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Rip time of a string or thread pulled from both sides

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Two bodies with masses m1 and m2 are connected by a massless thread that supports a maximum force |Ft|max . On mass m1 the force F1 = -β t x and on mass m2 the force F2 = 3 β t x (t: time, β: constant) acts. The motion is frictionless.

    2. Relevant equations

    When does the thread rip?

    3. The attempt at a solution

    According to Netwon's 3rd law if m2 pulls m1 with a force Ft (equal to the string tension), m2 is then pulled by m1 with a force equal and opposite -Ft. Therefore, Netwon's 2nd law for m1 is: (net force=mass x acceleration):

    Ft-F1 = m1*a

    and for m2:

    F2-Ft = m2*a

    The acceleration should be the same. We do not know which force "wins", in other words the direction of the acceleration. I guess it depends on the numerical values for m1 and m2 too.

    For example, let's fix the one-dimensional reference system with the positive x axis in the same direction of F2 as suggested by the problem text, but I could also do the other way. In this system, supposing a is in the positive direction, each vectorial equation becomes only one scalar equation, since we are in one dimension:

    Ft-F1 = m1*a
    F2-Ft = m2*a

    This is a linear system: two equations, two unknowns: Ft and a. I am only interested in finding out Ft. E.g. I solve the second equation for a, substitute into the first and solve for Ft. I get:

    Ft = (m2 F1 + m1 F2)/(m1 + m2)

    The tension is the weighted sum of the two forces, weighted by the mass values, but not the respective mass values, rather they are inverted: m2, not m1 multiplies F1, and similarly for F2. So a sort of inverted weighted sum.

    Since in our reference system:

    F1=-β t
    F2=3 β t

    we have:

    Ft = β t (3 m1 - m2) / (m1 + m2)

    The string will break at time t = trip when Ft is equal to Ftmax

    trip = Ftmax (m1 + m2) / (β (3 m1 - m2))

    which should be the solution to the problem. Is this the correct way to solve it?
     
  2. jcsd
  3. Oct 28, 2012 #2

    PeterO

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    I am reading through your post, and when I got to the sentence I made red above, I feel we do know which force "wins" since the size of F2 is 3 times the size of F1. I have not read past that point yet, since my suggestion here might enable you to finish the problem anyway.
     
  4. Oct 29, 2012 #3
    Hi Peter, yes, F2 is 3 times F1, but m2 could be more then 3 times m1, so that the acceleration would be in the direction of F1 and not the direction of F2. A force does not equate the acceleration, but there is a mass factor (this is Newton's 2nd law). I do not think we need to guess right to correctly solve the problem.
     
  5. Oct 29, 2012 #4

    PeterO

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    Er.. NO

    Suppose you instead glued these two masses together. The only difference would be that there was no string to snap/rip.
    And you are right - we don't need to guess to correctly solve the problem - we just have to realise, then correctly solve the problem.

    NOTE: If you are imagining that the string is at first slack, and the two masses accelerate apart, then as soon as the string goes tight, it will snap.
     
  6. Oct 29, 2012 #5
    Why not? What do you mean by realise? One can fix the reference system arbitrarily. Then you need to express the coordinate of vectors in this system. If they are unknown, you just use a variable, which at the end can be either positive o negative. So for instance, the acceleration is a vector a. Its module be a. If I solve the system for a, the sign of a will tell me my direction, depending on the value of other parameters.

    Why as soon as the string goes tight? Why couldn't it happen after a while, the time needed for the tension to reach the breaking value?
     
  7. Oct 29, 2012 #6

    PeterO

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    A string is inextensible, so when it goes tight, and the masses (attempt to) take on a common velocity, it will involve infinite acceleration - which would have to be caused by infinite force - but the thread does not have infinite strength
     
  8. Oct 29, 2012 #7

    PeterO

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    By realise I mean realise that the masses have the same acceleration - with a taut string between them - and thus the two masses and the string can be considered a single body.

    The mass is m1 + m2.

    That "body" has two forces acting on it. One in -x direction, the other in x direction. The force in the x direction is 3 times the size of the other, so the net force is -2F1 or 2/3F2 , which ever you prefer.

    From that you can calculate the acceleration.

    From the acceleration you can find the net force on m1.

    This is the net result of F1 and the tension in the string.

    Once that string tension reaches the maximum possible, the string breaks.

    That should be enough info to be going on with.

    EDIT: Perhaps you were working along those lines already. As I said, I stopped reading when I got to the sentence when you said you didn't know which direction the acceleration would be.
     
  9. Oct 30, 2012 #8
    Yes, Peter, there are two ways to solve this problem. One is the one you outlined. But one can also do my way, without regarding the two masses as a unique body, so without need to think about the verse of the acceleration. You just write Newton's 2nd law for both and you have a system of two equations. The solutions is, of course the same for both methods.

    BTW I made a mistake in my solution above: when I write equations of forces as vector, they should add, so no minus signs.

    Ft+F1 = m1*a

    F2+Ft = m2*a

    Then when I fix a coordinate system I see what the signs of the coordinate are, and in writing these I was correct. Here are them again:

    Ft-F1 = m1*a
    F2-Ft = m2*a

    but then, when I substitute the time-dependent values here, I have to use a positive values for both F1 and F2 since I agreed these are the modules of their respective vectors:

    F1=β t
    F2=3 β t

    So the final (correct) solution is:

    trip = Ftmax (m1 + m2) / (β (3 m1 + m2))

    the previous solution, with a minus sign was wrong.

    Of course if I solve the problem your way, I get the same expression.

    This comparison has been informative, so thanks, but also try to understand solutions different from yours (which may also be correct) if you want to be a good tutor. And the same, of course, goes for me: I have to think about all the different ways a problem can be solved, if I want to be a good student :)
     
  10. Oct 30, 2012 #9

    PeterO

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    Your method is indeed a common, logical and useful approach - but I was never passing comment on it - merely pointing out that your statement "we don't know which force "wins" " was in error. Had you not made that statement, I would have read through and perhaps found the incorrect sign.
     
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