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Tension of 2 angles

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball weighing 590-N is shown suspended from a system of cords. What are the tensions in the horizontal and angled cords?
    http://img207.imageshack.us/img207/2276/4figure39alt.gif [Broken]

    [tex]T_{50}=___[/tex]
    [tex]T_{90}=___[/tex]

    2. Relevant equations



    3. The attempt at a solution
    [tex]T_{1}cos(90)-T_{2}cos(50)=590[/tex]
    This problem is confusing me because of the 90 degree angle.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 11, 2009 #2

    rock.freak667

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    Homework Helper

    T1 acts in the x-direction already. So your "T1cos(90)" is really just supposed to be T1. Do you understand?
     
  4. Oct 11, 2009 #3
    So, [tex]T_{1} - T_{2}cos(50)=590[/tex]? Still, I don't get how to use that equation to find either [tex]T_{1} or T_{2}[/tex]. You can divide by cos(50) but that won't help.

    [tex]T_{2}cos(50)=T_{1}-590[/tex]......
    or
    [tex]T_{1}=T_{2}cos(50)+590[/tex]
     
  5. Oct 11, 2009 #4

    rock.freak667

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    Actually this is wrong, the vertical component is sin(50). The sum of the vertical components is zero and the sum of the horizontal component is zero.

    So your equation should be T1-T2cos(50)=0

    Now consider what the the vertical component of T2 should be equal to.
     
  6. Oct 11, 2009 #5
    Wait, ok so

    [tex]T_{2}=T_{1}cos(50)[/tex]
    so
    [tex]T_{1}sin(90)+(T_{1}cos(50))sin(50)-F_{g}=0[/tex] right?
    [tex]T_{1}=0.670F_{g}[/tex]
    [tex]T_{1}=590/.670=880.597N[/tex]

    This doesn't seem right, what did I do wrong?
     
  7. Oct 11, 2009 #6
    Alright, I'm gonna give you a few big hints :)

    I'll call T1 the horizontal cord

    and T2 the cord at an angle.

    Drawing a free body diagram @ the knot of all three cords. the forces i have... is a downward tension force from the weight that is equal to 590 N, a tension force T1 holding my knot in the horzontal direction, and a tension force T2 at an angle to my knot.

    Now because i know that a=0 (everything is in equilibrium (not moving)) so because of Newton's First Law, i know the sum of all the forces on my knot is = 0. You remember this?

    So now i deduce that the sum of the horizontal forces = 0, and the sum of the vertical forces = 0.

    T1 is acting horizontally on the knot. I know that T1 does NOT support ANY of the weight of the object, it merely PREVENTS it from moving horizontally. so i know that T1 is equal to another force equal and opposite to prevent the cord from pulling the weight into the wall. But when i look opposite to T1, i don't see another force pulling along the horizontal, BUT we have a cord (T2) that is on an angle to the horizontal. So i know that T2 supports ALL of the weight of the object(because T1 supports none) + pulls against T1 at an equal amount.

    So i know that the HORIZONTAL COMPONENT of T2 (ill call it [tex]T2_x[/tex]) is EQUAL to T1, because all the HORIZONTAL forces are equal to 0..

    Therefore [tex]T1 + T2_x = 0 \therefore T1 = -T2_x[/tex] so T1 exerts an equal and opposite force of [tex]T2_x[/tex]

    I also know i can solve for [tex]T2_x[/tex] because with the angle @ T2 50 degrees.

    i know that [tex] sin 50 = \frac{T2_x}{T2} \therefore T2_x = T2 \cdot sin 50 [/tex]

    But i dont know what T2 IS! so until i can find out T2, i can't find out T2x and until i find T2x i cant find t1.. get it? How do you find T2? What is [tex]T2_y[/tex] equal to? and how can you use that to get the values your looking for? Hope this helps..

    Senjai
     
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