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Tension of strings at different angles:

  1. Jul 10, 2006 #1
    Hi there- I've been trying to solve this problem for the last hour, and I still feel stuck! So far, I've only seen problems with equal angles between the strings and the horizontal- I have no idea how to solve a problem when there's two different angles!

    A 45-N lithograph is supported by two wires. One wire makes a 25 degree angle with the vertical and the other makes a 15 degree angle with the vertical. Find the tension in each wire.

    I thought that the Tensions of strings 1 and 2 should both add up to 45 N (because the object is in equilibrium), but apparently I'm wrong. Why is my logic off?
     
  2. jcsd
  3. Jul 10, 2006 #2

    nrqed

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    Forces are *vectors*!!

    You must decompose all the forces into their x and y components. Then you must impose that the net force along x is zero and the net force along y is zero. You will have two equations with two unknowns (being the magnitude of the two tensions in the strings).

    By the way, how would you do it if the two strings were both at 45 degrees (say)? Then the two euqations would be
    [tex] -T_1 cos(45) +T_2 cos (45) =0
    [/tex]
    and
    [tex] +T_1 sin(45) +T_2 sin(45) - mg =0 [/tex]
    Right? The first equation shows that T_1 = T_2 (same magnitude) and then the second equation helps you find the value of each.

    In your problem, the equations will be similar except that the angles will be different and you will find that T_1 and T_2 are not equal. You will have to solve two eqs with two unknowns.

    Patrick
     
  4. Jul 10, 2006 #3
    T2cos25-T1cos15= 0

    T2sin25+T1sin15-45=0

    Are these two equations & unknowns right? I keep getting T1 as being 34 N, but I know that cannot be right.
     
  5. Jul 10, 2006 #4
    They're not correct. Draw a free body diagram and check your directions. Are T2cos25 and T1cos15 the vertical components or the horizontal components of the string tensions?
     
    Last edited: Jul 10, 2006
  6. Jul 10, 2006 #5
    lol, ok, nevermind, I got it. I just switched the sin and cos, right? I think I drew my vertical and force vectors wrong...
     
  7. Jul 10, 2006 #6
    Thanks for the help- I really appreciate it!
     
  8. Jan 14, 2012 #7
    Say if I had both angles at 25 degrees each and I was solving for T1 could I express the equation as T1sin(25)x2=mg because both T1 and T2 are the same?
     
  9. Oct 16, 2012 #8
    Hi this came up in my search for am answer and I understand how you got to the answer but can't get why you used cos for the horizontal component of tension is that gives the adjacent component which is vertical
     
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