3.1 Suspended cable
A cable of constant line density is suspended between two vertical poles of equal height such that it takes the shape of a curve, . The origin of the curve is a point mid-way between the feet of the poles and is the height above the ground. If the cable is 600 metres long show that the distance between the poles is 55 metres to the nearest metre. Find the height of the centre of mass of the cable above the ground to the nearest metre.
Mathematical statement of the problem
We can draw a picture of the cable as in Figure 7 where and denote the end points.
For the first part of this problem we use the result found in HELM booklet 14 that the distance along a curve from to is given by
where in this case we are given and therefore .
If we take the distance between the poles to be then the values of in this integration go from to . So we need to find such that:
For the second part of this problem we need to find the centre of mass of the cable. From the symmetry of the problem we know that the centre of mass must lie on the -axis. To find the height of the centre of mass we need to take each section of the cable and consider the moment about the -axis through the origin. A section of the cable has mass where is the line density of the cable and is the length of a small section of the cable.
so the moment about the -axis will be
taking the limit as and using the fact that
we get that the moment about the -axis to be
This must equal the moment of a single point mass, equal to the total mass of the cable, placed at its centre of mass. As the length of the cable is 600 metres then the mass of the cable is 600 and we have
Dividing both sides of this equation by we get:
where we have already established the value of from Equation (1) so we can solve this equation to find .
We need to find so that
Rearranging the hyperbolic identity we obtain
so the integral becomes
Using the log identity for the function:
we find that m so the distance between the poles is 55 m to the nearest metre.
To find the height of the centre of mass above the ground we use
Substituting and therefore we get
From the hyperbolic identities we know that
so this integral becomes
So we have that
From the first part of this problem we found that so substituting for we find metres to the nearest metre.
We have found that the two vertical poles holding the cable have a distance between them of 55 metres and the height of the centre of mass of the cable above the ground is 150 metres.
Find the centre of mass of a lamina bounded by , for .