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GR0177 #72 |
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Problem
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This problem is still being typed. |
Mechanics }Forces
Before the string is cut, one has the basic static equilibrium condition, for each block  and  , where  refers to the upper block and  refers to the bottom block. Adding, one has  a = 0 = T-m_ug - m_b) .
After the string is cut, the tension goes to 0, but one has a non-zero net acceleration. The top mass has  , where the minus sign comes in since the question wants the downward acceleration. The bottom mass has  . But, immediately after the string is cut, the lower mass has 0 acceleration. Thus,  . Plugging this into the equation for the upper mass, one finds that  , as in choice (E).
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Alternate Solutions |
travis.nicholson
2006-10-27 13:24:38 | Try solving the problem like this:
 for the lower mass before the string breaks.
Therefore,  .
 for the upper mass after the spring breaks.
Using the fact that ,
Therefore, the downward acceleration is  |  | erc
2005-11-10 13:44:38 | Could one also do this as follows:
g)
ie gravity acting on masses, neglecting the mass of the spring.
Since the masses are identical, this is just
Then the force acting on the top mass is 
and this balances the tension, thus
and thus
 .
travis.nicholson
2006-10-27 13:06:21 |
When you say "Then the force acting on the top mass is  ," I assume that you began this sentence with "Then ..." in order to signify that you are talking about what happens after the string snaps. (My assumption arose from the fact that you have added a non-zero acceleration to your force equation.)
However, after the string snaps, it is not true that  because there is no tension after the string breaks.
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Comments |
StrangeQuark
2007-05-17 10:54:01 | I have always been bad at dealing with forces. Energy is my thing, so I did this problem using the Lagrangian Formalisium, I find this way I am less likely to drop minus signs or make these kinds of mistakes;
-(-mq_1g-mq_2g+\frac{1}{2}(q_2-q_1)^2k)
Now recall
==mg)
thus
==0)
lmontanari
2008-09-14 20:50:45 |
Shouldn't the T term in the Lagrangian be (1/2 m(dq/dx)^2 + 1/2 m(dq/dx)^2).
In which case, the solution whould be g, not 2g.
I don't think this is the correct value for the energy of this system.
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|  | travis.nicholson
2006-10-27 13:24:38 | Try solving the problem like this:
for the lower mass before the string breaks.
Therefore, .
for the upper mass after the spring breaks.
Using the fact that ,
Therefore, the downward acceleration is
jonny23
2007-06-14 12:15:18 |
suka
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jesford
2008-04-05 11:46:52 |
nice!
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| | travis.nicholson
2006-10-27 13:16:55 | The error in your reasoning occurs in the sentence "But, immediately after the string is cut, the lower mass has 0 acceleration. Thus,  ."
You have assumed that the lower mass has a acceleration of zero and concluded that . This assumption is not generally true (if the masses are different), nor, in my opinion, is it intuitively obvious enough (in the case of equal masses) to merely be stated.
However, it can be shown by analysis of in its equilibrium state that , and therefore the lower mass has an acceleration of 0 (but this reasoning cannot go in the other direction). | | erc
2005-11-10 13:44:38 | Could one also do this as follows:
ie gravity acting on masses, neglecting the mass of the spring.
Since the masses are identical, this is just
Then the force acting on the top mass is
and this balances the tension, thus
and thus
.
erc
2005-11-10 13:47:28 |
Ah...sorry that comment came out such a mess... evidently not all latex commands are allowed. \rightarrow should give an "implies" sign.
Apologies.
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yosun
2005-11-10 14:05:36 |
erc: all latex commands are allowed (math commands, that is). you just have to properly surround your equations with dollar-signs (one to start it and one to end it) or else the parser gets confused about which equations you want parsed. so  all work... among other things. i have manually corrected your post.
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travis.nicholson
2006-10-27 13:06:21 |
When you say "Then the force acting on the top mass is ," I assume that you began this sentence with "Then ..." in order to signify that you are talking about what happens after the string snaps. (My assumption arose from the fact that you have added a non-zero acceleration to your force equation.)
However, after the string snaps, it is not true that because there is no tension after the string breaks.
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