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Springs and strings

  1. Jan 9, 2012 #1
    In this question, doesn't the string pull on both A (downwards) and B (upwards)? So wouldn't the spring be doubly compressed? Does it matter whether or not B is fixed?
    Also I don't understand the line: "The spring force is greater than the gravitataional force of A. Thus this spring will not compress more." Shouldn't the spring be further compressed by gravity? Perhaps not because gravity also acts on the spring so both move downwards?
    Thanks! :)
     

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  3. Jan 9, 2012 #2

    Delphi51

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    The 20 N has nothing to do with the force on the second spring. Think of the two masses tied together as a single mass whose weight presses against the spring.
    (considering the 20 N is like considering interatomic forces among the molecules of the masses - they just don't matter)
    Your calc for the 0.49 cm is correct. But I have no idea why you multiply it by 2 in the last step.
     
  4. Jan 9, 2012 #3
    Hi Delphi!
    The writing in the thumbnail image is the official solution. I don't understand the compression between A and B (2 cm). Why doesn't the spring get further compressed by the weight of box A? Doesn't the string pull on both A and B?
    Thanks!

    P.S. Post #123! :)
     
  5. Jan 9, 2012 #4

    I like Serena

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    Hey sparkle123! Congrats on the 123! :smile:

    A and B are bound together by a massless string forcing the spring to give 20 N in force.
    That means the spring is compressed by (20 N)/(1000 N/m) = 2 cm.

    It also means that A is pushed up by the spring with 20 N.
    Since the weight of A is only 5 N, the weight of A only reduces the pull of the string by 5 N, so the string pulls by 15 N, keeping blocks A and B in equilibrium without changing their distance.

    Yes, the string pulls on both A and B, but only in the sense that action=-reaction.
    These forces are equal and opposite so the net force is zero.
    It has to be, otherwise the blocks or the string would start moving.
    Just like the spring pushes both on block A and block B equally and in opposite directions.
     
  6. Jan 9, 2012 #5
    Thanks I like Serena! That makes a lot of sense! Questions that remain are:
    Does the string pull on A by 15 N and pull on B by 20 N then?
     
  7. Jan 9, 2012 #6

    Delphi51

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    Sorry, Sparkle, I bungled it! I didn't notice that we were asked for the change in distance including A to B (2 cm). And I didn't see the plus sign in that final line. Of course the total distance is the 2 cm from A to B plus the 0.49 cm from B to C.
    Yes.
     
  8. Jan 9, 2012 #7

    I like Serena

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    No (sorry Delphi).
    In a massless ideal string the tensional force in the string is equal everywhere (between the points where it is attached).
    The force on the string by A has to cancel the force on the string by B, otherwise the string itself would not be in balance.

    So the string pulls on B by 15 N.
    The upper spring pushes B down by 20 N.
    The weight of B weighs it down by 5 N.
    The bottom spring pushes B up by 10 N.
    So all forces on B sum up to 0! Hurrah!
     
  9. Jan 9, 2012 #8
    Thanks I like Serena and Delphi51! Both of your helps are much appreciated!
     
  10. Jan 9, 2012 #9

    Delphi51

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    Thanks for catching that, ILS. Once again I missed seeing the obvious - I read 20 on each end. There must be something wrong with my eyes tonight. Glad you are here.
     
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