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Tension of wire attached to 45-kg ball.

  1. Jun 12, 2016 #1
    1. The problem statement, all variables and given/known data
    "A solid uniform 45-kg ball of diameter 32cm is supported against a verticle frictionless wall using a thin 30cm wire of negligible mass.

    A)Make a free body diagram for the ball and use it to find the tension in the wire.
    B)How hard does the ball push against the wall?



    2. Relevant equations
    ##\sum(Fx) = 0##
    ##\sum(Fy) = 0##
    ##a^2 + b^2 = c^2##



    3. The attempt at a solution
    k7Npvhc.jpg
    There is a picture of the situation.


    Here is another picture of my free body diagram.
    WJJzm9A.jpg

    Now, using the equilibrium equations, I got ##\sum(Fy) = -441.45 + TSinθ = 0##
    So, ##Tsinθ = 441.45## (Newtons).

    Now for the x components.
    ##\sum(Fx) = Tcosθ - N = 0##

    So, ##Tcosθ = N##

    Now I need to figure out what the angle is. The diameter of the sphere is 32cm, so the radius must be 16cm. The length of the wire is 30cm. I used the pythagorean theorem to find the height from the sphere to the wire.

    ##16^2 + b^2 = 30^2##
    ##256 + b^2 = 900##
    ##b^2 = 644##
    ##b = 25.38##

    Now ##\arcsin(25.38/30) = 57.78 degrees##

    This is where I start to have trouble. I figured I would find T by plugging that angle into ##Tsinθ = 441.45## and solving for T. This gave me an answer of 521.8 N. When I check the back of the book, the answer is actually 470 Newtons.

    Any help?
     
  2. jcsd
  3. Jun 12, 2016 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    So far, so good.
    This is where everything comes apart.

    You should make a detailed sketch of the triangle you are trying to solve. Even looking at the picture from the book should give you a hint that your first equation,
    ##16^2 + b^2 = 30^2##
    is wrong.

    Take another close look at the geometry of the sphere and see if you can't come up with a different (and correct) relationship.
     
  4. Jun 12, 2016 #3
    Okay, I looked and found the mistake. The 30 cm hypotenuse only went from the top to the center of the ball. I should have added an extra 16 cm to it to go through the entire sphere. I got the right answer. Thank you for your help!
     
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