Understanding which string breaks when one pulls on a hanging block from below

Question

Intro:

In completing Walter Lewin's 6th lecture on Newton's Laws, he presents an experiment (go to 42:44) which leaves me baffled.

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Experiment:

(I recommend watching the video; see link above.)

• There is a $2$ kg block with 2 identical strings attached to it: one at the top, the other at the bottom.
• The top string is attached to a "ceiling", and the bottom to a "floor".
• Professor Lewin "stretches" the system (by pulling on the bottom string) with the block not accelerating.
• One string snaps.

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Prediction:

• Initially, the top string has a tension of approximately $20$ N, to counter the force of gravity. The bottom string has no tension at all.
• Then, when Lewin pulls the bottom string, it gains some tension $n$ N. To counter act the force exerted by the bottom string, the top string exerts now $20 + n$ N.
• I assume that the string with more force will give out sooner, leading me to conclude that the top string will break.

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Results:

(This was conducted by Lewin, not me; see link above.)

• Trial 1: Bottom string breaks.
• Trial 2: Top string breaks.
• Trial 3: Bottom string breaks.

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Additional Notes:

The results don't seem consistent. If I was right, I'd expect all 3 experiments to be right; conversely, if I was wrong, I'd expect all 3 experiments wrong, with one exception: the results are more-less random and one result isn't preferred over the other.

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Question:

• Why was my prediction incorrect?
• Was there a flaw in my logic?
• Why were the results inconsistent?

Answer 1

Your predictions of the forces adding up is correct, if nothing accelerates. Because, think about it... you are adding up forces, right? That is what you do in Newton's 1st law. Which is the law that only applies when nothing accelerates.

What if you were told that you can't use Newton's 1st law in the second case? Is something accelerating in the second case?

Or in other words, is the string trying to accelerate something in the second case?

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Solution

If something should accelerate, we are in Newton's 2nd law. If not, Newton's 1st law. Let's write it out with the forces from each string and weight $w$ present:

$$- F_{up} +F_{down} + w=0\qquad \qquad - F_{up} +F_{down} + w=ma$$

(I hope it's okay I've put the y-direction downwards.)

• If you pull slowly down, no significant speeding up happens of the box. $F_{down}$ has some constant value. It all balances out. The 1st law.

• If you pull fast down, the box tries to speed up fast to follow along. That means large $a$. That requires large force to cause it. And the force, that tries to cause is the $F_{down}$.

Look at those two equations again. In the first case $F_{up}=F_{down}+w$, so the upper string breaks. In the second case $F_{up}=F_{down}+w-ma$. Hmm, here is being subtracted the part $ma$...

So, is $F_{up}$ becoming smaller? No, of course not, it has it's tension and only grows as you pull downwards. Rather $F_{down}$ becomes larger. Because it tries to cause the $a$.

And as you see, it tries to but simply can't apply enough force to cause that acceleration. The necessary force in the lower string is more than the strength of the string, so it breaks.

Answer 2

While I haven't seen the video, the description matches an old science trick using inertia: if you want the top string to snap, pull slowly. To snap the bottom string, pull suddenly - the inertia of the weight will “protect" the upper string for a brief moment.

Answer 3

This has a very simple explanation when it is analysed using Failure Mechanics or the study of how(why) things break. Things don't break because of reaction forces, they break because of the internal forces due to material deflection caused by the reaction forces.

The material deflection in this case is the change in the length of each string as the force is applied. This deflection is in actuality very small, but when the bottom string is quickly yanked the bottom string exceeds its maximum deflection before it can accelerate the 2kg block to the point where the top string reaches it's maximum deflection. The inertia of the block is simply too much to overcome and the bottom string breaks before the top string has time to deflect.

When the string is pulled slowly, it allows the time required for the 2kg block to deflect the top string. In this case the system does experience the force conditions the professor drew on the board. The tension in the top string is greater than the bottom by a magnitude of the block's weight, therefore the top string deflects further than the bottom until the point of maximum deflection where failure occurs in the top string.

Answer 4

This problem can be modeled quantitatively. But, in the description provided, if we assume that the upper string is inextensible, the problem is statically indeterminate. So, to overcome this, we can replace the upper string with a massless spring. In the initial state of the system, the force balance is:$$T_{T0}=mg$$where $T_{T0}$ is the initial tension of the top string. If x is the additional downward displacement that the mass experiences after tension is applied to the lower string, the tension in the upper string is given by:$$T_T=mg+kx$$where k is the (very high) spring constant. Let F(t) be the time dependent force that that the lower string exerts on the mass, where F(t) is zero up to time t = 0. So, the force balance on the mass at times greater than t = 0 is: $$mg+F(t)-(mg+kx)=m\frac{d^2x}{dt^2}$$or$$\frac{d^2x}{dt^2}+\omega^2 x=\frac{F(t)}{m}$$where $\omega^2=k/m$.

Let us next confine attention to the case where the applied force in the lower string is constant at $F(t) = F$. The solution for this case is:$$x=\frac{F}{k}(1-\cos{\omega t})$$ So, the time-dependent tension in the upper string is:$$T_T=mg+F(1-\cos{\omega t})$$and the tension in the lower string is $$T_L=F$$

Note that the maximum tension that the upper string can attain is mg + 2F, while the tension in the lower straing is F. So, if $F>T_{crit}$ (where $T_{crit}$ is the critical static tension for a string to break), the lower string will break first. But, even if $F<T_{crit}$ (so that the lower string does not break), the upper string can still break (after a very very short time) provided:$$F>\frac{T_{crit}-mg}{2}$$Finally, if $$F<\frac{T_{crit}-mg}{2}$$ neither string can break.

Answer 5

Suppose the instantaneous tensions in the upper and lower threads are $T_1$ and $T_2$. Then the equation of motion of the heavy block is
$Mg+T_2-T_1=Ma$
so
$T_2-T_1=M(a-g)$.

If $a>g$ then $T_2>T_1$.