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Centroids and Center of Gravity

  1. Oct 9, 2005 #1
    The homoegeneous wire ABCD is bent as shown and is supported by a pin at B. Knowing that l=8, determine the angle theta for which portion BC of the wire is horizontal.

    Below is the free body diagram with the forces exerted by the pin indicated by the two arrows on the wire.
    How do I approach this problem?

    The first part of the wire is a semi-circle with radius 6 and second part a horizontal line with length 6, and the third part a diagonal line iwth length 6.
     

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  3. Oct 9, 2005 #2

    Doc Al

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    I assume that if you had three point masses you could find the center of mass? If so, then treat this object as three pieces: replace each piece by its equivalent point mass at its center of mass. You'll have to integrate to find the center of mass of the half circle piece; but the two straight pieces should be easy. Express the center of the angled straight piece as a function of theta.

    For BC to be horizontal, where must the center of mass end up?
     
  4. Oct 9, 2005 #3
    If I want to set the center of gravity of the wire equal to the centroid of the wire, where should I set the origin to be?
     
  5. Oct 9, 2005 #4

    Doc Al

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    Not sure what you mean: The centroid is the center of gravity. Choose your origin wherever you like; the important thing is where the centroid has to be for the thing to be balanced as stated. (I'd pick that point as the origin.)
     
  6. Oct 9, 2005 #5
    The centroid of the entire wire has to be at the point where the forces of the pin is applied, right?
     
  7. Oct 9, 2005 #6

    Doc Al

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    The line of force must pass through the centroid. (In other words: The centroid must be on a vertical line directly under point B.)
     
  8. Oct 9, 2005 #7
    so the x coordinate of the center of mass of the entire wire must be 0 if the origin is chosen at the point where the vertical component of force exerted by the pin is applied?
     
  9. Oct 9, 2005 #8

    Doc Al

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    Yes. If the pin is at the origin, the centroid is on the -y axis.
     
  10. Oct 9, 2005 #9
    Is the centroid of the semi-circle -72 mm^2, the horizontal part 32mm^2, and the diagonal piece 8-3*cos(theta)mm^2? Then you set (x-cordinate of center of gravity)(Length of the wire)=sum of the centroids of the entire wire, which then becomes 0= sum of the centroids?
     
    Last edited: Oct 9, 2005
  11. Oct 9, 2005 #10

    Doc Al

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    Why are your units in mm^2? The centroid is a position; give the coordinates in the units of length. (No units were specified in your first post.)

    The centroid of a semi-circle is [itex]2 R /\pi[/itex] from the center (according to my handy math book).
     
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