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Tension forces problems

  1. Oct 13, 2003 #1
    I am having difficulty with tensions and pulleys. Can someone give a good explanation of how they work and what are some rules to solve the problems I have for homework. As far as I can see, the book and the instructor is not that good in describing this topic

    Here is one example of a problem:

    A picture hangs on the wall suspended by two strings as seen in the figure 6-20(see below for a written explanation). The tension in string one is 1.7N. (a) is the tension in string 2 greater than, less than or equal to 1.7N (explain) (b)verify your answer by calculating the tension on in string 2 (c) what is the weight of the picture?

    In the example there is a pictrue hanging in the middle of two strings, one string string one in 65degrees above the -X axis and string 2 is 32degrees above the X axis.

    Here is what I have done:

    T1x = 1.7(-cos115degrees)
    T1y = 1.7(sin115 degrees)
    and that is it...HELP!!
  2. jcsd
  3. Oct 13, 2003 #2
    More information....from what I am seeing

    Sum of Fx=t1(-cos115) + T2(cos32)

    Sum of Fy= t1(sin115) +T2(sin32) -mg + Normalforce.

    This seems that theses are the sum of the forces acting on the picture hanging on the wall, is this correct?

    How do I solve for the tension on the second string T2 if I do not have any information such as the wieght of the picture, or the force of T2??

    Thanks in advance
  4. Oct 13, 2003 #3
    I think I have a clue as to solve this problem.

    Since the forces acting on the picture =0 (because it is at rest) I can set the Sum of the Y forces equal to 0 and the sum of the Xforces equal to 0...like below

    sum Fx = T1(-cos115) + T2(cos32)
    1.7(-.422) + T2(.848)
    0 = -.717 + T2(.848) (solve for T2)
    .717 = T2(.848)
    T2 = .845 (not rounding up)

    Once I know T2 I can get the X and Y components of the tension T2

    T2x = .845(cos32) =.716
    T2y = .845(sin32) = .447 (not rounding up)

    Sum of Fy = T1(sin115) +T2(sin32) -mg + Normal force
    0 = 1.7(.906) + .845(.529) -m(9.81) + N
    0 = 1.54 + .447 -m(9.81) + N

    I got this far so how do I solve for the normal force and the weight of the picture??

    thanks in advance...
  5. Oct 13, 2003 #4
    it gets even better!!!

    I assume that there is no normal force because it is hanging, is that correct???

    so all I need to do is sum up the Y forces and solve for the mass...

    sumFy = T1y(sin115) + T2y(sin32) -mg
    0 = 1.7(.906) + .849(.529) -m(9.81)
    0 = 1.54 + .449 -m(9.81) (solve for m)
    0 = 1.99 -m(9.81)
    -1.99 = -m(9.81)
    m = .202

    Weight of picture is :

    .202(9.81) = 1.99

    Sound correct??
    I bow to the physics gods!!
  6. Oct 13, 2003 #5
    Glad we could help.
  7. Oct 13, 2003 #6
    I blew this one!!!the .848 is the force of the X component of T2.

    so for me to get the Y component of F2 I need to do...

    Tan32 = h/b

    tan32 = H/.848

    .624 = H/.848

    .624(.848) = h

    .529 = h = the Y component of T2

    Need to recalculate the rest of the problem.....

    T2^2 = .848^2 + .529^2

    T2^2 = .719 + .279

    T2^2 = .998

    T2 = .998 ???

    Sorry for the long post !

    When i get the information out of the problem Physics is cool!!
  8. Oct 13, 2003 #7
    No. You were right the first time. You're just getting confused by the rounding and the similarity between the value of cos32 and the magnitude of t2.

    Without rounding, you would do:
    t2 = 1.7*cos65/cos32
    and therefore the weight is give by
    mg = (1.7*cos65/cos32)*sin32 + 1.7*sin65
    mg = 1.99 N
  9. Oct 13, 2003 #8
    PS: .848 is not the x-component of t2.

    t2 = 1.7*cos65/cos32 = .847 N
    That is the full magnitude of t2. The x-component is .847*cos32 = .718 N
  10. Oct 13, 2003 #9
    Ok let me get this straight...and I felt good that I was able to work this through.

