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Find the position of the centroid of the shaded area

  1. May 28, 2015 #1
    1. Find the position of the centroid of the shaded area

    http://imgur.com/ieiSPPY

    2.

    The black triangle with the square cut out is where the centroid must be located.I know that the object is symmetrical and the triangle can be divided into parts with two smaller right angled triangles
    I have having immense difficultly studying centroids because my book does not explain how to equate them properly.

    3. I understand that the centroid of a triangle is where the medians intersect
    I believe the formula is the sum of the sides divided by the amount of sides ,I just do not know how to calculate the centroid of the shape once you remove the area of the square.I am uncertain about where I should place my reference point . I would appreciate any help/links/resources with understanding centers of gravity and centroids ,I am a new student of physics and I am very keen to learn. thank you.
     
  2. jcsd
  3. May 28, 2015 #2
    Does your book give you any formulas (equations) for determining where the centroid of an object is located? If so, please write them down for use so that we can better assess how we can help you.

    Chet
     
  4. May 28, 2015 #3
    my book only states that the centroid of a triangle is the point where the 3 medians intersect , it gives me no equations . I appreciate any help people are willing to give as there is no explanation in my book it just expects me to solve it and I am incredibly stressed
     
  5. May 28, 2015 #4
    What is the title of this book, and who is the author?

    Don't get too stressed out. We can help you. How much math have you had?

    Chet
     
  6. May 28, 2015 #5
    thank you , it is a study guide for my university so it doesn't state the author, I am in my first year of university and I have studied about one third of my coursework so I will understand any explanation of this problem, I just have no idea as to how to calculate the centroid when the triangle is of that shape. I can calculate centroids of rectangular objects and I understand moments and the principle of moments
     
  7. May 28, 2015 #6
    If the square were not cut out of the triangle, where would the centroid of the triangle be located relative to the right hand boundary of the triangle? If the triangle were taken away and the actual square were put back in the location from which it had been removed, where would the centroid of the square be located relative to the right hand boundary?

    Chet
     
  8. May 28, 2015 #7
    I am unsure how to measure the centroid when the square is still there because I know the center of gravity has changed , I have no idea how to solve this or what formulas to apply because my book didn't provide them . However, I do know how to calculate the centroid of a rectangular mass ,if anyone here has any resources they can share online I would appreciate it
     
  9. May 28, 2015 #8
    You haven't answered my questions in post #6. Until you can provide these answers, you will not be able to solve this problem.

    Chet
     
  10. May 28, 2015 #9
    do I measure from the right hand boundary line to the vertices of the triangle or to the centroid of the triangle and what formula do I apply, I appreciate your advice but I cant answer post 6 because I don't have the formula to do so , should I do a scale drawing?

    are there any resources you can suggest that I read because the subject of center of mass and centroids is summarized in 1 page in my book
     
  11. May 28, 2015 #10
    You have had plane geometry, correct? From what you learned in plane geometry, if s is the side of an equilateral triangle, how far from the base of the triangle do the three perpendicular bisectors meet? (The right boundary of your figure is the base of the triangle).
    The information in your book is adequate to solve this problem.

    Chet
     
  12. May 28, 2015 #11
    I took the perpendicular bisectors and found the centroid of the triangle and measured from the right(bottom right point A) to the centerand measured 74 mm, is that all I had to do? I have been so overworked lately that I think I may have overthought the question.I want to thank you for your patience
     
  13. May 28, 2015 #12
    Scaling it to the figure is not what I had in mind. The question expects you to calculate the answer. Lets start over.

    Look at the figure and tell me where you think the center of mass (i.e., the centroid) is located in the up-down (vertical y) direction:

    (a) at the top of the triangle
    (b) at the horizontal centerline of the triangle
    (c) at the bottom of the triangle

    Your next goal is to find out where the center of mass is located in the side-to-side (horizontal x) direction. You are aware that the triangle in your figure is an equilateral triangle, correct? Since you are interested in the location of the center of mass in the x direction, you are not going to be determining the distance of the center of mass from the right hand vertices of the triangle. You are going to focus on the distance from the right hand side of the triangle. Does this make sense?

    If s is the side of an equilateral triangle, what the the equation for the altitude of the triangle in terms of s? Focus on the altitude from the left vertex to the right hand side. Are you aware that the center of mass of an equilateral triangle is located at the 1/3-2/3 location along its altitude? In terms of s, how far is this from the right hand side of the triangle?

    Chet
     
  14. May 29, 2015 #13
    the object is symmetrical about A which cuts the object into two equal parts horizontally , the total area is equal to the area of the triangle minus the square 120x120/2 - 1600= 14400/2-1600=7200-1600=5600 , I then divide it by half because this is the area the shape =2800,

    the centroid is the distance from A to x , therefore the one side area y mm ----> (y)(1/2)x(120)x(y/120)=y^2/2

    now I have the total area and the moment , the centroid is measured from the vertex and by the principle of moments

    y^2/2=2800
    y^2=5600
    y=74.8mm

    is this correct

    I chose y/120 because I realised that the variable needs to be along the line of symmetry
     
  15. May 29, 2015 #14
    Are you sure about the geometry? Your figure indicates that the base of the triangle (the right hand side) is 120, and the altitude is also 120. Are you sure that it is not supposed to have all three sides of 120, and an altitude of 60√3?

    In this problem, what you need to do is calculate the moment of the triangle area about the right hand side (under the assumption that the square area is not cut out), and then subtract the moment of the square area about the right hand side. This will yield, by difference, the moment of the area of your actual shape about the right hand side. But the moment of the filled in isoceles triangle would have to be calculated using integration. That's why I was asking whether it's really supposed to be an equilateral triangle. Or, are you given any formulas for determining the center of mass of an isoceles triangle of arbitrary altitude?

    As your solution stands now, it is not correct.

    Chet
     
  16. May 29, 2015 #15
    in this book integration has not been covered yet so i believe i must solve it by using the whole area for one moment and the square as a separate area which has a moment in the opposite direction , the triangle is equilateral if i have to go by measurements alone.I will do as you have said
     
  17. May 29, 2015 #16
    Well, assuming it is an isoceles triangle, I did the integration for an isoceles triangle. The analysis shows that centroid of an isoceles triangle is located at a distance of 1/3 of the altitude up from the base. So, in your figure, since the altitude is 120, the centroid of the filled-in triangle is located at a distance of 40 units to the left of the right-hand boundary. Since the area of the triangle is 7200 square units, the moment of the filled-in triangle about the right hand boundary is (7200)(40) = 288000 cubic units. Now, relative to the right boundary, where do you think the centroid of the square is located? What is the moment of the square? What is the net moment?

    Chet
     
  18. May 29, 2015 #17
    i solved it , { 40x(1/2x120^2)-(20x40^2) } / {1/2 x 120^2)-(40^2)} =45.71 I took the moments of the square and the triangle from right to left and then took the difference of them and divided it by the area of the shape and my answer is 120-45.71=74.29 = 74.3 , i'm so thankful for all of your help
     
  19. May 29, 2015 #18
    Looks good.
     
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