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Math Help - work and energy problem

  1. #1
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    work and energy problem

    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

    giving
    epe= \frac{\lambda x^2}{2a}= \frac{\lambda a} {32}

    then PE+EPE=0 by conservation of energy

    gives

    \lambda =5rt(39)w

    now this is miles from the answer so where have i gone wrong? I can get the answer by resolving,but this question was in energy chapter so i want to do it by energy method. IF this is not possible can someone explain why this problem is not doable via enrgy?
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  2. #2
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    Re: work and energy problem

    Quote Originally Posted by jiboom View Post
    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

    giving
    epe= \frac{\lambda x^2}{2a}= \frac{\lambda a} {32}

    then PE+EPE=0 by conservation of energy

    gives

    \lambda =5rt(39)w

    now this is miles from the answer so where have i gone wrong? I can get the answer by resolving,but this question was in energy chapter so i want to do it by energy method. IF this is not possible can someone explain why this problem is not doable via enrgy?
    why are you using energy concepts for this problem? Hooke's law is more appropriate.
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  3. #3
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    Re: work and energy problem

    i have used hookes law and resolving to do the question, but its on a work and energy chapter so i really need to know where my mistake is as im so far off the answer.

    My understanding of energy must be awry as i can not see where i have gone wrong or missed maybe some other force that is contributing to the energy. Or i would like to know why energy is not the way to go so i can spot when to use energy and when not
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