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

AI Thread Summary
The discussion focuses on calculating the tension required for the crosspiece in an A-frame structure, where beams AB and AC each measure 5.00 m and have a mass of 375.0 kg. The tension (T) must be determined using the equation T = F*R*sinΘ, with considerations for forces acting on the frame and using point A as the pivot. An initial calculation led to a tension value of 4941.08 N, but a mistake was identified regarding the radius used in the calculations. The correct radius is 0.8 m, not 1.4 m, and the final tension value discussed is 8650 N. Accurate measurements and understanding of the geometry are crucial for solving the problem effectively.
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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