Static equilibrium problem: please help

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  • #1
ShawnD
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Here is a problem I have to solve
http://myfiles.dyndns.org/pictures/statics9-2.png

The load W is 1250lbs. The force exerted by the hydraulic thing GJ is 0 which makes HF a 2 force member.
What I am supposed to do is solve for the tension in spring DE



I just can't seem to get an answer for this problem. In all, there are 6 variables: Ax, Ay, Bx, By, ED, and FH.

So far I have tried 2 ways:

1. Create 6 equations to solve for the 6 variables. Every equation shows the sum of moments around a certain point for a certain member. My 6 equations were for the moments acting on A on ABC, where W is on ABC, B on ABC, B on BEF, E on BEF, F on BEF. When the 6 equations were put into a matrix on my calculator, the calculator could not get an answer. It gives answers like 1Ax - 18By = 1245451 and things like that; it doens't simplify it down to express only 1 variable.

2. Create 4 equations for member ABC to solve for the forces on ABC then work on BEF after. The equations were for moments acting on A, W, B and C. The equations were put into a matrix on my calculator. The calculator again could not get an answer for any variable.


Does anybody have an idea on how to approach this question?
 
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Answers and Replies

  • #2
NateTG
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The indication is that you don't have enough independant equations.

I'm not sure if the excercise can be completed without knowing the the angle that HF makes with BF or some equivalent information (i.e. the length of HF, or that BF is vertical)
 
  • #3
ShawnD
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We don't need the angle for HF because we know the ratios. It's vertical is 12", horizontal is (54 - 15) which is 39 and the hypotinuse is sqrt(1665). Vertical and horizontal forces from HF can be expressed as the ratio. The horizontal for example is (39/sqrt(1665))HF.

It is assumed that BF is vertical.
 
  • #4
NateTG
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Ok, then the problem is solvable.

You can determine the upward force on BF as a function of the spring tension, and the downward force on BF as a result of the weight.
 
  • #5
ShawnD
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I just just realized some basic stuff. On member ABC, the only horizontal forces are Ax and Bx. That means Ax = Bx. On member BEF, the only vertical parts are By and (12/sqrt(1665))HF which means By = (12/sqrt(1665)). That way, It reduces the number of variables to only 5. There are 5 joints to make equations for in ABC and BEF.
Right now I'm going to try summing moments around A and B on ABC then B, E, and F on BEF. I'll tell ya how it goes when I'm done.
 
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  • #6
ShawnD
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hmm i did something wrong....
 
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