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Problem on tension

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-1-27_19-17-45.png

    An ornament for a courtyard at a World's Fair is to be made up of four identical, frictionless metal spheres, each weighing ##2\sqrt{6}## ton-wts.The spheres are to be arranged as shown, with three resting on horizontal surface and touching each other; the fourth is to rest freely on the other three. The bottom three are kept from separating by spot welds at the points of contact with each other. Allowing for a factor of safety of 3, how much tension must the spot welds withstand?

    2. Relevant equations


    3. The attempt at a solution
    At first, I calculated the angle between the vertical axis and the line segment which connects the centres of the top and bottom spheres. Let, the angle be ##\theta##.;
    upload_2015-1-27_19-52-50.png
    In the picture, A, B , C and D are the centers of the spheres. D is the centre of the top sphere. Let the radius of a sphere be ##R## .
    So, AB = BC = CA = AD = 2R ;
    Applying pythagorean theorem for APC triangle, ## AP = \sqrt{3} R ##
    As, G is the centroid, AG : GP = 2 : 1 ;
    So, ## AG = \frac{2R}{\sqrt{3}} ##
    From, the triangle AGD, ##sin \theta = \frac{AG}{AD} = \frac{1}{\sqrt{3}} ##
    and, ## tan \theta = \frac{1}{\sqrt{2}} ##
    The vertically downward (component) force of the top sphere on each of the bottom spheres will be
    ##= \frac{1}{3} \cdot 2 \sqrt{6} = \frac {2\sqrt{2}}{\sqrt{3}} ##

    Now, let's consider the force on the bottom sphere having centre A :
    upload_2015-1-27_20-14-37.png
    So the vertical component is ## \frac {2\sqrt{2}}{\sqrt{3}} ##
    And so, horizontal component becomes ## = \frac {2\sqrt{2}}{\sqrt{3}} \cdot tan \theta = \frac{2}{\sqrt{3}}##
    Now, look at the forces on the bottom spheres in the horizontal plane.
    upload_2015-1-27_20-12-51.png
    This horizontal force on A, can be divided into two components, along CA and BA ;
    To, get the tension in AC, I took the component along AC which is ##= \frac{1}{2} \frac{2}{\sqrt{3}} cos 30 = \frac{1}{2} ##
    As, the safety factor is 3, the tension becomes ##\frac{3}{2}##;
    But it doesn't match the answer.
     
  2. jcsd
  3. Jan 27, 2015 #2

    TSny

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    Your work looks good to me all the way to the last part where you are considering the horizontal forces on the bottom spheres. Draw a free body diagram of one of the bottom spheres showing all of the horizontal force components acting on the sphere.
     
    Last edited: Jan 27, 2015
  4. Jan 27, 2015 #3

    CWatters

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    Check the line that begins...

    When you resolve a force into two orthogonal components the original force vector is the hypotenuse not the side of the triangle.
     
  5. Jan 27, 2015 #4
    I confirm your result for the horizontal force. For the horizontal force balance, I get 2Tcos(30)=(horizontal force). This gives T = 2/3. If I multiply by the factor of safety, I get 2.

    Chet
     
  6. Jan 28, 2015 #5
    Your suggestion was helpful to find out my mistake.
     
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