Calculating extensions in elastic string

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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1
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Could anyone please help me with the following question?



enter image description here



Here's my failed line of reasoning so far:



My diagram:



enter image description here



You can see that I am assuming that P moves down to lie vertically under it's original position, is even this correct?



Otherwise:



I've tried a few ways, here's the most naive, thanks for any help:



Let extension in AP = $x_1$ and extension in PB be $x_2$ then:



$L_1=frac320,cos,theta$



and so



$x_1=frac320,cos,theta-frac320$



and



$L_2=frac120,sin,theta$



and so



$x_2=frac120,sin,theta-frac120$



Hence:



$fracx_1x_2=fracfrac60-60,cos,theta400,cos,thetafrac20-20,sin,theta400,sin,theta=frac60-60,cos,thetacosthetatimesfracsin,theta20-20,sin,theta$



which can be simplified to:



$frac(3-3,cos,theta),sin,theta(1-sin,theta),cos,theta$



Which isn't the required answer. Thanks for any help.




classical-mechanics




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  • It could be that your assumption that the new point is exactly below the old point, is not right. The force equilibrium in the horizontal tells that the forces have to be equal on both sides, which leads to that the (absolute) elongation has to be equal on both sides.
    – Matti P.
    Jul 16 at 11:58














up vote
1
down vote

favorite












Could anyone please help me with the following question?



enter image description here



Here's my failed line of reasoning so far:



My diagram:



enter image description here



You can see that I am assuming that P moves down to lie vertically under it's original position, is even this correct?



Otherwise:



I've tried a few ways, here's the most naive, thanks for any help:



Let extension in AP = $x_1$ and extension in PB be $x_2$ then:



$L_1=frac320,cos,theta$



and so



$x_1=frac320,cos,theta-frac320$



and



$L_2=frac120,sin,theta$



and so



$x_2=frac120,sin,theta-frac120$



Hence:



$fracx_1x_2=fracfrac60-60,cos,theta400,cos,thetafrac20-20,sin,theta400,sin,theta=frac60-60,cos,thetacosthetatimesfracsin,theta20-20,sin,theta$



which can be simplified to:



$frac(3-3,cos,theta),sin,theta(1-sin,theta),cos,theta$



Which isn't the required answer. Thanks for any help.




classical-mechanics




share|cite|improve this question



















  • It could be that your assumption that the new point is exactly below the old point, is not right. The force equilibrium in the horizontal tells that the forces have to be equal on both sides, which leads to that the (absolute) elongation has to be equal on both sides.
    – Matti P.
    Jul 16 at 11:58












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Could anyone please help me with the following question?



enter image description here



Here's my failed line of reasoning so far:



My diagram:



enter image description here



You can see that I am assuming that P moves down to lie vertically under it's original position, is even this correct?



Otherwise:



I've tried a few ways, here's the most naive, thanks for any help:



Let extension in AP = $x_1$ and extension in PB be $x_2$ then:



$L_1=frac320,cos,theta$



and so



$x_1=frac320,cos,theta-frac320$



and



$L_2=frac120,sin,theta$



and so



$x_2=frac120,sin,theta-frac120$



Hence:



$fracx_1x_2=fracfrac60-60,cos,theta400,cos,thetafrac20-20,sin,theta400,sin,theta=frac60-60,cos,thetacosthetatimesfracsin,theta20-20,sin,theta$



which can be simplified to:



$frac(3-3,cos,theta),sin,theta(1-sin,theta),cos,theta$



Which isn't the required answer. Thanks for any help.




classical-mechanics




share|cite|improve this question











Could anyone please help me with the following question?



enter image description here



Here's my failed line of reasoning so far:



My diagram:



enter image description here



You can see that I am assuming that P moves down to lie vertically under it's original position, is even this correct?



