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Razzputin

  • one year ago

Tutorial on finding the Centroid/Center of mass of a complex object. UNISA Students come here.

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  1. UnkleRhaukus
    • one year ago
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    \[\ast\]

  2. Razzputin
    • one year ago
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    In this tutorial we will find the Centroid about this complex extruded shape. All cutout pieces shall be taken as negative.|dw:1363796078055:dw|

  3. Razzputin
    • one year ago
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    Split the complex shape into its component shapes.|dw:1363796223427:dw|

  4. jores
    • one year ago
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    interesting

  5. Razzputin
    • one year ago
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    Find the area of each of these simple shapes.\[A1 = \frac{ 1 }{ 2 }B \times \tau H\]\[A2 = L \times B\]\[A3 = L x B\]\[A4 = L x B\]

  6. jores
    • one year ago
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    yea

  7. Razzputin
    • one year ago
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    \[A1 = 10m ^{2}\]\[A2 = 20m ^{2}\]\[A3 = 7.5m ^{2}\]\[A4 = 2m ^{2}\] The total are of these 4 will be \[AT = 39.5m ^{2}\]

  8. Razzputin
    • one year ago
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    Now if we take the origional drawing we see that the shapes are in very definate places, and they are at definate points above and away from the x and y axis.|dw:1363796713376:dw|

  9. Razzputin
    • one year ago
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    To determine the centroid of each shame we must include these distances from the axis as well. So for example the y and x centroid of the cut out square in the center will be. \[y _{Cutout} = \frac{ 3m }{ 2 } + 2m\]\[y_{Cutout} = 3.5m\] \[x_{Cutout} = \frac{ 2.5m }{ 2 } + 0.5m\]\[x_{Cutout} = 1.75m\]

  10. Razzputin
    • one year ago
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    yust to make sure|dw:1363797121334:dw|

  11. jores
    • one year ago
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    so we include those formula for the cutout!!!!!

  12. Razzputin
    • one year ago
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    The centroid of the triangular shape will thus be.\[y_{Triangle} = \frac{ 5m }{ 2 } + 2m\]\[y_{Triangle} = 4.5m\] \[x_{Triangle} = \frac{ 2m }{ 2 } + 5m\]\[x_{Triangle} = 6m\]The centroid of the large rectangle will be\[y_{LRec} = \frac{ 5m }{ 2 } + 2m\]\[y_{LRec} = 4.5m\] \[x_{LRec} = \frac{ 4m }{ 2 }\]\[x_{LRec} = 2m\] The centroid of the small rectangle is\[y_{sRec} = \frac{ 2m }{ 2 }\]\[y_{sRec} = 1m\] \[x_{sRec} = \frac{ 1m }{ 2 }\]\[x_{sRec} = 0.5m\]

  13. Razzputin
    • one year ago
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    yes jores we do, we wer not doing that at UNISA, the textbook did not include that piece of information. You have to find the centroid of the shape and add it to the distance from the x or y axis.

  14. Razzputin
    • one year ago
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    So now the final equation to find the centroid of the complex shape will be\[Ay_{Complex}=\frac{ A1y_{Triangle}+A2y_{LRec}−A3y_{Cutout}+A4y_{sRec}}{ AT } \]\[Ax_{Complex}=\frac{ A1x_{Triangle}+A2x_{LRec}−A3x_{Cutout}+A4x_{sRec}}{ AT } \]we minus A3 as it is the cutout shape and therefore is not actually there. The numerical values are as such.\[Ay_{Complex}=\frac{ (10)(4.5)+(20)(4.5)−(7.5)(3.5)+(2)(1) }{ 39.5 }\]\[Ay_{Complex}= 2.80m\] \[Ax_{Complex}=\frac{ (10)(6)+(20)(2)−(7.5)(1.75)+(2)(0.5) }{ 39.5 }\]\[Ax_{Complex}= 2.22m\]

  15. Razzputin
    • one year ago
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    To conclude it will be so.|dw:1363798358172:dw|

  16. Razzputin
    • one year ago
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    Questions are free now.

  17. Razzputin
    • one year ago
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    Don't forget to drop a medal in the box on your way out if you think this helped you? Also don't be afraid to fan me as well.

  18. jores
    • one year ago
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    where did you get this formula yCutout=3m2+2m

  19. Razzputin
    • one year ago
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    I found it while looking on the internet. it is the centroid of the cutout area being 3m over 2 to find the centroid + the distance from the edge of the cutout closest to the y axis to the axis its self. You follow? Should I draw it?

  20. Razzputin
    • one year ago
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    and sorry for my poor spelling i was typing in a rush. It must not be shame it must be shape. I also mistyped just as yust.

  21. jores
    • one year ago
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    oh ok

  22. jores
    • one year ago
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    thanx sheldont i get it

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