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Math Help - hookes law,

  1. #1
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    hookes law,

    in need of big help with this question: cant get any where with it

    a particle of weight w is attached to a point c of an unstretched elastic string AB where AC=4a/3 CB=4a/7. the ends A and B are then attached to the extremeties of a horizontal diameter of a fixed hemispherical bowl of radius a and the particle rests on the smooth inner surface, the angle BAC being 30, show that the modulus of elasticity of the string is W and the reaction of the bowl on the particle is mg
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  2. #2
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    Re: hookes law,

    have gotten a bit closer now, i post my try in the hope someone can spot an error or finish the problem

    i have joined C to O, centre of hemisphere so A0=0B=OC=a. this gives some isoc trinagle which lead to ACB=90,ABC=60.

    resoloving vertically

    Tsin 30+Tsin 60=w
    T=2W/(rt(3)+1)

    now
    sin 60=AC/2a so AC=art(3) hence extension is art(3)-4a/3=a(21rt(3)-28)/21

    cos 60=BC/2a so BC=a hence extension is a-4a/7=a(21-12)/21=9a/21

    giving total extension a[21rt(3)-28+9]/21=a[rt(3)-1]

    hence using

    \lambda= \frac{Ta}{x}
    \lambda= \frac{2W [29a]}{21[rt(3)+1][a[rt(3)-1]]}

    = \frac {29W}{21}

    which is close but no cigar. How can i rid myself of the 29/21?
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  3. #3
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    Re: hookes law,

    Quote Originally Posted by jiboom View Post
    have gotten a bit closer now, i post my try in the hope someone can spot an error or finish the problem

    i have joined C to O, centre of hemisphere so A0=0B=OC=a. this gives some isoc trinagle which lead to ACB=90,ABC=60.

    resoloving vertically

    Tsin 30+Tsin 60=w
    T=2W/(rt(3)+1)

    now
    sin 60=AC/2a so AC=art(3) hence extension is art(3)-4a/3=a(21rt(3)-28)/21

    cos 60=BC/2a so BC=a hence extension is a-4a/7=a(21-12)/21=9a/21

    giving total extension a[21rt(3)-28+9]/21=a[rt(3)-1]

    hence using

    \lambda= \frac{Ta}{x}
    \lambda= \frac{2W [29a]}{21[rt(3)+1][a[rt(3)-1]]}

    = \frac {29W}{21}

    which is close but no cigar. How can i rid myself of the 29/21?
    oh dear,that post is so wrong! there are 2 glaring numerical errors, one borne out of sheer desperation to get an anwser (the 9-28=21 bit)

    anyway i now think i have a solution, if someone can be so kind to confirm what i have done is correct.

    I have now said that the tensions are different either side of the hanging weight,letting t1 be tension from A to c t2 tension from b to c
    then i have not resolved vertically,but perpendicular to the reaction force to get

    w/2+t1sin 30-t2sin60=0

    then using lambda= T1(natural length of AC)/(extension of AC)
    and lambada=t2(natural length of bc)/(extension of bc)

    i can find lambda

    is this ok to treat either side of the weight as seperate strings with different tensions,and using not the natural length of the whole string but just the parts AC and BC?




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