    By summing up all of the X components T2x, T1x, Nx, MGx I get the equation
    T2x = T2(cos32)
    T1x = 1.7(cos115)
    Nx = 0 (no X movement for the normal force)
    mgx = 0 (weight is only in the Y direction)
    SumFx = 0 (net force = 0 due to no motion F=m(0) = 0 )

    sumFx = T2x(cos32) + 1.7(cos115) + 0 + 0
    0 = T2x(.848) + 1.7(-.042) + 0 + 0
    0 = T2x(.848) + (-.0714)
    .714 = T2x(.848)

    T2x =.842 (may be off by a smidge due to rounding up)

    then I would take the T2x and using the angle of the T2 I can solve for the T2y and then eventually T2...correct??

    tan32 = T2y/.842

    .624 = T2y/.842

    T2y = .624(.842)

    T2y = .526

    T2^2 = T2x^2 +T2y^2

    T2^2 = .848^2 + .624^2

    T2^2 = .719 + .389
    T2^2 = 1.038
    T2 = 1.018 (a little different from the first calc due to rounding and maybe a mistake but is the format to solve this correct??

  11. Oct 13, 2003 #10
    sorry about that, I only posted it once but somehow it posted twice...very odd???
  12. Oct 13, 2003 #11

    T2x = T2(cos32)
    T1x = 1.7(cos115)
    Nx = 0 (no X movement for the normal force)Forget about normal force. It has no relevance to this problem, so just leave it out.
    mgx = 0 (weight is only in the Y direction)
    SumFx = 0 (net force = 0 due to no motion F=m(0) = 0 )

    In the following 5 lines, the red x doesn't belong there.
    sumFx = T2x(cos32) + 1.7(cos115)
    0 = T2x(.848) + 1.7(-.042) That should be 1.7(-.4226)
    0 = T2x(.848) + (-.0714) Should be (-.718)
    .714 {should be .718} = T2x(.848)
    T2x =.842 (may be off by a smidge due to rounding up)
    T2 = .847 Note: NOT T2x

    You may be confusing x:multiplication operator with x:subscript.

    Then, everything from that point on that is based on tan32 is wrong.

    As I said before, the solution you posted at 2:46 PM was correct, except for minor rounding errors (and the fact that you did some extra work in converting force to mass & then converting back to force) which actually netted out to the correct result anyway.
    Last edited: Oct 13, 2003
  13. Oct 13, 2003 #12
    I am sorry but as you can see I am a rookie at physics :(

    I guess I dont quite understand why the X component of T2 is the vector T2 (the tension on the second string)?

    Like I said I am a rookie at this but how can you solve for T2 without getting an X and Y component? My line of thinking was that if I solved string 2 X component (by summing all of the X components of both strings) I could then solve for the magnitude of the tension on the second string T2.

    I know that there is some super reason why you are right but is it possible to explain why?

    Thanks for your guidance.
  14. Oct 13, 2003 #13
    I don't understand what happened. You had the right answer, but then around 3:00 you started thinking your way up a dead end.

    T1 is a vector directed up & to the left. T1*cos115 is its x component. T1*sin115 is its y component.

    T2 is a vector directed up & to the right. T2*cos32 is its x component. T2*sin32 is its y component.

    w is a vector directed straight down. (w = -mg)

    So the equation in the x-direction is:
    T1*cos115 + T2*cos32 = 0

    The equation in the y-direction is:
    T1*sin115 + T2*sin32 + w = 0

    That's all there is in this problem.

    You seemed to understand this when you set up the equations initially. Then it seems you got mixed up when you put in numbers.

    When you solve the first equation, you should get T2 = .84718
    That is the FULL MAGNITUDE OF T2, NOT the x-component of T2. Do you see that? (The x-component of T2 is .84718*cos32=.71845) OK?
    Last edited: Oct 13, 2003
  15. Oct 14, 2003 #14
    I see now!!! I am solving for T2 which is T2 and the X component of T2 would be T2(cos32).

    Thanks for the help!

    Now I need to work on those damn pulley problems and the centrepial force problems....
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