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Trouble understanding energy concepts

  1. Sep 4, 2011 #1
    a particle of weight W is attached by 2 light inextensible strings each of length a to 2 fixed points,A and B, distant a apart in a horizontal line. write down tension in either string.

    one of the strings is now replaced by an elastic string of the same natural length and it is found that in the new position of equilibrium this string has stretched to a length 5a/4. prove modulus of elasticity of this string is 7W/(rt(39)

    first part is done.

    for second part:

    i have triangle with sides a,a,5a/4
    let fixed points be A and B and weight at C so AB=BC=a AC=5a/4

    using cos rule i find cos CAB =5/8 so
    sin CAB= rt(39)/8
    define PE=0 at the line AB so here pe=epe=ke=0

    at equilibrium

    h/(5a/4)=sin CAB

    so h=5a(rt{39})/32}
    hence PE=-(W5a(rt(39))/32

    natural length of string =a
    at equilibrium length =5a/4 so extension= a/4

    epe=[itex]\lambda[/itex]x^2/2a = [itex]\lambda[/itex]a/32

    then PE+EPE=0 by conservation of energy



    as can be seen, this is nowhere near the answer. Now i can do this by reolving foces but the question is from an energy work chapter so as i cant do it i have not understood energy very well. Can anyone explain my errror? Have i missed work done, energy, or other forces?
    If the problem is too messy to do by energy can someone explain how they know this so in future i will be able to spot when to stick with resolving.

    this is not a one off question as im also having problems with the following too. i wont post all my working, but i am miles away from the answer here too. any hints with this will be appreicated.

    BAC is a rigid fixed rough wire and BAC is 60 degrees,. P and Q are 2 identical rings of mass m connected by a light elastic string of naturla length 2a and modulus of elasticity mg. If P and q are in equilibrium when PA=AQ=3a and the perp to PQ from A splits BAC is half,find the least coefficient of friction between the rings and the wire.
    in the diagram they have wire in shape of a triangle with base BC (BC not joined) and apex A

    P is 3a down from A along AB Q is 3a down from A along AC

    here i have let PE=0 at the top (point A) so EPE+PE=0

    worked out h at equilibrium point

    worked out PE and EPE at equi point.

    work done by friciton +(PE+EPE at equilibrium)=(PE+EPE at top)=0

    from work done =forcexdistance i can work the friction force

    then F=[itex]\mu[/itex]R should give the answer but NO :(
  2. jcsd
  3. Sep 5, 2011 #2


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    I don't know where your formulas are coming from, but since the extension of the string is given, this problem is best solved by resolving forces with the system in its equuilibriun possition, then solving for E using the axial deformation equation.. which requires the area of the string, which is not given, so either you or the problem forgot to state it. Energy methods are best used when calculating joint dispacements.
  4. Sep 5, 2011 #3
    my formulae for sin and cos are correct, since using them when resolving gives me the correct answer.

    my h is just the height taken from the line AB to the equilibrium point, which i can calculate using the sin and cos.
    this gives me the PE at the equilibrium point

    for the epe i know the modulus [itex]\lambda[/itex]

    natural length = a

    extension = stretched- natual=5a/4-a=a/4

    so epe= [itex]\lambda[/itex](a/4)^2/2a
  5. Sep 5, 2011 #4


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    You are running into problems using energy methods. You calculate the final PE at the equilibrium point with reference to AB (which should be negative), , but if you choose AB as your zero reference, then the initial PE of the hanging weight is not zero, since it hangs well below AB, for what it is worth. And you say you know the modulus λ... ,is that E?.....which you are trying to solve, and which cannot be solved unless the cross section area of the wire is given as A. Now when you say epe, i assume you mean internal strain energy? The problem can be solved using energy methods, but you must be careful about how you approach it that way.
  6. Sep 5, 2011 #5
    I see the concept of energy as simply describing force macrophysically. Make sense?
  7. Sep 6, 2011 #6
    do i not have a choice where i start? im choosing to hold the weight so it is in line with AB.
    As AB is a and the strings are of length a each if i start with the weight at a/2 from point A then both strings will be slack so no EPE (elastic potential energy). then i release and then find the height from AB to the point of equilibrium.

    A similar question the book had the same choice, but it was a weight a the midpoint of 1 string. they defined the point of first instantaneos rest as PE=0 level then worked out the height from the top where the weight had fallen from.

