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Homework Help: Tension between two blocks

  1. Jun 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Two blocks of weight 50N and 200N are connected by a cord rest on two inclined planes .determine the maximum tension in the cord when limiting friction condition develop in both blocks
    R1=Reaction of weight 50N
    R2=reaction of weight 200 N



    2. Relevant equations
    for block of mass 50 N moving upward
    µ1R1+wsin45-T1=0
    for block of mass 200N assuming moving downward
    T2+ µ2R2-W2sin30=0


    3. The attempt at a solution:Its just that I am not able to understand why there is different tensions in the cord .Doesnt the tension should be equal if the cord is weightless?
     

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  3. Jun 24, 2015 #2

    Simon Bridge

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    What happens if there is no friction?

    (By limiting friction" do you mean that the blocks are stationary or moving at a constant speed?)
     
  4. Jun 24, 2015 #3
    by limiting friction i mean the bodies have just started to move overcoming friction .if there is no friction and if we consider tension same the the bodies will move in the direction of of 200N
     
  5. Jun 24, 2015 #4

    Simon Bridge

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    They have just overcome the static friction - so they are in motion, and the coefficients are for kinetic friction?
    Or they are just about to overcome static friction - and the coefficients are of static friction?
     
  6. Jun 24, 2015 #5
    They have just overcome motion hence they are coefficient of static friction
     
  7. Jun 24, 2015 #6

    Simon Bridge

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    OK - what would be the tension if there were some acceleration?
    Does that tension depend on the acceleration?
     
  8. Jun 25, 2015 #7
    i am just confused in this part i think the tension will remain same in this case.Yes i think it depends since f=m*a and tension is also one kind of forcwe
     
  9. Jun 25, 2015 #8

    haruspex

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    Yes, the tensions must be the same if the cord is massless and there's no friction at the pulley. What makes you think they're different?
    That's not what limiting friction usually means. I would take it to be at maximum static friction.
    That's inconsistent. If they have overcome friction then it's kinetic; if they haven't overcome it's static.
     
  10. Jun 25, 2015 #9
    Yes it takes maximum static friction its just that in my textbook there are two different tension so i got confused and
     
    Last edited: Jun 25, 2015
  11. Jun 25, 2015 #10

    haruspex

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    The tensions are actually different, or merely it is using different variables to represent them? If the second, maybe there's an equation later that says they are equal.
     
  12. Jun 25, 2015 #11
    They are represented by different variables in the text book. but now i think since we have to find maximum tension in this question they are represented differently.But i am not so sure
     
  13. Jun 25, 2015 #12

    haruspex

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    Well, as I wrote, unless there is friction on the rope at the apex the tensions will surely be equal.
     
  14. Jun 25, 2015 #13
    Then i think its a mistake of textbook since it hasnt said any tension in apex
     
  15. Jun 25, 2015 #14

    Simon Bridge

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    Why would it? The text has not mentioned lots of things - that does not mean that it is wrong.

    i.e. the text has not said anything about the tension half way between the left block and the apex ... but we can tell some things about that tension by thinking about the physics involved. Just cause it is not written down don't mean its not true.

    If you consider the point on the rope exactly at the apex, you can see there must be some tension pointing each direction away from the apex. Since the rope is not accelerating, you can deduce that the magnitude of the tensions must be equal.

    Your book has not said that the tensions T1 and T2 are not equal.
    Different variable names may still have the same physical value.

    Try redoing the calculation assuming that T1=T2 - be careful about the directions of your vectors.

    Notes:
    ##\sin(45)=\cos(45)=1/\sqrt{2}##, ##\sin(30)=1/2##, ##\cos(30)=\sqrt{3}/2##.
     
  16. Jun 25, 2015 #15

    haruspex

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    I said friction at apex, not tension. And as Simon and I are both trying to explain to you, it is not wrong to use different symbols for entities that could in general be different but in the present case turn out to be the same.
    It sounds like you are walking through a solution in your textbook. Is that right? Or is it just that your textbook provides a diagram with labels T1 and T2 in it?
     
  17. Jun 25, 2015 #16
    Sorry i was trying to write friction at the apex.I am seeing solution in text book and it clearly has two tension value
     

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  18. Jun 26, 2015 #17

    Simon Bridge

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    I'd like to see the exact wording of the question.

    If you put a nail through the rope at the apex, fixing the rope at that point, it is easy to see why there would be a different tension on each side of the nail.
     
  19. Jun 26, 2015 #18

    haruspex

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    Ok, I think I understand what is going on here.
    The T1 calculated is an upper limit on the tension for static equilibrium. If the tension exceeds this then block 1 will slide up. The T2 calculated is a lower limit on the tension for static equilibrium. If the tension is less than this then block 2 will slide down.
    If the calculated upper limit were greater than the calculated lower limit then it would be stable. The actual tension would be anywhere in that range. (It can happen that it simply is not possible to determine the actual tension.). As it is, the lower limit comes out higher than the upper limit, so it is not stable.
    But if I'm right, the final line in the textbook makes me feel the author has not understood the problem.
     
  20. Jun 26, 2015 #19
    thanks for explaining.But i still have one confusion.How does the tension have upper and lower limits doesnt it should be equally distribured in the chord?
     
  21. Jun 26, 2015 #20
    I have written the exact question and there is no wording about fixing the rope
     
  22. Jun 26, 2015 #21

    haruspex

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    There is only one actual tension throughout the cord. The calculation is finding an upper and lower bound for its value, on the assumption that there is static equilibrium.
    If the upper bound is the greater then there is static equilibrium and the actual tension is somewhere in the range (but there is no way to find out where in the range).
    If the lower bound is the greater then we have a contradiction, so the assumption of static equilibrium must be false.
     
  23. Jun 27, 2015 #22

    haruspex

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    In my view, the approach taken in the textbook is confusing, and even seems to have arrived at the wrong conclusion.
    I would take the tension to be T both sides, then write down the constraints for static equilibrium each side.
    ##T<=m_1g\sin(\theta_1)+m_1g\cos(\theta_1)\mu_s##
    ##T>=m_2g\sin(\theta_2)-m_2g\cos(\theta_2)\mu_s##
    Whence
    ##\mu_s(m_1\cos(\theta_1)+m_2\cos(\theta_2))>=m_2\sin(\theta_2)-m_1\sin(\theta_1)##
    If that condition is satisfied, there is a T that produces static equilibrium (but we might not be able to say exactly what it is).

    If the condition is not satisfied (as here) then it will slide. If we wish to find the tension in the cable we must now introduce an unknown acceleration and use the kinetic friction coefficient.
     
  24. Jun 27, 2015 #23

    SammyS

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    Your image of the text book "solution" certainly helped. Can you similarly supply an image of the problem statement from the textbook?

    Lacking that at the moment ...
    This is my take on this problem in light of the supplied "book" solution.

    Both tension, T1 and T2 are arrived at assuming that the fiction forces are to the left, thus opposing motion to the right.

    The value of T1 that's calculated is the maximum value of tension, T, that will produce static equilibrium. (If we consider this to be a kinetic situation, then T1 is the minimum value of tension, T, that will enable sustained motion to the right.)

    T2 is a bigger puzzle.
    The value of T2 that's calculated is the minimum value of tension, T, that will produce static equilibrium. Any smaller value for T2 will give motion to the right. If T is greater than this (up to a point), then the static friction will simply be less than μ2R2, R being the normal force. (T2 can be as large as 120.78 N for static equilibrium.)

    Considering the values of T1 and T2, there is no possibility of this system being in static equilibrium.
     
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