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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In this tutorial we will find the Centroid about this complex extruded shape. All cutout pieces shall be taken as negative.dw:1363796078055:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Split the complex shape into its component shapes.dw:1363796223427:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Find 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[A1 = 10m ^{2}\]\[A2 = 20m ^{2}\]\[A3 = 7.5m ^{2}\]\[A4 = 2m ^{2}\] The total are of these 4 will be \[AT = 39.5m ^{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now 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
 3 years ago
Best ResponseYou've already chosen the best response.0To 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
 3 years ago
Best ResponseYou've already chosen the best response.0yust to make suredw:1363797121334:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so we include those formula for the cutout!!!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0To conclude it will be so.dw:1363798358172:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Questions are free now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Don'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
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get this formula yCutout=3m2+2m

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0and 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
 3 years ago
Best ResponseYou've already chosen the best response.0thanx sheldont i get it
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