Centroid of an Area - Finding X & Y Coordinates

In summary: I guess this is easy for me to see.ae(a/2) + be(b/2+a) + cd/2(c/3+a+b)-πr2//4(a+r/3) = x(ea+be+cd/2+πr2/4)I just check and recheck my expression...That may work for simple expressions, but in this case, the OP has made multiple errors in their calculations, making it difficult to pinpoint where the mistake is coming from. It would be more helpful for them to double-check their work and possibly seek assistance with the calculations, rather than relying solely on symbolic expressions.
  • #1
werson tan
183
1

Homework Statement


total area = (75x175) +(100x175) +(0.5x50x175) -(π/4 x100x100) = 27964

my centroid for y = (75x175)(87.5) + (100x175)(87.5) +(0.5x150x75)(25)
-(π/4 x100x100) (175 - (4(100)/3π) ) / 27964 = 97.13

my centroid for x = (75x175)(37.5) +(100x175) (125) + (0.5x50x175)(225)-(π/4 x100x100)(175 - (4(100)/3π) ) / 27964 = 37.26 , but the ans given is x = 102 , y= 62.5 , which part of my working is wrong ?
p/s : i have divided the areas into 4 parts , the first is the (75x175) reactangle , second is (100x175) trianlgle , fourth is -(π/4 x100x100) , the third is the triangle (0.5x50x175)

Homework Equations

The Attempt at a Solution

 

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  • #2
werson tan said:

Homework Statement


total area = (75x175) +(100x175) +(0.5x50x175) -(π/4 x100x100) = 27964

You have a mistake here. The triangle has dimensions of base = 150 x height = 75
my centroid for y = (75x175)(87.5) + (100x175)(87.5) +(0.5x150x75)(25)
-(π/4 x100x100) (175 - (4(100)/3π) ) / 27964 = 97.13
You have the correct dimensions of the triangle here and the correct moment arm, but you are dividing by the incorrect area.
my centroid for x = (75x175)(37.5) +(100x175) (125) + (0.5x50x175)(225)-(π/4 x100x100)(175 - (4(100)/3π) ) / 27964 = 37.26 , but the ans given is x = 102 , y= 62.5 , which part of my working is wrong ?
You have used the wrong dimensions for the triangle here, but you have the correct x-bar distance.
You are also dividing by the incorrect area again.
p/s : i have divided the areas into 4 parts , the first is the (75x175) reactangle , second is (100x175) trianlgle , fourth is -(π/4 x100x100) , the third is the triangle (0.5x50x175)

You could have combined rectangles I and II into one large rectangle and saved some calculation.

You should always double-check your work to eliminate silly mistakes, like what happened with the dimensions of the triangle.

Note: My calculations show x-bar = 118.28 mm; y-bar = 56.52 mm
 
Last edited:
  • #3
SteamKing said:
You have a mistake here. The triangle has dimensions of base = 150 x height = 75

You have the correct dimensions of the triangle here and the correct moment arm, but you are dividing by the incorrect area.

You have used the wrong dimensions for the triangle here, but you have the correct x-bar distance.
You are also dividing by the incorrect area again.You could have combined rectangles I and II into one large rectangle and saved some calculation.

You should always double-check your work to eliminate silly mistakes, like what happened with the dimensions of the triangle.

Note: My calculations show x-bar = 118.28 mm; y-bar = 56.52 mm
My ans is x = 108.3 , y = 61.3 after changing 50 to 150, is my ans wrong?
 
  • #4
werson tan said:
My ans is x = 108.3 , y = 61.3 after changing 50 to 150, is my ans wrong?

It's not clear. Why don't you post your revised calculations?
 
  • #5
SteamKing said:
It's not clear. Why don't you post your revised calculations?
total area = (75x175) +(100x175) +(0.5x50x75) -(π/4 x100x100) = 35464

my centroid for y = (75x175)(87.5) + (100x175)(87.5) +(0.5x150x75)(25)
-(π/4 x100x100) (175 - (4(100)/3π) ) / 35464 = 61.36

my centroid for x = (75x175)(37.5) +(100x175) (125) + (0.5x50x175)(225)-(π/4 x100x100)(175 - (4(100)/3π) ) / 35464 = 108.3
 
  • #6
werson tan said:
total area = (75x175) +(100x175) +(0.5x50x75) -(π/4 x100x100) = 35464
You have the wrong dimensions for the end triangle here. It is L = 150 mm, H = 75 mm
my centroid for y = (75x175)(87.5) + (100x175)(87.5) +(0.5x150x75)(25)
-(π/4 x100x100) (175 - (4(100)/3π) ) / 35464 = 61.36
You have used the correct dimensions of the end triangle here to calculate the moment, and you are dividing by the incorrect area carried over from the step above.
my centroid for x = (75x175)(37.5) +(100x175) (125) + (0.5x50x175)(225)-(π/4 x100x100)(175 - (4(100)/3π) ) / 35464 = 108.3
For some reason, you are showing the wrong dimensions of the end triangle here in calculating the moment. The triangle should be L = 150 mm, H = 75 mm
Also, you are dividing by the incorrect area calculated in step one.

