Problems with finding moment of inertia of triangle

Click For Summary
SUMMARY

The discussion focuses on calculating the moment of inertia of a thin triangular sheet. The key formulas derived are x1 = ay/h and x2 = by/h, where x1 and x2 represent the bases of smaller triangles similar to larger triangles with bases a and b, respectively. The relationship is established through the properties of similar triangles, confirming that the ratios of corresponding sides are equal. This method effectively aids in determining the moment of inertia for the given triangular configuration.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Knowledge of similar triangles and their properties
  • Familiarity with basic geometry and ratios
  • Ability to apply algebraic manipulation in geometric contexts
NEXT STEPS
  • Study the derivation of moment of inertia for various geometric shapes
  • Explore applications of similar triangles in physics and engineering
  • Learn about integration techniques for calculating moments of inertia
  • Investigate the role of moment of inertia in structural engineering design
USEFUL FOR

Students and professionals in physics, engineering, and mathematics who are involved in mechanics and structural analysis will benefit from this discussion.

knight92
Messages
100
Reaction score
0
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 don't get, it seems to be some sort of ratio but I can't work out how or why its like that. Can anyone help please ?
 

Attachments

  • Src1.jpeg
    Src1.jpeg
    5.1 KB · Views: 574
Engineering news on Phys.org
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.
 
xandro101 said:
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 :)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K