Solve Statics Question: Vertical Reactions at A,D,F

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The discussion focuses on calculating vertical reactions at points A, D, and F in a system of three wooden beams under vertical forces. The equations derived are: (1) F = 2/3(P + E) = 2B, (2) D = 2/3C = 2E, and (3) A = 2/3B = 2C. A key point of confusion is why force P does not factor into the equation for D, which is clarified by isolating beam DCE in a Free Body Diagram, demonstrating that P does not affect the calculations for D.

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Homework Statement



d7ebb587e9b4ddd7d3e68141e14356e4.jpg


This is the diagram of the question. There are 3 wooden beams nailed together and supported at A,D, and F. Assuming only vertical forces, what are the vertical reactions at A, D, and F?
The solution uses this diagram to make three basic equations:
smallunit.jpg

(1) for BEF, F=2/3(P+E)=2B
(2) for ECD, D=2/3C=2E
(3) for ABC, A=2/3B=2C

For (2), why is it D=2/3C=2E and not D=2/3C=2(P+E)?
Why doesn't P come into the equation? Isn't it acting at point E? I understand the solution completely except for this one point. Thank you.

Using algebra I can get the solution of D=1/13P, F=9/13P, A=3/13P
But only if D=2E rather than D=2(P+E)
 
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Someone suggests that force P may be only applied to beam BEF, not on beam DCE. Would that make it work?
 
JoshuaR said:
Someone suggests that force P may be only applied to beam BEF, not on beam DCE. Would that make it work?
That is not quite correct, force P is applied as shown, to the system of beams, at the intersection of the 2 beams BEF and DCE. Draw a Free Body Diagram of beam DCE. Isolate it such that you cut the beam with your Free Body Diagram immediately to the right of joint E, such that P, the applied force, does not enter into the diagram, and therefore the given solution appears correct.
 

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