forces acting on the ball. Take a look at the components of both forces, if any.
     
  8. May 3, 2013 #7

    BruceW

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    yeah, for this problem you should just use an inertial reference frame (since that is what I would guess you are used to). So think of the motion of the ball with respect to the ground, not with respect to the plane. And think of how the acceleration of the ball must be related to the acceleration of the plane ?
     
  9. May 4, 2013 #8
    I'm sorry but I can't do it, I stuck with the first steps (drawing the forces), I really appreciate that someone give me the forces.
     
  10. May 4, 2013 #9
    Please don't give up. Here are some hints:

    1. If the plane is accelerating, and the ball is stationary relative to the plane, then relative to the ground, the ball is accelerating at the same rate as the plane.
    2. Think of the ball hanging from the string at an angle of θ=30 degrees to the vertical. There are two forces acting on the ball: A vertical force of gravity through the center of mass, pointing downward toward the ground, and a tension force pointing upward at an angle of 30 degrees to the vertical. If T is the tension in the string, (in terms of T and θ), what is the component of the tension force on the ball vertically upward? What is the component of the tension force on the ball horizontally?
    3. Is the ball accelerating vertically upward relative to the ground? If not, the forces on the ball in the vertical direction must be in balance. From this you can determine the tension in the cord.
    4. Is the ball accelerating horizontally relative to the ground? What is the net component of force acting on the ball horizontally? From newton's second law, how is this net component of force related to the acceleration of the ball?
     
  11. May 4, 2013 #10
    Oh, thanks a lot.

    T is the tension and P is the gravity:

    Vertically : T sin(30°)=P.
    Horizontally : T Cos(30°)= m a.

    Using the first equation, we have : m g tan(30°)=m a, which means that : a=g tan(30°).
     
  12. May 4, 2013 #11

    BruceW

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    looks good to me. nice work!
     
  13. May 4, 2013 #12
    Good, except for one thing, In the first two equations, you have the sine and cosine reversed. Astonishingly, even though you did this, you somehow got the final answer correct.
     
  14. May 4, 2013 #13

    BruceW

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    ah, yeah, the string is meant to be 30 degrees with respect to the vertical. but his equations would imply 30 degrees with respect to the horizontal, and rearranging them give a tan(30) = g , but he wrote them the wrong way around, which made them the same as the correct answer. hehe two wrong make a right?
     
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