How Do You Calculate the Tension in Multiple Wires Supporting Artifacts?

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Homework Help Overview

The problem involves calculating the tension in three wires supporting two artifacts in a museum display. The context is set within the framework of static equilibrium, where the sum of forces in both the vertical and horizontal directions must equal zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the tensions in the wires and the weights of the artifacts. There are attempts to apply equilibrium equations, with some confusion about the components of forces and the setup of the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and clarifications regarding the forces acting on the system. Some participants express uncertainty about their calculations and the relationships between the tensions, while others confirm that they are on the right track.

Contextual Notes

There is mention of a diagram that could clarify the setup, and participants are grappling with the implications of equilibrium in the context of multiple forces acting on the system. The weights of the artifacts and their corresponding tensions are central to the discussion.

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Homework Statement



HI! ok i drew the problem out.

tension.jpg



Question:

Two artifacts in a museum display are hung from vertical walls by very light wires. Find the tension in each of the three wires.


Homework Equations



[tex]\Sigma[/tex]Fy=0
[tex]\Sigma[/tex]Fx=0


The Attempt at a Solution



Ok so i know the tension in T1 is the same as the weight of the second artifact which is 314N
I just get all confused trying to figure out the other two

[tex]\Sigma[/tex]Fy= -175N + 175cos53 = 0

which came to N = .60 ??

and for [tex]\Sigma[/tex]Fx i only got 175sin90 ??

i just need help trying to figure out where to go or how to set it up I'm so lost =(
 
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I can help you solve this, but you need to show the diagram slightly better. Does the 175N correspond to T3?
What do you mean by N? is it a variable or the unit of Newtons?
If the latter, why were you solving for N?
 
Hi! thank you!

the 175 N is the weight of the first object so yes the unit of Newtons

i have no idea why i was solving for N
i didnt know what to solve for if it equals 0 that's why.
 
Okay, here are some hints.

The sum of forces in the x-axis will equal zero. T1 has no x components. Therefore the x components of T2 must equal the x components of T3.
You should now be able to write a relation between these two.

Now you can see that T1 is only holding the weight of the 32kg mass. However, must the y component of T3 effectively support both weights?

Give that a try.
 
ok

so for [tex]\Sigma[/tex]Fx = -T2 + 175sin53 = 0

so T2 = 139.76 ?

[tex]\Sigma[/tex]Fy = 489cos53 = 294.28 ?

am i suppose to divide something by something?

like the weight divided by sin53 ??
 
Why do you say that the sum of forces in the y direction is only 489cos53. I can see two component forces acting there.
The sum of forces in y must equal zero. The two weights that are affecting the system have forces 175N and 314N. We know this.

Try again please.
 
This is where i get confused because i thought they are all at equilibrium because the weight is -314N and the T1 is 314N right? so they equal 0

then for the -175N is there another tension force going up?
so wouldn't they equal zero as well?

but then i know there's two more tension forces to the sides

i don't know I'm so lost right now.
am i even on the right track?
 
Yes you are definitely on the right track. The two side forces are just there to cancel each other out in a manner of speaking. If T3 was vertical, T2 would not exist.

Look at it like this: Say the 32 kg mass was attached to the roof of something by T1. Tension of cable 1 would be 32 * 9.81 right? The net force = 0

This system is in equilibrium, so essentially, for T1 this is exactly what is happening.
That has T1 out of the equation.
The total force of the weights in the system = 314 + 175 = 489
 

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