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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 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).
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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
----------------------------------------------------------------------
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
Last edited: