What is the tension required for the crosspiece in an A-frame structure?

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Homework Help Overview

The discussion revolves around determining the tension required for the crosspiece in an A-frame structure, specifically focusing on the forces acting on the beams and the calculations involved in finding the tension.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the structure and the equations used to calculate tension. Questions arise regarding the values used in the calculations, particularly the lengths and radii involved.

Discussion Status

The discussion is ongoing, with participants clarifying values and addressing mistakes in previous calculations. There is a focus on understanding the setup and ensuring the correct parameters are used in the equations.

Contextual Notes

Participants note discrepancies in values from different versions of the problem, which may affect the calculations. There is an emphasis on ensuring that assumptions about lengths and forces are accurate.

iPromise
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Homework Statement



You are designing the crosspiece for the A-frame structure in the figure below. Beams AB and AC are 5.00 m long and have a mass of 375.0 kg each. How much tension must the crosspiece EF withstand? Assume that the mass of the crosspiece and the friction at points B and C are negligible.

http://oi57.tinypic.com/2zfq4y8.jpg

Homework Equations



T = F*R*sinΘ
∑T = 0

The Attempt at a Solution



I took the first half of the frame into consideration. I chose point A to be my pivot point. There are three forces acting on the first-half of the frame:

http://oi59.tinypic.com/fu7sw0.jpg

My initial equation:

(N)(4.75)(cosΘ) - (Mg)(2.375)cosΘ = (T)(1.4)

Θ = sin-1 ( 2.2 / 4.75 )
= 27.59

N = the force of gravity of the first half of the structure
= Mg = 335 * 9.81

I solve for T:

T = 4941.08

The answer: 8650
 
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iPromise said:
(N)(4.75)(cosΘ) - (Mg)(2.375)cosΘ = (T)(1.4)
Where are you getting 4.75 and 1.4?
 
4.75 is the length of the ladder. 1.4 was the radius from the pivot point. That was my mistake, 1.4 is NOT the radius, 0.8 is.
 
iPromise said:
4.75 is the length of the ladder.
Isn't it given as 5 m?
 
Sorry I was thinking of the last practise version I did. Each version has different values.
 

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