Problems with finding moment of inertia of triangle

  • Thread starter knight92
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  • #1
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Hi guys I am trying to find the moment of inertia of a thin triangle sheet as shown in the attached file, I couldn't so I looked at the solution and it said that x1 = ay/h and x2 = by/h. That is the only part I dont get, it seems to be some sort of ratio but I cant work out how or why its like that. Can anyone help please ?
 

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  • #2
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You have to look carefully. You will see that there are that each big triangle on the left and right side have bases a and b respectively. Now there two other smaller triangles on the left and right side with bases x1 and x2. The triangle with base x1 has a height y and is similar to the big triangle with base a and a height h. If u use the ratio of similar triangles you get x1/y=a/h which gives x1= ay/h. This method also applies to the right handside to give x2= by/h. I hope you get it now.
 
  • #3
101
0
You have to look carefully. You will see that there are that each big triangle on the left and right side have bases a and b respectively. Now there two other smaller triangles on the left and right side with bases x1 and x2. The triangle with base x1 has a height y and is similar to the big triangle with base a and a height h. If u use the ratio of similar triangles you get x1/y=a/h which gives x1= ay/h. This method also applies to the right handside to give x2= by/h. I hope you get it now.
Yes I get it now. Thanks mate :)
 

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