Physics Help Forum Simple Tension Problem

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 Jul 28th 2010, 04:22 PM #1 Junior Member   Join Date: Jul 2010 Posts: 3 Simple Tension Problem First off, thank you for taking the time to read this post. I'm in between semesters right now so I'm trying to reinforce my knowledge by watching class lectures on MIT.edu. At the end of a particular lecture for basic physics the teacher did an experiment. The experiment consists of a mass on a string with another string hanging below the mass. The professor then exerts a force on the bottom string, by pulling, until one of the strings break. He did this experiment three times, twice the bottom string broke and once the top. This was at the end of class and he asked his students to think about this problem after class, never giving the reason. I have attempted to figure this out using logic and what I know about newton's laws, etc. Any way I look at it, the tension in the top string has to be greater then the tension in the bottom string by the factor mg of the hanging mass. I noticed when the bottom string broke he pulled quickly on the bottom string and when the top string broke he pulled slow. So is this a problem involving impulse? Logically it seems like time wouldn't be a factor because the forces exerted on both strings should be instantaneous for the most part. I'm very confused.
 Jul 29th 2010, 01:49 AM #2 Junior Member   Join Date: Jul 2010 Posts: 3 YouTube - Lec 6 | 8.01 Physics I: Classical Mechanics, Fall 1999 The experiment starts at about 42:00.
 Jul 29th 2010, 07:10 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 eloiamb: your logic is correct, for a static system. Under static conditions the tension in the upper string (call it T1) is greater than the tension in the lower string (T2) by mg. Draw a fee body diagram and under equilibrium you have: T1 = T2 + mg That's why when he pulled slowly the upper string was the one that broke firts. When I say "pulling slowly," and "equilibrium" what I mean is we can ignore any acceleration of the mass. The strings may stretch before breaking, which means there is a displacement of the mass downward, but we ignore the acceleration of the mass during that stretching. But if you pull the lower string quickly, the acceleration of the mass must be considered. Draw a free body diagram and you get: ma = T2 + mg -T1 Rearrange: T1 = T2 + mg - ma So you see that the acceleration of the mass downward actually reduces the tension in the upper string. Hence the bottom string may indeed break first.
 Jul 29th 2010, 11:32 AM #4 Junior Member   Join Date: Jul 2010 Posts: 3 Wow, that is not intuitive at all but very intriguing. I have got to get it through my head that I cannot logically solve every physics problem. Thank you so much ChipB.
Jul 29th 2010, 12:19 PM   #5
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Join Date: Jun 2010
Location: Morristown, NJ USA
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 Originally Posted by eloiamb Wow, that is not intuitive at all but very intriguing. I have got to get it through my head that I cannot logically solve every physics problem. Thank you so much ChipB.
You're welcome! In thinking about his some more here's another explanatoin, which is essentailly the same as before but is perhaps expalined in a more "intuitive" way (no formulas ths time):

Suppose the strings are such that they break when stretched a certain amount - say 1 inch. In other words, as long as the amount of stretch in a string is less than 1 inch it doesn't break. Further, suppose the mass has enough weight to stretch the upper string 1/2 inch. In the static case, pulling down on the lower string with enough force to stretch it 1/2 inch will cause the upper string to be stretched an additional 1/2 inch, which means the mass has moved 1/2 inch downward. The total stretch of the upper string is now 1 inch, and the upper string breaks. In essence, as far as the upper string is concerned it breaks because the mass moved downward that additional 1/2 inch. The upper string doesn't care why the mass moved - the fact that it moved is what the upper string is concerned with.

But in the case where you pull really fast, the mass doesn't have enough time to move 1/2 inch. The inertia of the mass causes a time delay from the time you tug on the string until it can mive that far. So the upper string won't feeling enough extra tension to break for perhaps a second or so. Consequently the bottom string can be pulled past its breaking point before the mass moves 1/2 inch, and so the lower string breaks. In effect the mass buffers the tension for a split second.

A real world example of this phenomenon is what happens with the suspension of your car. You have a spring that connects the chassis to the hub and wheel, the hub and wheel have mass, and then the rubber of the tire acts like another spring between the road and the wheel. Under the static condition the tire and the spring each "feel" the full weight of the car and are compressed accordingly. But when you're driving and hit a pothole the tire takes a much greater shock from the impulse than does the spring, because the mass of the hub & wheel prevents the transfer of force from the tire to the spring. The result is the tire takes a huge jolt, and you have a blow out. So when you're driving and feel a bump from hitting a pot hole, remember that the tire has just felt a much greater bump than you did.

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