- #1
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Homework Help Overview
The discussion revolves around the calculation of a force labeled R1 within the context of moments of forces, specifically related to isosceles triangles in a structural setup.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants are questioning the correctness of R1 calculations and the definitions of R1 as a force versus a moment. There are inquiries about the horizontal distances and the geometry of the triangles involved. Some participants are attempting to clarify the setup and the angles involved in the calculations.
Discussion Status
The discussion is active with participants providing feedback on calculations and geometry. Some guidance has been offered regarding the assumptions made about the structure and the need to clarify lengths and angles in the triangles. There is no explicit consensus yet, but participants are engaging with the problem constructively.
Contextual Notes
There are indications of difficulties in communication due to handwriting and the use of the editor. Participants are also working with specific measurements and angles that may not be fully detailed in the discussion.
- #2
adjacent
Gold Member
- 1,552
- 62
I don't understand what's R1,Is it the force or the moment?Can you please write the calculations in the editor?It's difficult to read itanthonyk2013 said:
- #3
- #4
- 42,887
- 10,521
I don't think you have the horizontal distances right. Triangles E and G are isosceles, yes? How are you calculating the horizontal displacement to the vertical load?
- #5
anthonyk2013
- 125
- 0
yes E and G are isosceles and if you split D in half you get two more isosceles, draw a vertical line R1 to the pin on the left the distance is 2 square root of 3.
10KN acts down from B on that vertical line and on graph paper its 1/2 root of 3.
10 KN acts down from CH to vertical line and the distance is 1 1/2 root 3.
should I be using the force acting at 45deg at B along the line EF?
10KN acts down from B on that vertical line and on graph paper its 1/2 root of 3.
10 KN acts down from CH to vertical line and the distance is 1 1/2 root 3.
should I be using the force acting at 45deg at B along the line EF?
- #6
- 42,887
- 10,521
you seem to be assuming that the vertical line through the top left pin is halfway to the midpoint of the whole structure. It isn't. Fill in as many lengths and angles as you can in triangles D and E.anthonyk2013 said:
- #7
anthonyk2013
- 125
- 0
Thanks, I'll give that a go.
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