## Razzputin Group Title Tutorial on finding the Centroid/Center of mass of a complex object. UNISA Students come here. one year ago one year ago

1. UnkleRhaukus

$\ast$

2. Razzputin

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

Split the complex shape into its component shapes.|dw:1363796223427:dw|

4. jores

interesting

5. Razzputin

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

yea

7. Razzputin

$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

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

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

yust to make sure|dw:1363797121334:dw|

11. jores

so we include those formula for the cutout!!!!!

12. Razzputin

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

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

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

To conclude it will be so.|dw:1363798358172:dw|

16. Razzputin

Questions are free now.

17. Razzputin

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

where did you get this formula yCutout=3m2+2m

19. Razzputin

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

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

oh ok

22. jores

thanx sheldont i get it