Otherwise:



I've tried a few ways, here's the most naive, thanks for any help:



Let extension in AP = $x_1$ and extension in PB be $x_2$ then:



$L_1=frac320,cos,theta$



and so



$x_1=frac320,cos,theta-frac320$



and



$L_2=frac120,sin,theta$



and so



$x_2=frac120,sin,theta-frac120$



Hence:



$fracx_1x_2=fracfrac60-60,cos,theta400,cos,thetafrac20-20,sin,theta400,sin,theta=frac60-60,cos,thetacosthetatimesfracsin,theta20-20,sin,theta$



which can be simplified to:



$frac(3-3,cos,theta),sin,theta(1-sin,theta),cos,theta$



Which isn't the required answer. Thanks for any help.





classical-mechanics





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 10:51









gnitsuk

525




525











  • It could be that your assumption that the new point is exactly below the old point, is not right. The force equilibrium in the horizontal tells that the forces have to be equal on both sides, which leads to that the (absolute) elongation has to be equal on both sides.
    – Matti P.
    Jul 16 at 11:58
















  • It could be that your assumption that the new point is exactly below the old point, is not right. The force equilibrium in the horizontal tells that the forces have to be equal on both sides, which leads to that the (absolute) elongation has to be equal on both sides.
    – Matti P.
    Jul 16 at 11:58















It could be that your assumption that the new point is exactly below the old point, is not right. The force equilibrium in the horizontal tells that the forces have to be equal on both sides, which leads to that the (absolute) elongation has to be equal on both sides.
– Matti P.
Jul 16 at 11:58




It could be that your assumption that the new point is exactly below the old point, is not right. The force equilibrium in the horizontal tells that the forces have to be equal on both sides, which leads to that the (absolute) elongation has to be equal on both sides.
– Matti P.
Jul 16 at 11:58










2 Answers
2






active

oldest

votes

















up vote
0
down vote













Well, here's my other attempt which does not assume that P moves down vertically:



I'll be using Hooke's law $T=fraclambda xa$



Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.



$h=L_2,cos,theta=L_1,sin,theta$



and so



$L_1=fracL_2,cos,thetasin,theta$ - call this "Equation A"



$L_2=fracL_1,sin,thetacos,theta$ - call this "Equation B"



Resolving forces horizontally:



$T_1,cos,theta-T_2,sin,theta=0$ - Call this "Equation 1"



and Hooke's law gives us:



$T_1=fraclambda(L_1-frac320)frac320$



and



$T_2=fraclambda(L_2-frac120)frac120$



And so Eqaution 1 gives us:



$frac203lambda(L_1-frac320),cos,theta-20lambda(L_2-frac120),sin,theta=0$



Substituting fot $L_1$ from Equation A gives:



$lambda(frac203timesfracL_2,cos,thetasin,theta-1),cos,theta=lambda(20L_2-1),sin,theta$



Which rearranges to:



$L_2=frac3,sintheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



Similarly substituting for $L_2$ from Equation B leads to:



$L_1=frac3,costheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



Now we are asked to find $fracL_1 - frac320L_2-frac120$



Which is equal to:



$frac60,cos,theta(cos,theta-sin,theta)-3(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)/frac60,sintheta(cos,theta-sin,theta)-(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)$



Which simplifies to:



$frac9,sin^2theta-3,sin,theta,cos,theta3,sin,theta,cos,theta-cos^2theta$



Which is not the required answer.






share|cite|improve this answer




























    up vote
    0
    down vote













    We have



    $$
    |AP|_0 = 15\
    |BP|_0 = 5\
    r = frac_02\
    |AP| = 2rcostheta\
    |BP| = 2rsintheta
    $$



    then



    $$
    fracAPBP = frac20 cos (theta )-1520 sin (theta )-5 = frac4 cos (theta )-34 sin (theta )-1
    $$



    the solution follows as in
    Calculation of the modulus of elasticity of a stretched string



    NOTE



    with $lambda$ the elastic modulus,



    $$
    lambdaleft(fracAPAPright)costheta = lambdaleft(fracBPright)sintheta
    $$



    (In equilibrium the horizontal projection for $F_AP$ and $F_PB$ are equal) then



    $$
    fracAPBP=fracsinthetacosthetafracAP
    $$



    or as expected



    $$
    frac4 cos theta-34 sintheta-1 = 3left(fracsinthetacosthetaright)
    $$






    share|cite|improve this answer























    • Thanks very much for your work. I'll try an learn from it! Mitch.
      – gnitsuk
      Jul 18 at 9:21










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    2 Answers
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    2 Answers
    2






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    up vote
    0
    down vote













    Well, here's my other attempt which does not assume that P moves down vertically:



    I'll be using Hooke's law $T=fraclambda xa$



    Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.



