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Logistics

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Hey guys this is my question

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A long thin rod is clamped vertically at its lower end and a mass

[tex] EI\frac{d^2x}{dy^2} = Mg(a-x)[/tex]

where

[tex] x = a [1- \frac{sin\omega(L-y)}{sin\omega L}][/tex]

where [tex]\omega^2 = \frac{Mg}{EI}[/tex]. (Hint: Let

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Now I have got this here:

[tex] z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}[/tex]

Thus:

[tex] EI\left(-\frac{d^2z}{dy^2}\right) = Mgz[/tex]

[tex] \frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0[/tex]

I'm thinking that it's correct so far, How would I solve that?. ie. Continue on from here

Could someone help me out please

Thanks

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A long thin rod is clamped vertically at its lower end and a mass

*M*is attached to its upper end. The coordinates (*x, y*) of any point on it satisfy the equation[tex] EI\frac{d^2x}{dy^2} = Mg(a-x)[/tex]

where

*E, I*and*a*are constants. Given that*x = 0*when*y = 0*and*x = a*when*y = L*show that[tex] x = a [1- \frac{sin\omega(L-y)}{sin\omega L}][/tex]

where [tex]\omega^2 = \frac{Mg}{EI}[/tex]. (Hint: Let

*z = a - x*).----------------------------------------------------------------------

Now I have got this here:

[tex] z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}[/tex]

Thus:

[tex] EI\left(-\frac{d^2z}{dy^2}\right) = Mgz[/tex]

[tex] \frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0[/tex]

I'm thinking that it's correct so far, How would I solve that?. ie. Continue on from here

Could someone help me out please

Thanks

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