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Static Equilibrium Ball Problem

  1. Sep 16, 2007 #1
    I have been working on this problem all day and I cannot figure out how to solve it hopefully somebody has an idea on how to get me moving in the right direction.


    1. The problem statement, all variables and given/known data

    Two Identical 15.0-Kg Balls, each 25.0 cm in diameter are suspende by two 35.0-cm wires. The entire Apparatus is supported by a single 18.0 - cm wire and the surfaces of the balls are perfectly smooth. (A) Find the tension in each of the three wires (B) How Hard does each ball push on the other one.

    [​IMG]



    2. Relevant equations

    Pythagorean theorem? None given

    3. The attempt at a solution

    I tried to find the angle near the top and use cos and sin to find the equation but I got nowhere near the answer. I got the answer for the top wire holding the other to but I need to find the tension in the two identical wires still. I keep getting 105N and the answer is 152 for both wires. The answer to the second part is 40 and I got 36. I think where I did wrong was in the angle at the top but I confered with a friend and they got 20.92 as well.

    thanks in advance for any help with this problem
     
  2. jcsd
  3. Sep 16, 2007 #2

    PhanthomJay

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    I agree with your angle, but otherwise, I don't get either your answer or the book's. Let's start witht the top wire tension. What did you get for that? Should be about 30(9.8) = 294N.
     
  4. Sep 16, 2007 #3

    Astronuc

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    Top wire bears the weight of the two balls, so the tension in each of the two diagonal wires must be greater than half the tension in the single wire.

    If the diagonal wires are connected as shown in the image, one can find the triangle and apply the Pythagorean theorem by virtue of symmetry in the problem.

    The triangle has a hypotenuse of 35 cm and a base of the diameter (12.5 cm) of the ball. Draw a free body diagram with the tension up along the 35 cm wire, and the weight acting downward. Since the weight of one ball is 147 N, then the tension in the 35 cm wire has to be greater than 147 N.

    Now if the 35 cm wire acts from the junction of the three wires to the center of mass of each ball, then the base of the triangle is smaller if measured at the attachment point of the 35 cm wire and ball, but that triangle is similar to the one with a hypotenuse of the wire+diameter of the ball and a base = diameter of ball.
     
  5. Sep 16, 2007 #4

    PhanthomJay

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    Yes, sorry, i missed that, the angle is smaller than 20.4 degrees, and the book answer looks good....the balls rotate and the wires act thru the center of mass.
     
  6. Sep 16, 2007 #5
    I got that the top angle should be 16 degrees but I don't know how. We never went over center of mass so I am assuming you arn't supposed to find it that way. Thank you very much for you help. How do you find the angle?
     
  7. Sep 17, 2007 #6

    Astronuc

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    Does one mean the angle between the two diagonal (35 cm) wires? That angle is about 30.5°.

    To find the angle between the diagonal wire and vertical, use simple trig.

    As I mentioned previously, it will depend on what one assumes for the orientation of the balls. I initially took the 35 cm as hypotenuse and the 12.5 cm as the base of a right triangle. One can then take sin-1 (12.5 cm / 35 cm) and obtain 20.9°. Using that angle T cos (20.9°) = 147 N (weight of one ball) gives on a tension T of 157.4 N, which is not ~152. The angle is based on the assumption that the balls are oriented such that the connection between 35 cm wire and ball is vertically above the center as shown in the drawing.

    Now if one assumes that the 35 cm wire is oriented linearly from the connection with the vertical wire to the center of the ball, that changes the geometry of the wire. The base of a right triangle with hypotenuse of 35 cm is less than 12.5 cm (diameter of ball), but it is similar to the right triangle formed with a hypotenuse = 35 cm + 12.5 cm, the horizontal leg (base) of 12.5 cm (diameter of ball), and the vertical leg (between juntion of vertical and diagonal wires and the contact point of the two balls).

    Then on uses sin-1 (12.5 cm / 47.4 cm) = 15.26°, which is half the angle between diagonal wires, or the angle of the diagonal wire with respect to vertical.

    Then T cos (15.26°) = 147 N yields T = 152.4 N

    One can see the importance of orientation of the wires in the physical situation.
     
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