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Centroid homework

  • Thread starter Jbreezy
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  • #1
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Homework Statement



Wondering if I did this right. Find the centroid of the region shown , not by integration, but by locating the centroids of the rectangles and triangles and using additivity of moments.

Homework Equations



I will give you the coordinates since I can't draw it.
For the rectangle (-1,0), (0,0) ,(-1,2), (0,2) and for the triangle I have (0,0), (2,0) and (0,2)

The Attempt at a Solution



So I did x Bar = M(triangle) + M(square) / (Area triangle + area square)
So to get M(triangle) I did xbar(area triangle) = (1)(4/2) = 2

I did this for the square too. I got (-1) When I plugged it into my above formula I got x bar = (1/3)

So, I did the same thing but with y. So I got y bar = 4/3

Does anyone agree? I pretty much did it like an example in class.

Thanks!!!
 

Answers and Replies

  • #2
SteamKing
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Your triangle has a base of 2 and a height of 2. According to you, the centroid of this triangle is 1 unit from the origin. Does this look right? What is the centroid of a triangle? There is a formula which you should be using.

Trying to follow your work above is confusing. You should organize your calculations better.

Remember, for a plane figure, the centroid requires 2 numbers to locate it.
 
  • #3
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I don't know what it is for a triangle. I think it like 1/3 something. I don't know.
 
  • #4
SteamKing
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Well, you could look it up, if it's not too much trouble. But what you guessed definitely ain't it.
 
  • #5
HallsofIvy
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The centroid of a "simplex" in n dimensions (triangle in 2 dimensions, tetrahedron in 3 dimensions, a polyhedron with n+1 vertices in n dimensions) has coordinates the average of the coordinates of the vertices. If a triangle in the plane has vertices [itex](x_1, y_1)[/itex], [itex](x_2, y_2)[/itex], and [itex](x_3, y_3)[/itex], then its centroid is at
[tex]\left(\dfrac{x_1+ x_2+ x_3}{3}, \dfrac{y_1+ y_2+ y_3}{3}\right)[/tex]

The centroid of more complex figures can be found by "triangulation"- dividing the figure into triangles, finding the centroid of each triangle, then taking the "weighted" average, weighted by the areas of the triangles. That is what your text means by "additivity of moments".
 
  • #6
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Using your formula the centroid the triangle was (2/3,2/3)
and the centroid I got for the square was (-1/2), 1
This seems reasonable to me. Then I summed the x components and the y components and divided each by the total area of both objects which is 4. So for my final answer I ened up with (5/12, 1/24) Seems pretts low.
 
  • #7
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So I know my two centroid of the individual objects are OK. But I'm confused on the next part.
 
  • #8
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Solved it. Thx
 

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