Why does the string break?

At the end of this lesson. http://mfile.akamai.com/7870/rm/mitstorage.download.akamai.com/7870/8/8.01/f99/videolectures/wl99lec6-300k.rm
Why does the lower string break, and then the upper and finally the lower one again?
Has it anything to do with the speed he is pulling with?

Look at the picture below. The acceleration of the following systems are different (1 =1.4 m/s^2 and 2 = 3.27m/s^2). Is this any related to the result with the string?
Also, why does it give different answers for the two systems in the picture?

Cheers
 

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Doc Al

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I am unable to view the lesson in the link you provided, so I can't comment on it. (Why not just describe it?)

danielI said:
Also, why does it give different answers for the two systems in the picture?
The string tension is different for each system.

In the left picture, a 200 kg mass hangs from the string. What's the tension in the string? (Hint: It's not equal to the weight of the mass!)
 
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Great video lecture.
Anyway, the answer involves inertia.

When pulled fast, the lower string breaks because it cannot move the 2kg mass downwards as fast as the downward extension of the string. Thus, something must "give", in this case the string.

However, when pulled slowly, the 2kg mass is pulled about as fast as the lower string is being pulled, but the top string is being critically extended.
 
Doc Al, My first theory regarding the picture was that 1 will accelerate but not system 2 since the 150kg cylinder is being pulled with a constan force. But this is wrong.

I know how to calculate system 1.

T - 150*9,81 = 150a
200*9,81 - T = 200a
+
=> a = 1.401m/s^2 => T = 1681,65N

And the second one ... Oh, I just came up with the answer as I speak (write).:rofl:
T = 200*9,81.

pallidin,
My first guess was that when pulling fast his hand is accelerating, while it isn't (almost) accelerating when pulling the thing slowly. So, am I correct to say that if he accelerates faster that g the lower one will break (because of inertia). Did I understand you?

Cheers!
 

Doc Al

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danielI said:
Doc Al, My first theory regarding the picture was that 1 will accelerate but not system 2 since the 150kg cylinder is being pulled with a constan force. But this is wrong.
Why in the world would you think that just because the 150 kg mass is being pulled with a constant force that it wouldn't accelerate? (By the way, in both cases the 150 kg mass is being pulled with a constant force. :wink: )

I know how to calculate system 1.

T - 150*9,81 = 150a
200*9,81 - T = 200a
+
=> a = 1.401m/s^2 => T = 1681,65N

And the second one ... Oh, I just came up with the answer as I speak (write).:rofl:
T = 200*9,81.
Looks like you figured it out.

pallidin,
My first guess was that when pulling fast his hand is accelerating, while it isn't (almost) accelerating when pulling the thing slowly. So, am I correct to say that if he accelerates faster that g the lower one will break (because of inertia). Did I understand you?
The acceleration due to gravity has nothing to do with this.

By pallidin's description, I understand that this is the standard demo of a mass hanging from a string with a second string hanging down from the mass. Here's another way to look at it. (Just another way of saying what pallidin already explained.) When you pull the string slowly, the mass has plenty of time to move and stretch the top string: the tension in the top string equals the tension in the bottom string plus the weight of the mass. So the top string reaches its breaking point first.

But when you pull the string sharply, it will stretch to its breaking point long before the mass can move enough to stretch the top string (due to the mass's inertia).
 
The reason I thought so was because the 200kg mass accelerates and so would the 150kg mass.
While the 200*9,81 force in system 2 doesn't do that. Ofcourse there is some logical error with my reasoning.

Also, here are all the videos.
http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm [Broken]

The lectures are good for self-study and fun to watch. Like at the end of lecture 8 (Friction) :rofl:
 
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