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Finding Tension in Wire

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data

    HI!! ok i drew the problem out.

    [​IMG]


    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.


    2. Relevant equations

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


    3. 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 =(
     
  2. jcsd
  3. Oct 3, 2008 #2
    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?
     
  4. Oct 3, 2008 #3
    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 thats why.
     
  5. Oct 3, 2008 #4
    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.
     
  6. Oct 3, 2008 #5
    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 ??
     
  7. Oct 3, 2008 #6
    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.
     
  8. Oct 3, 2008 #7
    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 wouldnt they equal zero as well?

    but then i know theres two more tension forces to the sides

    i don't know i'm so lost right now.
    am i even on the right track?
     
  9. Oct 3, 2008 #8
    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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