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Equilibrium of rigid bodies

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    why 2/3 is required in this calculation as circled ? I have no idea.
    I know that 4.5/2 is the distance of the centroid of the bar from point A .

    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2015 #2
    Because it is not a constant load, the load increases as you move to the right. Do you know how to draw Shear/Moment diagrams?
     
  4. Sep 24, 2015 #3

    SteamKing

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    The calculation shown is not about the centroid of the bar is located; it's about where the centroid of the triangular load is located from the left support.
     
  5. Sep 25, 2015 #4
    well , can you explain what the 2/3 means here?
     
  6. Sep 25, 2015 #5

    SteamKing

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    The centroid of a right triangle is located 2/3 of the length of the base from the pointy end.

    https://en.wikipedia.org/wiki/List_of_centroids
     
  7. Sep 25, 2015 #6
  8. Sep 25, 2015 #7

    SteamKing

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    Is 2 kN/m x 4.5 m total the force? Wouldn't that be true only if the 2 kN/m distributed load was applied evenly over the entire length of the beam?
     
  9. Sep 25, 2015 #8
    you mean the different points at entire length of rod have different forces? if so, why shouldn't the force = 2x4.5x2/3 ?
    since the force involved only from point A to centroid ...
     
  10. Sep 25, 2015 #9

    SteamKing

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    You've apparently missed some basic instruction in how distributed loads are represented. Here are two commonly encountered types of distributed loadings:

    distributed-loading.jpg
    Unlike with a concentrated load, the individual arrows in the figures on the left do not represent individual loads; they are merely diagrammatic. The equivalent concentrated load and its location are given in the corresponding figures to the right.

    The first type is the uniformly distributed load. This is a loading which has a constant amount of force applied per unit length along the beam. In the diagram, this load is w kN/m, for example.
    The Total Load = w * L and its center is located at L/2 from one end of the beam.

    The second type is the uniformly varying load. This is a loading which has a zero amount of force applied per unit length at the right end of the beam, and this loading increases in a linearly increasing fashion until it reaches say w kN/m at the left end of the beam.
    The Total Load = (1/2)*w*L and its center is located at (2/3) of the length of the beam from the end where w = 0.
     
  11. Sep 25, 2015 #10
    ok , why The Total Load = (1/2)*w*L is located at (2/3) of the length of the beam from the end where w = 0.
    why not w*L as in the first case ?
     
  12. Sep 25, 2015 #11

    SteamKing

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    Because most of the total load is located toward the end opposite of the end of the beam where w = 0.

    If you took each beam as loaded in the diagrams, and placed a fulcrum under the location of the center of the total load, the beam would be balanced, which indicates that the moments of the load on either side of the fulcrum produce equal and opposite rotation of the beam. The moments cancel out, the rotations cancel out, and the beam is balanced around the point.

    It's like playing on a see-saw: as long as the load on each end of the seesaw is the same and is located the same distance away from the middle, the see-saw is balanced. Add some more load to one side, and the see-saw wants to sink on that side.

    I can't believe you've never tried to balance anything, a ruler, a stick, a bat, a baton.
     
  13. Sep 25, 2015 #12
    so , at the distance 2L/3 from the position where w= 0 , the force only become wL/2 ?
    Is there any mathematical proof for this ? it's hard for me to fully understand it
     
  14. Sep 25, 2015 #13

    SteamKing

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    No, the total load over the entire length of the beam = wL/2.

    The total load is the area under the distributed load diagram, which in this case is a triangle. What's the area of a right triangle where the base is L units long and the height is w units?

    When the loading has a uniform distribution, the distributed load diagram is a rectangle of length L and height of w. What's the area of this rectangle?

    Those are your mathematical proofs.
     
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