    $h=L_2,cos,theta=L_1,sin,theta$



    and so



    $L_1=fracL_2,cos,thetasin,theta$ - call this "Equation A"



    $L_2=fracL_1,sin,thetacos,theta$ - call this "Equation B"



    Resolving forces horizontally:



    $T_1,cos,theta-T_2,sin,theta=0$ - Call this "Equation 1"



    and Hooke's law gives us:



    $T_1=fraclambda(L_1-frac320)frac320$



    and



    $T_2=fraclambda(L_2-frac120)frac120$



    And so Eqaution 1 gives us:



    $frac203lambda(L_1-frac320),cos,theta-20lambda(L_2-frac120),sin,theta=0$



    Substituting fot $L_1$ from Equation A gives:



    $lambda(frac203timesfracL_2,cos,thetasin,theta-1),cos,theta=lambda(20L_2-1),sin,theta$



    Which rearranges to:



    $L_2=frac3,sintheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



    Similarly substituting for $L_2$ from Equation B leads to:



    $L_1=frac3,costheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



    Now we are asked to find $fracL_1 - frac320L_2-frac120$



    Which is equal to:



    $frac60,cos,theta(cos,theta-sin,theta)-3(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)/frac60,sintheta(cos,theta-sin,theta)-(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)$



    Which simplifies to:



    $frac9,sin^2theta-3,sin,theta,cos,theta3,sin,theta,cos,theta-cos^2theta$



    Which is not the required answer.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Well, here's my other attempt which does not assume that P moves down vertically:



      I'll be using Hooke's law $T=fraclambda xa$



      Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.



      $h=L_2,cos,theta=L_1,sin,theta$



      and so



      $L_1=fracL_2,cos,thetasin,theta$ - call this "Equation A"



      $L_2=fracL_1,sin,thetacos,theta$ - call this "Equation B"



      Resolving forces horizontally:



      $T_1,cos,theta-T_2,sin,theta=0$ - Call this "Equation 1"



      and Hooke's law gives us:



      $T_1=fraclambda(L_1-frac320)frac320$



      and



      $T_2=fraclambda(L_2-frac120)frac120$



      And so Eqaution 1 gives us:



      $frac203lambda(L_1-frac320),cos,theta-20lambda(L_2-frac120),sin,theta=0$



      Substituting fot $L_1$ from Equation A gives:



      $lambda(frac203timesfracL_2,cos,thetasin,theta-1),cos,theta=lambda(20L_2-1),sin,theta$



      Which rearranges to:



      $L_2=frac3,sintheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



      Similarly substituting for $L_2$ from Equation B leads to:



      $L_1=frac3,costheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



      Now we are asked to find $fracL_1 - frac320L_2-frac120$



      Which is equal to:



      $frac60,cos,theta(cos,theta-sin,theta)-3(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)/frac60,sintheta(cos,theta-sin,theta)-(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)$



      Which simplifies to:



      $frac9,sin^2theta-3,sin,theta,cos,theta3,sin,theta,cos,theta-cos^2theta$



      Which is not the required answer.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Well, here's my other attempt which does not assume that P moves down vertically:



        I'll be using Hooke's law $T=fraclambda xa$



        Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.



