How do I solve for the equilibrium of a cable system?

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To solve for the equilibrium of a cable system, it's essential to establish that the net force on the point in question is zero, leading to two equations based on the x and y components. The forces exerted by the post and the cable can be expressed in terms of their angles, allowing for the calculation of their magnitudes. Participants in the discussion noted discrepancies in their calculations, with one confirming that the correct force exerted by the post is approximately 2600 N. Clarification was provided on how to eliminate one variable by substituting it into the other equation to find the solution. The conversation emphasized the importance of careful algebraic manipulation to arrive at the correct answer.
brentwoodbc
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This question BUGS me

I've tried it a whole lot of times, my teacher explained it to me at least 2 times but he came up with the wrong answer.

I had a couple class mates explain it but it was an Extremely long/condusing way of doing it.

I know the x components are equal so there is no x force

so you are finding the Y component only.

I do all the trig to find the y components and its the wrong answer.





Anyone have a good way to solve this?

thanks
 
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Easy. You are assuming a static situation, i.e. point x is in equilibrium. This means that the net force on it is zero. That condition gives you two equations (one for the x components, and one for the y components). You also have two unknowns: the force exerted by the post on point x (Fpost), and the force exerted by the angled cable on point x (Fcable).

Two equations, two unknowns, all you have to do is solve:

\sum F_x = 2200\textrm{N} - F_{\textrm{post}}\cos(75^{\circ}) - F_{\textrm{cable}}\cos(58^{\circ}) = 0\sum F_y = F_{\textrm{post}}\sin(75^{\circ}) - F_{\textrm{cable}}\sin(58^{\circ}) = 0

Edit: I got 2551 N, which is closest to 2600 N. But I did it hastily and might have made algebra mistakes. What's the answer?
 
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Yes its 2600

I am still kind of confused.

so you find the ratio between Fpost and F cable In one of the equations and plug it into the other?

then what does the force on x =? post -/+ cable?
 


brentwoodbc said:
Yes its 2600

I am still kind of confused.

so you find the ratio between Fpost and F cable In one of the equations and plug it into the other?

then what does the force on x =? post -/+ cable?

Yes.

Doesn't resolving the second equation with the first substitution yield the right result?
 


I have Fc=2895 and Fp=2541 F_p is the answer correct?
 


brentwoodbc said:
Yes its 2600I am still kind of confused. so you find the ratio between Fpost and F cable In one of the equations and plug it into the other?

Yes. One way of solving a system of two equations in two unknowns is to use the first equation to eliminate one variable by expressing it in terms of the other, and then substitute that relation into the second equation to solve for the second variable.

brentwoodbc said:
then what does the force on x =? post -/+ cable?

What do you mean?

The net force on point x is zero, because it is not moving! We used this fact to solve the problem.

The force on point x DUE TO the post (which is what you are trying to solve for) is just Fpost, by definition.
 


brentwoodbc said:
I have Fc=2895 and Fp=2541 F_p is the answer correct?

Well I already told you that I got Fp = 2551 N, so I guess our answers differ by 10 Newtons.
 


Thank you.
 
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