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

Click For Summary

Homework Help Overview

The discussion revolves around solving for the equilibrium of a cable system, specifically focusing on the forces acting on a point in a static situation. Participants are exploring the relationships between the forces exerted by a post and a cable, as well as the conditions for equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equilibrium conditions, noting that the net force must be zero. They explore the equations derived from the x and y components of the forces and question the methods of solving for the unknown forces.

Discussion Status

There is ongoing exploration of the relationships between the forces, with some participants expressing confusion about the methods of substitution and the implications of their calculations. Multiple interpretations of the problem are being considered, and some guidance on solving systems of equations has been provided.

Contextual Notes

Participants mention discrepancies in their calculated values for the forces, indicating potential algebraic mistakes or misunderstandings in the setup of the equations. There is a lack of consensus on the correct values, and the discussion reflects varying levels of confidence in the approaches taken.

brentwoodbc
Messages
62
Reaction score
0
peeee.jpg

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
 
Physics news on Phys.org


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?
 
Last edited:


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.
 

Similar threads

Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
8
Views
5K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K