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Demonstration and Newton's Laws

  1. Oct 9, 2004 #1
    A stone hangs by a fine thread from the ceiling and a section of the same thread dangles from the bottom of the stone. If a person gives a sharp pull on the dangling thread, the bottom string will break, but if he slowly pulls the bottom string the top string will break.

    After seeing this demonstration on video, I am still confused on the principles that are involved here. Exactly why do these things occur (the breaks in the string at various times) in reference to the difference in tensions and a change in acceleration?

    I attempted to do a force diagram to help with this question:
    ---| Tension 1
    ---| Tension 2 (applied force) Another arrow (force) also points downward for mg.

    My teacher mentioned the formula T2-T1 + mg = ma, but I do not understand why mg is not -mg (T2 - T1 - mg = ma).

    Pertaining to the reasons why, I did some readings on the subject and they mentioned that the top thread experiences the tension due to the weight of the rock in addition to the force exerted by pulling the bottom thread. The bottom thread breaks due to the fact that the rock has inertia resisting a sudden change of its velocity. I still don't truly get the difference in tensions and change in acceleration that lead to the threads breaking.

    I would really like to understand this experiment. Please explain the things that I mentioned to me in a simple but thorough manner. (I'm sorry if I sound really ignorant.)

    Thank you.
    Last edited: Oct 10, 2004
  2. jcsd
  3. Oct 9, 2004 #2


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    Homework Helper

    You're dealing with vectors. One way to deal with them in a one-dimensional situation is with + and - signs. Note that T2 is expressed as positive while T1 is negative. What does this tell you about which was is positive and which negative? And, given that, which way should the weight act?

    Remember - the direction associated with positive is arbitrary.
  4. Oct 10, 2004 #3
    Now as for the breaking of the string,
    lets say that the strings have a breaking tension T' > T2 and T1

    we know that,
    T2-T1 + mg = ma
    rearranging this a bit we get,
    (T2+mg)-T1 = m (dv/dt)

    if dv/dt is large , it implies
    (T2+mg) is much larger than T1
    mg is a constant so T2 is very large
    the chances that T2 reaches T' first before T1 is very high
    implying that the lower string will break first.

    if dv/dt is small, then a bit of rearranging gives,
    T1 = T2 + mg - ma
    if T2 goes nearly equal to T'-mg+ma, T1 would have nearly gone to T'
    i.e T1 would reach T' before T2 does ..
    which indicates that the top string would break first..

    -- AI
  5. Oct 10, 2004 #4
    I don't understand the second part. Where did T'-mg+ma come from? (a = dv/dt still, right?) Why was the presumption T' > T1 and T2 (T1 + T2) made?

  6. Oct 10, 2004 #5
    T2 is the tension which is developed due to the action of pulling it
    so i can pull the string with a force T'-mg+ma (and yes a=dv/dt still!)
    which would mean T2 = T'-mg+ma.

    Why the presumption T' > T1 and T2 ?
    if T1 were greater than T', the upper part would have broken even before i pulled it.
    i made T' > T2 specifically to mean that the person who is pulling the string would need to exert considerable amount of force before the string gives way. i.e the string is not very weak.

    -- AI
  7. Oct 11, 2004 #6
    Why is the derivative of velocity (dv/dt) used in place of a? I know a = dv/dt, but why is this definition used here?

    Also how does T2 = T' - mg + ma come from T1 = T2 + mg - ma? In order to show that a (dv/dt) is small, it becomes -ma?

    Last edited: Oct 11, 2004
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