# Statics Centroid Problem

1. Dec 11, 2008

### MCS5280

1. The problem statement, all variables and given/known data

Determine The Centroid of The Area

2. Relevant equations

x_centroid = [(Area of each piece)(Centroid of each piece)]/(Area of each piece)
3. The attempt at a solution

This seems like a really basic problem, I am just reviewing for my statics final on saturday. The only problem is that I keep getting a different answer than the book for the X centroid. I broke the shape up into different pieces (a square, rectangle and semi-circle) and am treating the cutout circle as negative area. I tried it using a different shapes and got the same answer. What am I doing wrong?

My work:

X_centroid = $$\frac{(40 mm)(6400 mm^2)+(120 mm)(9600 mm^2)+(120 mm)(1/2)(1600\pi mm^2)-(120 mm)(400\pi mm^2)}{(6400 mm^2)+(9600 mm^2)+(1/2)(1600\pi mm^2)-(400\pi mm^2)}$$

#### Attached Files:

• ###### Picture 1.png
File size:
14.3 KB
Views:
1,075
Last edited: Dec 11, 2008
2. Dec 12, 2008

### PhanthomJay

I am not following your numbers. The centroid must have an x and a y coordinate. To determine the y dimension, choose a convenient reference axis, like the bottom of the shape. The y centroid, as measured up from the base, is then determimed by adding up (or subtracting) the products of each individual area shape times the distance from its individual centroid to the base, then dividing that sum by the total area of the combined shape. To find the x coordinate, you proceed in a similar manner, except choose the right edge as the refrence axis about which you sum moment areas.

3. Dec 12, 2008

### MCS5280

My calculations are for the X coordinate of the centroid with reference being the left edge of the shape. I divided it up into an 80x80 square for the left part, an 80x120 rectangle for the right with a semi-circle of with radius 40mm ontop of it. The cutout is a circle with radius 20mm. Its just the sum of the [(X centroid of each piece)*(Area of each piece)]/(Total Area of shape). I've had a couple of friends do this one and they got the same exact answer as well. So either we are all making the same mistake or the book is wrong.

4. Dec 12, 2008

### PhanthomJay

OK, I get your answer for the x coordinate. I haven't done the math for the y coordinate. You need both values to define the centroid.

5. Dec 12, 2008

### MCS5280

I got the Y centroid coordinate to work out to the books answer, so that one is all good. Its just that the X centroid coordinate that I have calculated is 90.33 mm, while the book says it should be 87.3 mm. So is the book wrong? (I don't think rounding error could account for a 6 mm difference in the answers)

6. Dec 12, 2008

### PhanthomJay

if me, you, and your 2 friends get the same answer, I think we've got the book answer outnumbered by 4:1.

7. Dec 12, 2008

### mathmate

Make it 5 to 1!

8. Dec 12, 2008

### MCS5280

Ok thanks for the help, just wanted to make sure I wasn't making a stupid mistake. Its so nice that my \$120 textbook can't answer its own problems correctly.