Suspension Bridge
A catenary is a chain that is curving because of its own weight as opposed to a bridge chain, which is a parabola. It too curves under its own weight, but a much greater influence is the weight of the bridge deck. The source and magnitude of the weight is not important, but its distribution makes a big difference.
Look at this sketch, and consider the distribution of the mass on a suspension bridge. The road deck hangs on vertical cables suspended from the main cables. The cables themselves may be large, but their mass is insignificant when compared with that of the deck, so disregard the mass of the cable. Suppose that the deck is horizontal, and suppose that its cross section is uniform throughout the length of the bridge. Also, the vertical cables are so close together that we can assume that the weight of any part of the deck is transferred to the part of the main cable directly above it. These conditions are reasonably close to the real thing.
Consider only that part of the cable in the interval between O and P. It is motionless. More importantly, it is not accelerating. That means that the net force on it is zero. In the study of physics, forces may be modeled with a free body diagram. Each force is identified with a vector arrow. The direction of the arrow is parallel to the direction of the force, and the length of the arrow is proportional to the magnitude of the force.
There are three forces acting on the cable interval OP. One force is the tension from the cable to the left. Call this force . It acts in a horizontal direction to the left. There is also a force from the cable on the right, exerted at point P. This force will be called . Its direction is what we need to determine, because that will tell us the slope of the cable at P. The third force is the weight on the cable, . Since this is a gravitational force, its direction can only be vertical and downward.
The net force can be determined by adding the vectors. Graphically, this is accomplished by aligning the vectors point to tail. We already know that the net force is zero, so the last vector must end where the first one began. By doing this in the illustration on the right, we can see the direction of , which gives us the slope at P.
slope at P =
In this equation W and T are magnitudes only, not vectors. But what are W and T? Well, T is clearly a constant. It is the tension at point O, which is the same no matter what point is chosen for P. W is the weight on the cable between O and P. This is the weight of the roadway directly below that length of cable. How can it be determined?
Remember that one of our conditions was that the deck has a uniform cross section throughout. That means that one meter of road on one part of the bridge would have to weigh the same as one meter of road on any other part of the bridge. Let μ represent the linear weight density of the deck. That is, multiplying μ by the length of any interval on the deck will give the weight of that interval.
So what is the length of the road from O to P? It is not the distance between those two points. The cable is curved, but the road below it runs horizontally. The length of that stretch of road is simply the difference in the x-coordinates of those points. Point O is the origin, so all we need is the x-coordinate of point P.
W=µx
slope at P =
This is where the calculus begins. Let y be the height of the cable. The slope of the cable then must be the first derivative of y.
So y is an antiderivative of the function μx/T.
, for some constant C.
We know that the origin is on the curve. Substituting zero for x and y and solving for C, we find that C = 0.
Since μ/(2T) is a constant, this is a quadratic equation, a parabola.