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elle
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cable problem
Hi,
I'm doing some self study on the subject of cables and I am curious about the reversed process of working out the linear mass density when given the shape of the cable (instead of the usual case of determining the shape of the cable given the density). The text I am currently reading explains an example but I would like to know how it works for other examples.
Say we have a flexible chain hanging between fixed ends separated by a distance r_0. The weight of a small element ds located at a position along the chain defined by the angle \theta is \lambdagds where \lambda is the linear mass density of the cable and g the acceleration due to gravity.
I don't have a scanner so I can't scan the diagram of forces, but considering the forces on the element, the equilibrium equations give:
(1) tangential component: dT = \lambda g ds sin \theta
(2) normal component: Td \theta = \lambda g ds cos \theta
dividing (1) by (2) we get
\frac {dT}{T} = tan \theta d\theta
i.e. a separable O.D.E thus integrating both sides we get:
T = \frac {T_0}{cos \theta}
where T_0 equaling the cable tension when \theta = 0
Now suppose I want to find the linear density for the shape of a parabola y = x^2. I have taken the following approach: (please correct me if I am wrong)
The attempt at a solution
From (2), rearranging we get T \frac{d\theta}{ds} =\lambda g cos \theta
Curvature \kappa is given by \frac{d \theta}{ds} or in the form \kappa = \frac{y''}{(1+y'^2)^{3/2}}
So we now have
T \kappa = \lambda g cos \theta (*)
differentiating our equation for the parabola y=x^2, y'=2x, y''=2 and substituting these into our \kappa equation and rearranging (*) for \lambda:
\lambda = T(\frac{\frac{2}{1+(2x)^2)^{3/2}}}{g cos \theta}
Can someone please tell me if I am taking the right approach? Also is there any way to simplify what i have obtained for \lambda?
Many many thanks in advance!
Hi,
I'm doing some self study on the subject of cables and I am curious about the reversed process of working out the linear mass density when given the shape of the cable (instead of the usual case of determining the shape of the cable given the density). The text I am currently reading explains an example but I would like to know how it works for other examples.
Homework Statement
Say we have a flexible chain hanging between fixed ends separated by a distance r_0. The weight of a small element ds located at a position along the chain defined by the angle \theta is \lambdagds where \lambda is the linear mass density of the cable and g the acceleration due to gravity.
I don't have a scanner so I can't scan the diagram of forces, but considering the forces on the element, the equilibrium equations give:
(1) tangential component: dT = \lambda g ds sin \theta
(2) normal component: Td \theta = \lambda g ds cos \theta
dividing (1) by (2) we get
\frac {dT}{T} = tan \theta d\theta
i.e. a separable O.D.E thus integrating both sides we get:
T = \frac {T_0}{cos \theta}
where T_0 equaling the cable tension when \theta = 0
Now suppose I want to find the linear density for the shape of a parabola y = x^2. I have taken the following approach: (please correct me if I am wrong)
The attempt at a solution
From (2), rearranging we get T \frac{d\theta}{ds} =\lambda g cos \theta
Curvature \kappa is given by \frac{d \theta}{ds} or in the form \kappa = \frac{y''}{(1+y'^2)^{3/2}}
So we now have
T \kappa = \lambda g cos \theta (*)
differentiating our equation for the parabola y=x^2, y'=2x, y''=2 and substituting these into our \kappa equation and rearranging (*) for \lambda:
\lambda = T(\frac{\frac{2}{1+(2x)^2)^{3/2}}}{g cos \theta}
Can someone please tell me if I am taking the right approach? Also is there any way to simplify what i have obtained for \lambda?
Many many thanks in advance!
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