You're still not checking your work to eliminate silly mistakes, like using the incorrect dimensions of the end triangle.

Please re-do your calculations.

P.S.: I found a mistake in my own calculations. My calculations for the centroid location now agree with the values shown for x-bar and y-bar in the image attached to the OP.
 
  • #7
(75x175) +(100x175) +(0.5x150x75) -(π/4 x100x100) = 35464

my centroid for y = (75x175)(87.5) + (100x175)(87.5) +(0.5x150x75)(25)
-(π/4 x100x100) (175 - (4(100)/3π) ) / 35464 = 61.36

my centroid for x = (75x175)(37.5) +(100x175) (125) + (0.5x150x75)(225)-(π/4 x100x100)(175 - (4(100)/3π) ) / 35464 = 108.3
i made the typo..
 
  • #8
werson tan said:
(75x175) +(100x175) +(0.5x150x75) -(π/4 x100x100) = 35464
This calculation is still incorrect. Did you even bother to check the arithmetic after changing the dimensions of the triangle?
my centroid for y = (75x175)(87.5) + (100x175)(87.5) +(0.5x150x75)(25)
-(π/4 x100x100) (175 - (4(100)/3π) ) / 35464 = 61.36

my centroid for x = (75x175)(37.5) +(100x175) (125) + (0.5x150x75)(225)-(π/4 x100x100)(175 - (4(100)/3π) ) / 35464 = 108.3

The centroidal locations will not be correct until you calculate the correct area for the figure. Get the first step right.
 
  • #9
Sometimes the easiest way to solve a problem is to translate it into symbolic expressions and equations and then apply algebraic solution methods.

Less typo and use the help of computer for final algebraic solution. This also helps you to make sure your expression is correct without worrying about your calculation error.
 
  • #10
azizlwl said:
Sometimes the easiest way to solve a problem is to translate it into symbolic expressions and equations and then apply algebraic solution methods.

Less typo and use the help of computer for final algebraic solution. This also helps you to make sure your expression is correct without worrying about your calculation error.
It's not clear how this would help the OP. The problem cannot be solved by algebraic manipulation; it must be solved by the old-fashioned direct grunt work of arithmetic.

Besides, if you can't get the arithmetic to work, how are you going to get the algebra right? :wink:
 
  • #11
I guess this is easy for me to see.
ae(a/2) + be(b/2+a) + cd/2(c/3+a+b)-πr2//4(a+r/3) = x(ea+be+cd/2+πr2/4)
I just check and recheck my expression without those dizzying numbers.

A bit of cheating, use Math app to calculate after assigning the value. I'm not a teacher but I guess he likes to see the working not the final answer.
 
  • #12
azizlwl said:
I guess this is easy for me to see.
ae(a/2) + be(b/2+a) + cd/2(c/3+a+b)-πr2//4(a+r/3) = x(ea+be+cd/2+πr2/4)
I just check and recheck my expression without those dizzying numbers.

A bit of cheating, use Math app to calculate after assigning the value. I'm not a teacher but I guess he likes to see the working not the final answer.
Whatever floats your boat.
 

Related to Centroid of an Area - Finding X & Y Coordinates

Question 1: What is the centroid of an area?

The centroid of an area is the geometric center of a 2-dimensional shape. It is the point where all the individual points of the shape would balance if it were cut out of a sheet of material.

Question 2: How is the centroid of an area calculated?

The centroid of an area can be calculated using the formula (Xc, Yc) = (∫∫(xdA), ∫∫(ydA)) / (∫∫dA), where x and y are the coordinates of each individual point in the area and dA is the differential area element. This can also be represented as (Xc, Yc) = (X̅, Y̅), where X̅ and Y̅ are the average of the x and y coordinates, respectively.

Question 3: Why is finding the centroid of an area important?

Finding the centroid of an area is important in various fields, such as engineering, architecture, and physics. It helps determine the stability and balance of structures, as well as the distribution of forces and moments acting on an object. It is also used in designing and analyzing various shapes and structures.

Question 4: What are the applications of finding the centroid of an area?

The applications of finding the centroid of an area include calculating the center of mass of an object, determining the neutral axis of a cross-section, analyzing the stability of structures, and designing various objects and structures.

Question 5: Are there any special cases in finding the centroid of an area?

Yes, there are two special cases in finding the centroid of an area. The first one is when the shape has a line of symmetry, in which case the centroid will lie on that line. The second case is when the shape has two or more lines of symmetry, in which case the centroid will be at their intersection point.

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