        $h=L_2,cos,theta=L_1,sin,theta$



        and so



        $L_1=fracL_2,cos,thetasin,theta$ - call this "Equation A"



        $L_2=fracL_1,sin,thetacos,theta$ - call this "Equation B"



        Resolving forces horizontally:



        $T_1,cos,theta-T_2,sin,theta=0$ - Call this "Equation 1"



        and Hooke's law gives us:



        $T_1=fraclambda(L_1-frac320)frac320$



        and



        $T_2=fraclambda(L_2-frac120)frac120$



        And so Eqaution 1 gives us:



        $frac203lambda(L_1-frac320),cos,theta-20lambda(L_2-frac120),sin,theta=0$



        Substituting fot $L_1$ from Equation A gives:



        $lambda(frac203timesfracL_2,cos,thetasin,theta-1),cos,theta=lambda(20L_2-1),sin,theta$



        Which rearranges to:



        $L_2=frac3,sintheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



        Similarly substituting for $L_2$ from Equation B leads to:



        $L_1=frac3,costheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



        Now we are asked to find $fracL_1 - frac320L_2-frac120$



        Which is equal to:



        $frac60,cos,theta(cos,theta-sin,theta)-3(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)/frac60,sintheta(cos,theta-sin,theta)-(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)$



        Which simplifies to:



        $frac9,sin^2theta-3,sin,theta,cos,theta3,sin,theta,cos,theta-cos^2theta$



        Which is not the required answer.






        share|cite|improve this answer













        Well, here's my other attempt which does not assume that P moves down vertically:



        I'll be using Hooke's law $T=fraclambda xa$



        Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.



        $h=L_2,cos,theta=L_1,sin,theta$



        and so



        $L_1=fracL_2,cos,thetasin,theta$ - call this "Equation A"



        $L_2=fracL_1,sin,thetacos,theta$ - call this "Equation B"



        Resolving forces horizontally:



        $T_1,cos,theta-T_2,sin,theta=0$ - Call this "Equation 1"



        and Hooke's law gives us:



        $T_1=fraclambda(L_1-frac320)frac320$



        and



        $T_2=fraclambda(L_2-frac120)frac120$



        And so Eqaution 1 gives us:



        $frac203lambda(L_1-frac320),cos,theta-20lambda(L_2-frac120),sin,theta=0$



        Substituting fot $L_1$ from Equation A gives:



        $lambda(frac203timesfracL_2,cos,thetasin,theta-1),cos,theta=lambda(20L_2-1),sin,theta$



        Which rearranges to:



        $L_2=frac3,sintheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



        Similarly substituting for $L_2$ from Equation B leads to:



        $L_1=frac3,costheta(cos,theta-sin,theta)20,cos^2theta-60,sin^2theta$



        Now we are asked to find $fracL_1 - frac320L_2-frac120$



        Which is equal to:



        $frac60,cos,theta(cos,theta-sin,theta)-3(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)/frac60,sintheta(cos,theta-sin,theta)-(20,cos^2theta-60,sin^2theta)20(cos^2theta-60,sin^2theta)$



        Which simplifies to:



        $frac9,sin^2theta-3,sin,theta,cos,theta3,sin,theta,cos,theta-cos^2theta$



        Which is not the required answer.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 16 at 14:15









        gnitsuk

        525




        525




















            up vote
            0
            down vote













            We have



            $$
            |AP|_0 = 15\
            |BP|_0 = 5\
            r = frac_02\
            |AP| = 2rcostheta\
            |BP| = 2rsintheta
            $$



            then



            $$
            fracAPBP = frac20 cos (theta )-1520 sin (theta )-5 = frac4 cos (theta )-34 sin (theta )-1
            $$



            the solution follows as in
            Calculation of the modulus of elasticity of a stretched string



            NOTE



            with $lambda$ the elastic modulus,



            $$
            lambdaleft(fracAPAPright)costheta = lambdaleft(fracBPright)sintheta
            $$



