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Homework Help: Calculate tensiøn in strings.

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    A weight with mass m is strung up by two strings which are attached in A and B. The mass m is considered a point.
    Calculate the tension i in string A and in string B.

    Figure is here http://tinypic.com/r/1y1jrr/6

    The mass m is 11,1kg
    Distance a is 3,2m
    Distance b is 3,4m
    Distance c is 2,6 m
    Distance d is 3,7m

    (I'm translating directly from Danish, so I might not have gotten all the terms into correct English...)

    3. The attempt at a solution

    The first thing I did was to calculate the force G for the weight using G= mg
    = 11,1*9,81 = 1089N.

    I then tried to find the angles between the strings and the imaginary x-axis by using the measurements provided.

    The angle between string A and the x-axis(angle v1): invers tan 3,2/2,6≈51°

    The angle between string A and the x-axis(angle v2): invers tan 6,6/3,7≈61°

    I then set up two equations, one for the horisontal force and one for the vertical force:

    Horisontally: A*cosv1-B*cosv2=0
    Vertically: A*sinv1+B*sinv2=108,9

    And this is where I get stuck... I don't even know if I'm going about this the right way so I'd appreciate it if someone could have a look at it and point me in the right direction.
  2. jcsd
  3. Feb 9, 2013 #2
    So far you have done everything correctly. Why are you stuck? You should just solve the system now.
  4. Feb 9, 2013 #3
    I wasn't sure I'd done it correctly since I ended up with two equations with two unknowns(up until now I've done tasks where either A or B would have been given from the beginning.)

    To continue on from my first post, would this be the correct way to solve the system?

    A*sin v1+B*sinv2= 108,9


    (add up A)

    1,4064A = 108,9 -->A=774kN

    Then put value for A into the first equation which gives B = 100,5kN
  5. Feb 10, 2013 #4
    This is not entirely correct. If you simply add together the two equations, you won't eliminate B; you will have

    1.4064 A + 0.424 B = 108.9, which leads you nowhere.

    What you should do is multiply the first equation by 0.8746/0.4848 and then add the resultant equation to the second one; the result will not have B. Can you see why?
  6. Feb 10, 2013 #5
    Ah, I see my mistake. In order to do what I did the two B values would have had to be the same value in order to cancel each other out?(one being - and the other being +)

    I'm not sure if I understand you correctly, but do you mean using 0,8746/0,4848 as a fraction like this:

    [itex]A*0,6293-B*04848=0 *\frac{0,8746}{0,4848}[/itex] and then cancel 0,4848 against 0,4848?

    I'm pretty sure what I just wrote is "illegal" in the world of math, but I'll leave it for now as I don't have anything better to write. I know most calculators could solve equation systems like these, but I would like to learn how to do it by hand as well, so I'd be grateful if anyone could recommend some good videos on the topic.
  7. Feb 10, 2013 #6
    I am not sure about videos, but what you have here is a system of linear algebraic equations - the simplest case, two equations for two unknowns. I am pretty sure your high school curriculum should cover this.

    When you multiply an equation such as aA - bB = 0 by some c, you end up with the equation acA- bcB = 0. In this case, c = 0.8746/0.4848, and, indeed, it would transform b = 0.4848 into bc = 0.8746. But mind what happens to a: it becomes ac.
  8. Feb 11, 2013 #7
    Thank you so much for your help. I appreciate it.
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