Demonstration and Newton's Laws

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SUMMARY

The discussion centers on the physics of tension in strings and the application of Newton's laws, particularly in the context of a stone suspended by two strings. When a sharp pull is applied to the lower string, it breaks first due to the inertia of the stone, which resists sudden changes in velocity. The tension equations T2 - T1 + mg = ma illustrate the relationship between the forces acting on the system, where T2 represents the tension from the pull, T1 is the tension in the upper string, and mg is the weight of the stone. The analysis concludes that the breaking of the strings depends on the rate of change of acceleration (dv/dt) and the comparative breaking tensions of the strings.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Basic knowledge of tension and forces in physics
  • Familiarity with vector analysis and force diagrams
  • Concept of inertia and its effect on motion
NEXT STEPS
  • Study the principles of tension in strings using "Physics of Strings" resources
  • Learn about Newton's Second Law and its applications in real-world scenarios
  • Explore the concept of inertia and its mathematical representation in "Dynamics of Motion"
  • Investigate vector analysis in physics, focusing on force diagrams and their applications
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of tension and forces in mechanical systems.

Soaring Crane
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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
--rock
---|
---| 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:
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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.
 
AN INFORMAL ANALYSIS :
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
 
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?

Thanks.
 
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
 
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?

Thanks.
 
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