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enter image description here

Please excuse the crude drawings

The question is, if I burn the thread t, how will the reading in the spring balance(S) change? Ie, increase, decrease, or remain the same?

My approach:

When we burn the thread, the objects will move, but the tension should increase a bit. ( I can't explain this, I only get this intuitively.) and, since the spring balance reads the double of the tension, I think this should make it a lot more than the initial reading. Since if the tension in the intact string changes by $\triangle T$, the reading should increase by $2 \triangle T$. That's all I can think of. So as you see, my approach is very intuitive, and hence I am asking this in order to get a logical and stronger reasoning

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  • $\begingroup$ The system will never be under equilibrium. The $m+M$ mass will accelerating down, in which case you will not get a constant spring reading. $\endgroup$ – udiboy1209 Jul 28 '13 at 12:45
  • $\begingroup$ It was in equilibrium before burning. That is given. $\endgroup$ – Saurabh Raje Jul 28 '13 at 12:50
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    $\begingroup$ That I agree, but it cannot achieve equilibrium after burning. $\endgroup$ – udiboy1209 Jul 28 '13 at 12:51
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    $\begingroup$ @SaurabhRaje Sorry, but I think your intuition is wrong. Since the center of mass will be falling, the net force applied by the spring will decrease! $\endgroup$ – Ali Jul 28 '13 at 12:56
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Let me assume that the spring's constant is so high that the change in its length will be negligible. Now we will have a simple pulley's system, which is straight-forward to solve.

pulley system

After solving the equations, the tension in the ropes connecting the masses will be

$$T=\frac{2M(M+m)g}{2M+m}$$

So the tension in the spring will be $2T$, which is less than its initial value:

$$2T=\frac{4M(M+m)g}{2M+m} = \frac{(2M+m)^2-m^2}{2M+m}g\le (2M+m)g $$

Update:

As requested, I will derive the first equation:

  • Newton's second law $$M \ddot{y_1}=Mg-T \\ (m+M)\ddot{y_2}=(M+m)g-T $$
  • Constraint $$y_1 + y_2 = \text{const.} \Rightarrow \ddot{y_1}+\ddot{y_2}=0$$

Now substituting from the first two equations into the third:

$$2g-\frac{T}{M}-\frac{T}{M+m}=0 \Rightarrow T\left(\frac{1}{M}+\frac{1}{M+m} \right)=2g \\ T \frac{2M+m}{M(M+m)}=2g \Rightarrow T = \frac{2M(M+m)g}{2M+m} $$

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  • $\begingroup$ pls elaborate how you got the first equation? $\endgroup$ – Saurabh Raje Jul 28 '13 at 13:42
  • $\begingroup$ @SaurabhRaje That's a classic question which can be found in whatever introductory mechanics textbook you pick up. $\endgroup$ – Ali Jul 28 '13 at 13:51
  • $\begingroup$ @SaurabhRaje I will add something about it. $\endgroup$ – Ali Jul 28 '13 at 14:47
  • $\begingroup$ Thanks a lot! Really clarified my thoughts.....another example of how intuition leads wrong... $\endgroup$ – Saurabh Raje Jul 28 '13 at 16:37

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