            (In equilibrium the horizontal projection for $F_AP$ and $F_PB$ are equal) then



            $$
            fracAPBP=fracsinthetacosthetafracAP
            $$



            or as expected



            $$
            frac4 cos theta-34 sintheta-1 = 3left(fracsinthetacosthetaright)
            $$






            share|cite|improve this answer























            • Thanks very much for your work. I'll try an learn from it! Mitch.
              – gnitsuk
              Jul 18 at 9:21














            up vote
            0
            down vote













            We have



            $$
            |AP|_0 = 15\
            |BP|_0 = 5\
            r = frac_02\
            |AP| = 2rcostheta\
            |BP| = 2rsintheta
            $$



            then



            $$
            fracAPBP = frac20 cos (theta )-1520 sin (theta )-5 = frac4 cos (theta )-34 sin (theta )-1
            $$



            the solution follows as in
            Calculation of the modulus of elasticity of a stretched string



            NOTE



            with $lambda$ the elastic modulus,



            $$
            lambdaleft(fracAPAPright)costheta = lambdaleft(fracBPright)sintheta
            $$



            (In equilibrium the horizontal projection for $F_AP$ and $F_PB$ are equal) then



            $$
            fracAPBP=fracsinthetacosthetafracAP
            $$



            or as expected



            $$
            frac4 cos theta-34 sintheta-1 = 3left(fracsinthetacosthetaright)
            $$






            share|cite|improve this answer























            • Thanks very much for your work. I'll try an learn from it! Mitch.
              – gnitsuk
              Jul 18 at 9:21












            up vote
            0
            down vote










            up vote
            0
            down vote









            We have



            $$
            |AP|_0 = 15\
            |BP|_0 = 5\
            r = frac_02\
            |AP| = 2rcostheta\
            |BP| = 2rsintheta
            $$



            then



            $$
            fracAPBP = frac20 cos (theta )-1520 sin (theta )-5 = frac4 cos (theta )-34 sin (theta )-1
            $$



            the solution follows as in
            Calculation of the modulus of elasticity of a stretched string



            NOTE



            with $lambda$ the elastic modulus,



            $$
            lambdaleft(fracAPAPright)costheta = lambdaleft(fracBPright)sintheta
            $$



            (In equilibrium the horizontal projection for $F_AP$ and $F_PB$ are equal) then



            $$
            fracAPBP=fracsinthetacosthetafracAP
            $$



            or as expected



            $$
            frac4 cos theta-34 sintheta-1 = 3left(fracsinthetacosthetaright)
            $$






            share|cite|improve this answer















            We have



            $$
            |AP|_0 = 15\
            |BP|_0 = 5\
            r = frac_02\
            |AP| = 2rcostheta\
            |BP| = 2rsintheta
            $$



            then



            $$
            fracAPBP = frac20 cos (theta )-1520 sin (theta )-5 = frac4 cos (theta )-34 sin (theta )-1
            $$



            the solution follows as in
            Calculation of the modulus of elasticity of a stretched string



            NOTE



            with $lambda$ the elastic modulus,



            $$
            lambdaleft(fracAPAPright)costheta = lambdaleft(fracBPright)sintheta
            $$



            (In equilibrium the horizontal projection for $F_AP$ and $F_PB$ are equal) then



            $$
            fracAPBP=fracsinthetacosthetafracAP
            $$



            or as expected



            $$
            frac4 cos theta-34 sintheta-1 = 3left(fracsinthetacosthetaright)
            $$







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 16 at 18:25


























            answered Jul 16 at 16:24









            Cesareo

            5,7922412




            5,7922412











            • Thanks very much for your work. I'll try an learn from it! Mitch.
              – gnitsuk
              Jul 18 at 9:21
















            • Thanks very much for your work. I'll try an learn from it! Mitch.
              – gnitsuk
              Jul 18 at 9:21















            Thanks very much for your work. I'll try an learn from it! Mitch.
            – gnitsuk
            Jul 18 at 9:21




            Thanks very much for your work. I'll try an learn from it! Mitch.
            – gnitsuk
            Jul 18 at 9:21












             

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            Color the edges and diagonals of a regular polygon

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            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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            Relationship between determinant of matrix and determinant of adjoint?

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            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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