anonymous
  • anonymous
Tutorial on finding the Centroid/Center of mass of a complex object. UNISA Students come here.
Mathematics
jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
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anonymous
  • anonymous
In this tutorial we will find the Centroid about this complex extruded shape. All cutout pieces shall be taken as negative.|dw:1363796078055:dw|
anonymous
  • anonymous
Split the complex shape into its component shapes.|dw:1363796223427:dw|

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anonymous
  • anonymous
interesting
anonymous
  • anonymous
Find the area of each of these simple shapes.\[A1 = \frac{ 1 }{ 2 }B \times \tau H\]\[A2 = L \times B\]\[A3 = L x B\]\[A4 = L x B\]
anonymous
  • anonymous
yea
anonymous
  • anonymous
\[A1 = 10m ^{2}\]\[A2 = 20m ^{2}\]\[A3 = 7.5m ^{2}\]\[A4 = 2m ^{2}\] The total are of these 4 will be \[AT = 39.5m ^{2}\]
anonymous
  • anonymous
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|
anonymous
  • anonymous
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. \[y _{Cutout} = \frac{ 3m }{ 2 } + 2m\]\[y_{Cutout} = 3.5m\] \[x_{Cutout} = \frac{ 2.5m }{ 2 } + 0.5m\]\[x_{Cutout} = 1.75m\]
anonymous
  • anonymous
yust to make sure|dw:1363797121334:dw|
anonymous
  • anonymous
so we include those formula for the cutout!!!!!
anonymous
  • anonymous
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\[y_{sRec} = \frac{ 2m }{ 2 }\]\[y_{sRec} = 1m\] \[x_{sRec} = \frac{ 1m }{ 2 }\]\[x_{sRec} = 0.5m\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.\[Ay_{Complex}=\frac{ (10)(4.5)+(20)(4.5)−(7.5)(3.5)+(2)(1) }{ 39.5 }\]\[Ay_{Complex}= 2.80m\] \[Ax_{Complex}=\frac{ (10)(6)+(20)(2)−(7.5)(1.75)+(2)(0.5) }{ 39.5 }\]\[Ax_{Complex}= 2.22m\]
anonymous
  • anonymous
To conclude it will be so.|dw:1363798358172:dw|
anonymous
  • anonymous
Questions are free now.
anonymous
  • anonymous
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.
anonymous
  • anonymous
where did you get this formula yCutout=3m2+2m
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
thanx sheldont i get it

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