# B Which thread snaps first ?

1. Jul 16, 2016

### Vibhor

Hello ,

Please see the attached image , in first case thread A is pulled with gradually increasing force while in second case an impulsive force is applied on thread A ? Which one breaks first in the two cases ?

I think in the first case , thread B breaks . In second case , thread A breaks first .

This is my reasoning -

Consider the mass and apply Newton's II law to it .Downwards is taken positive .

∑F = ma
FA - FB + mg = ma

FA - FB = m(a-g)

Case 1 ) When the thread is pulled slowly, acceleration 'a' would be negligible and we get ,

FA - FB = -mg < 0
FA < FB

Hence B breaks first .

Case 2) But if A is pulled very fast than 'a' would be very large and we get ,

FA - FB = m(a-g) >0 . Now here I am assuming that 'a' > g ( Is this correct ??)

FA > FB

Hence A breaks first .

Is the above reasoning correct ? I am a little unsure in Case 2) where I am considering 'a' to be very large when the sudden force is applied on thread A . Is it right to take 'a' very large ??

Many Thanks

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2. Jul 16, 2016

### Daymare

Yes,what you have done is correct.

3. Jul 16, 2016

### Daymare

When you pull the string suddenly what happens is that the system wont be in equilibrium.Also note that when you pull the string suddenly the rock accelerates for a very short time(in this time we can assume that the both the strings won't be following hooke's law) the acceleration is for a very short time but is so huge that the force is very high on string A and it breaks.An interesting case would be when a=g.Then the force on both strings are the same and we can expect both of them to break together(assuming they are identical strings).

4. Jul 16, 2016

### andrewkirk

I think the answer is correct.
We can think of it in terms of shock waves. An increase of force travels through matter as a shockwave and it propagates at the speed of sound in that material. So an increase of force travels through the lower thread at the speed of sound in thread, say $CT$, until it hits the block, then travels through the block at speed $CB$, the speed of sound in the block. Let the length of the lower thread be $L$ and the thickness of the block be $D$. Let the time rate of increase of force be $J$.

Then the time for a shock wave to travel from the bottom of thread A to the bottom of thread B is $\left(\frac L{CT}+\frac D{CB}\right)$. It follows that the tension at the bottom of thread B minus the tension at the bottom of thread A is
$$mg-J\left(\frac L{CT}+\frac D{CB}\right)$$
If $J$ is small this will be positive, so that the tension is higher at the bottom of the upper thread, so that thread breaks first. If $J>\frac{mg}{\frac L{CT}+\frac D{CB}}$ it is negative so that the the tension is higher at the bottom of the lower thread, so that thread breaks first.

5. Jul 16, 2016

### Vibhor

But do you think the reasoning in the OP is also correct ??

6. Jul 16, 2016

### Daymare

If your string A breaks it means,force on it was larger than that on B.This would imply a>g,hence I think your reasoning is correct.

7. Jul 16, 2016

### Vibhor

I am still unable to convince myself fully that acceleration of mass would be very large when A is pulled suddenly . Do you mind explaining ?

8. Jul 16, 2016

### Daymare

I think the difficulty here lies in imagining anything accelerating greater than g without moving too much.Good way of looking at this would be to imagine that when you pull the string suddenly the bottom and lower string elongating a bit.The force causes the string A to elongate and break faster than that of B;once string A breaks no more force acts and the system is again in equilibrium.

9. Jul 16, 2016

### andrewkirk

I don't think acceleration of the block is the key issue, or at least, it's not the easiest way to understand the problem. The comparison of the speed of propagation of the shockwave to the rate of increase in force is the key issue, as per the above equation. If a sufficiently savage jerk were applied to one end of a thread that was unattached at the other end, it would break.