    My lambda is the modulus of elasticity
  8. Sep 7, 2011 #7


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    I don't believe the problem assumed that the particle was released from AB, since if it were, the string would stretch a distance greater than a/4, and oscillate (assuming no dampening) about the at rest equilibrium position where the stretch is a/4. and at such point, the acceleration would be zero but the velocity would be non-zero. Instead, you must assume that the weight is slowly lowered (by holding it), until the systen reaches its equilibrium position. If you insist on using energy methods, you should use work-energy principles involving changes in internal energy. I don't recomend such an approach here.
  9. Sep 8, 2011 #8
    your post maybe highlights my problem here. Im assuming, when i see such questions, that when im told a particle is at equilibrium point, it means the particle was attached and released then does the oscilliations unitl it come to a stop and this point where it stops is the equilibrium point.

    so in the above question this allows me to use conservation becasue no other forces act so at eaqch point of the motion

    PE+KE+EPE=0 (as i have PE+KE+EPE=0 at line AB)

    is this not how equilibrium points work and why the other question i alluded to asked about first position of instantaneous rest? does energy get "Lost" by the system eac time it reaches the bottom of an oscillation and begins to go up again?
  10. Sep 8, 2011 #9


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    Yes, energy gets lost through damping (internal or external losses due to friction and other forces), and as such, you cannot use conservation of energy, because mechanical energy is not conserved. Likewise, if you assume the equilibrium point to be the at rest position, you need to slowly lower the particle with your hands to that position, applying a nonconservative force, so conservation of mechanical energy fails you again. So you either have to solve this problem using force resolution, or using, the harder way in this example, work-energy methods.
  11. Sep 11, 2011 #10
    i get what you are sayiong about energy loss, but where is it lost in my problem?

    i think i totally misunderstood equilibrium point,as i thought that was a point of no motion. I see now that although my weight is at equilibrium it could still have velocity and so KE.

    can you maybe eplain how my weight can oscillate about the equilibirum point?
    I cant see how it can do this motion. AM i correct in thinking at the top of an oscillation V=0 but we will have accelertation and so the big difference in the question asking about point of first rest.
  12. Sep 11, 2011 #11


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    If you are talking about the original problem, it makes no mention of dropping the particle from rest or asking for the 'first instant of rest'. Let's not confuse this problem with a differnt problem, such as attachimg a weight to a spring and dropping it from rest to calculate its displacement at the instant of first rest, where its dis[lacemnt is twice it's dis[placemnt at equilibrium..
    A particle can be in equilibrium when it is totally at rest, moving in a straight line at constant speed, or instantaneously at a point of no acceleration but still at some speed. In all cases, there is no net force acting on the particle, and there is no acceleration of the particle at the equilibrium point.[/quote]
    can you maybe eplain how my weight can oscillate about the equilibirum point?[/quote] It can oscillate if it is dropped or forced. If it is allowed to slowly come to its at reast position by lowering it slowly with your hand or other device, until it doesn't move anymor, it will not oscuillate. See "Harmonic Motion" or "Damped Harmonic Motion".
    Yes. At the top and bottom of the oscillatory motion, the speed is 0, but there is max acceleartion. At the equilibrium point, v is max but a is 0 (dv/dt = 0). But this problem has nothing to do with oscilations.
  13. Sep 15, 2011 #12
    this is true, but i was trying to invoke energy methods by assuming it was at the top line then dropped. i can see why now i cant use conservation in ths case as it will have KE, but you mention positioning it at equilibrium point from the AB line. I still cant use conservation here so where is energy lost? you mention internal energy changes, what are they?
  14. Sep 17, 2011 #13


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    you might want to consider a simpler example of an extensible string hung from the ceiling. The string, being elastic, can be considered as spring with a spring constant k. Now attach a weight, mg, to it, and hold it there so it does not move and the string does not stretch. Now slowly lower it with your hand, palm up, until the weight comes to a stop, and release your hand. The weight is now in its 'at rest' (v = 0) equilibrium position. The force in the string is mg, per Newton 1, and the stretch (displacement) of the string is mg/k, per Hooke's Law.

    Now same setup, using work energy methods, that is

    Wnc + initial PEstring + initial PEgravity + initial KE = final PEstring + final PEgravity + final KE.

    Here, Wnc is the work done by your hand while lowering the weight, and represents the 'lost' energy of the system. Since the initial force applied by your hand is mg, and after you lower the weight to its equilibrium position the force you apply at that point is 0, then the average force is mg/2, and the work done is mgx/2. Substituting into the above equation, then

    -mgx/2 + 0 + 0 + 0 = 1/2kx^2 - mgx + 0.

    Solving, x = mg/k...the same result as above...the hard way.

    Note that the strain energy (internal energy) in the elastic string is 1/2kx^2. Where k is AE/L.
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