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Tension and Friction [Tension in a horizontal direction] Help.

  1. Sep 26, 2009 #1
    My question is if I actually did the said problem correctly and if there is perhaps an easier way of coming across an answer. Thanks!

    1. The problem statement, all variables and given/known data
    There are two blocks present (A and B) A is connected to B via a rope and the applied force is affecting B (to the right).

    Find the force of tension in the string if the coefficient of friction is u=0.13
    mA = 40 kg
    mB = 25 kg
    Fapplied = 150 N
    u = 0.13

    2. Relevant equations
    See below.


    3. The attempt at a solution
    Tapplied – Tblock – μmBg = mBa

    Tblock – μmAg = mAa
    Tapplied – μmBg – μmAg = a
    mA + mB
    Tblock – μmAg = mAa

    Tblock – μmAg = mA(Tapplied – μmBg – μmAg)
    mA + mB
    Tblock = mA(Tapplied – μmBg – μmAg) + mAg
    mA + mB
    Tblock = 40(150 – (0.13)(25)(-9.8) – (0.13)(40)(-9.8)) + (40)(-9.8)
    40 + 25
    Tblock = 40(150 + 31.85 + 50.96) + (-392)
    65
    Tblock = 40(232.81) + (-392)
    65
    Tblock = 40(3.58) + (-392)
    Tblock = 143.3 + (-392)
    Tblock = -248.73
     
  2. jcsd
  3. Sep 26, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Caerus! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    It would be a lot easier to write the third one Tapplied = (a + µg)(mA + mB). :wink:

    (oh, and you missed out a µ in the middle of your calculations, which is why your 392 is so large :rolleyes:)
     
  4. Sep 26, 2009 #3
    Woah, that did help thanks.

    Question: My answer, for the tension of the rope with friction, was identical to the tension of the rope without friction. Does that mean that friction has no direct effect on the tension of the rope. I assume so.
     
  5. Sep 26, 2009 #4

    tiny-tim

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    Let's write a + µg = b, to save space.

    The the first two equations are Tblock = Tapplied - bmB

    and Tblock = bmA

    then eliminating b (which eliminates µ also, of course) gives Tblock = … ? :smile:
     
  6. Sep 26, 2009 #5
    I appreciate the help tiny-tim! I solved my problem and everything has become much clearer to me. My final question however, and I hope you can assist me, are there any other cases in which the force of tension is zero except for when the force applied is zero? I think it has to do with the coefficient of friction.
     
  7. Sep 26, 2009 #6

    tiny-tim

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    ?? :confused: how can the tension be zero if the back block is moving?
     
  8. Sep 26, 2009 #7
    I suppose I should have phrased the question a little better...

    If I'm not mistaken, say my force applied was 0 N. Which would obviously mean the tension between the two blocks would be zero but would there be any other possibility for the tension to equal 0 even when there is an applied force?
     
  9. Sep 26, 2009 #8

    tiny-tim

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    If the applied force is small enough (but non-zero), then it's possible for the blocks not to move, and for the tension to be zero … what would be the maximum applied force for that? :smile:
     
  10. Sep 26, 2009 #9
    Hmm, let's see. I know that 1 N would bring it awfully close but as for anything else. I cannot think of one. I do however think that if the coefficient of friction was a large enough number to cancel out the applied force, then the force of tension would equal zero. Right?
     
  11. Sep 26, 2009 #10

    tiny-tim

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    You can't change the coefficient of friction, but you can change the applied force … how small would be applied force have to be? :smile:
     
  12. Sep 26, 2009 #11
    Yes, but in a situation where I am able to decide the coefficient of friction could it not alter the equation in such a way that tension equals zero? An applied force of 0 N would equal zero tension. I cannot think of any others.
     
  13. Sep 27, 2009 #12
    i have the exact same assignment can any1 help me?
     
  14. Sep 27, 2009 #13
    Do you by any chance live in Ontario, pheonix4u? Possibly even Toronto? What are you having problems with?
     
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