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Homework Help: Elastic Spring

  1. Jan 26, 2005 #1
    Hi There,
    Iam having trouble with this question:

    A block B of mass 5 kg is fastened to one end of each of two springs. The other ends of the springs are attached to fixed points A and C, 4 metres apart on a smooth horizontal surface, as shown in the diagram <the link contains the file>.

    http://s3.yousendit.com/d.aspx?id=2ZAI1AKPYV05L1MUR0PJY6RRMP

    Spring AB has natural length 2 metres and modulus of elasticity 30 N, while BC has natural length 1 metre and modulus 40 N (you may assume that the springs meet at the centre of B).

    (i)Find the length of AB when B is in its natural equilibrium position.

    now if extension = length of the spring - natural length
    Iam i required to find out the length of the spring?
    If i am then using hooke's law T=kx, 5g=30x/2 i get x= 3.26 as the extension.
    :confused:
    I would appreciate some help
    thanks in advance
     
  2. jcsd
  3. Jan 26, 2005 #2

    quasar987

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    Not an easy one. There are a few things to think about. Here's how I would proceed.

    1° The total system, while stretched, measures 4 m, but the relaxed lenght of AB is 2m, and the relaxed lenght of BC is 1m.. SOoo if we call [itex]\Delta l_{AB}[/itex] and [itex]\Delta l_{BC}[/itex] the lenght by which each spring is streched, we get the equation

    [tex]4 = 2 + 1 + \Delta l_{AB} - \Delta l_{BC}[/tex]

    (Why [itex]- \Delta l_{BC}[/itex] ? Because the [itex]\Delta l[/itex]s represent stretching in two oposite directions, so [itex]\Delta l_{BC}[/itex] itself will be negative. Thefor, if we want the sum to add up to 4, we must substract [itex]\Delta l_{BC}[/itex], not add it.)

    Or, more simply,

    [tex]\Delta l_{BC} = \Delta l_{AB} - 1 \ \ (*)[/tex]


    2° In the case we're interested in, the mass B is in equilibrium position. This is fancy words to say that [itex]\Sigma F = 0[/itex], so we also have the equation

    [tex]0 = -k_{AB}x_{AB} - k_{BC}x_{BC}[/tex]

    Or, since at equilibrium position, these x represent what we have called [itex]\Delta l_{AB}[/itex] and [itex]\Delta l_{BC}[/itex], we get, making use of relation (*):

    [tex]0= -30\Delta l_{AB} - 40(\Delta l_{AB} - 1)[/tex]

    Allowing you to find what you were looking for.


    Side note: We could also have supposed [itex]\Delta l_{AB}[/itex] and [itex]\Delta l_{BC}[/itex] to be both positive, but it would have made a mess because [itex]\Delta l_{AB}[/itex] and [itex]\Delta l_{BC}[/itex] still represent stretching in two oposite directions: towards the right for AB, and towards the left for BC. So, this means that we would be considering a stretching of the spring AB to the right as positive, left as negative. But also considering a stretching of spring BC to the left as positive, right as negative! The consequence of this subtelty would be apparent in the equation of motion [tex]0 = -k_{AB}x_{AB} + k_{BC}x_{BC}[/tex] where we would have taken opposite signs for the force of the different springs. Again, that is because our setting [itex]\Delta l_{AB}[/itex] and [itex]\Delta l_{BC}[/itex] as positive implied that displacement to the right of the mass means a positive [itex]x_{AB}[/itex] but a negative [itex]x_{BC}[/itex]. But the forces by both springs are not opposite; they are in the same direction. Hence we must adjust the sign of k.
     
  4. Jan 26, 2005 #3
    cheers thx for sharing your ideas
    as a sidenote should the -30 be -15 if k= lambda/l
     
  5. Jan 26, 2005 #4

    quasar987

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    what are lambda and l?
     
  6. Jan 26, 2005 #5
    l is the natural length
    lambda is modulus of elasticity
     
  7. Jan 26, 2005 #6

    quasar987

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    I don't really know what the modulus of elasticty is. I thought it was the same as the spring constant k.

    If you know what the relation between the modulus of elasticity and the spring constant is, then there shoudln't be a problem.

    Out of curiosity, what is the relation between the modulus of elasticity and the spring constant ?

    Thanks
     
  8. Jan 26, 2005 #7
    spring constant depends on the length of the spring, if the 2 springs made by the same material, the longer one has a smaller spring constant

    the elasticity modulus is the property of a material, it doesn't not depend on its physical size...
     
  9. Jan 26, 2005 #8

    quasar987

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    How are they related mathematically in temrs of k, lambda and relaxed lenght l ?
     
  10. Jan 26, 2005 #9
    k=lambda/l
     
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