ÐÓ°ÉÊÓƵ

Applications of Differential Equations

A Suspended Wire

Theoretical Introduction

The problem we're about to tackle is that of finding the mathematical equation which describes the shape of aÌýhanging cable, such as a telephone wire hanging between telephone poles, or an electrical cable suspended between pylons. It may seem at first that the solution is obvious--after all, the shape formed looks very much like aÌýparabola. Sadly our instincts lead us astray in making this assumption! Perhaps even more surprising than this is the fact that the derivation of the correct formula involves the use ofÌýdifferential equations.

Setting up the Mathematical ModelÌý

So let's begin the mathematical derivation. Before we can even think about getting a formula describing the shape in question we need to impose some kind of coordinate system on the problem in order to put it into a mathematical context. Perhaps the easiest choice we could make is to impose a Cartesian coordinate system over the profile of the wire such that ³Ù³ó±ðÌýlowest point on the wire passes through the originÌýof the system, as shown here:

Now let's concentrate on a small segment of the wire, and analyze ³Ù³ó±ðÌýforcesÌýthat this segment is experiencing. For ease of analysis, let's consider a short section that stretches from the origin to a point on the wire somewhere in the first quadrant. We'll refer to the two end points asÌýPÌý²¹²Ô»åÌýQ. So now we have a picture that looks like this:

A Little Physics

We're going to need just a little physics at this point, because we're now going to look at ³Ù³ó±ðÌýtensionÌýexperienced by the wire at the two end points,ÌýPÌý²¹²Ô»åÌýQ, of our segment of interest. I think you'll agree that the tension at each point can be thought of as aÌývectorÌýwhose direction isÌýtangentÌýto the curve of the wire, and pointedÌýawayÌýfrom the segment. If we call these two tension vectorsÌýT_PÌý²¹²Ô»åÌýT_QÌýthen our picture becomes:

OK, now a little more physics. Since the wire is in anÌýequilibrium state, (it isn't moving), then we can conclude that the forces acting on the wire are in equilibrium themselves. (If they weren't there would be an overall non-zero force acting on the wire in some direction, which should result in acceleration, according to Newton's second law of motion.) So, all forces involved here are balanced by other, equal and opposite forces. At this point then, it may pay us to start breaking down the vectors in our picture into their individual horizontal and vertical components, then use the "balanced" nature of the forces to produce some equations that may be useful.

Decomposing the Tension Vectors

T_PÌýis obviously 100% horizontal component, i.e. it's vertical component is zero, so we only need to bother with decomposingÌýT_Q. This decomposition may be seen in the picture below, where the triangle on the right has legs whose lengths are the horizontal and vertical components ofÌýT_QÌýthat we seek.

Note that in order to discuss the decomposition of vectors it is necessary to have theirÌýmagnitudes, (their lengths.) These I have denoted in the picture by using absolute values around the vector name, i.e. |T_P| is the length, or magnitude ofÌýT_P, and |T_Q| is the length, or magnitude ofÌýT_Q.

Note also that it only takes a small amount of simple, right triangle trigonometry to label the horizontal and vertical components ofÌýT_QÌýonce the angle and hypotenuse of the triangle have been labeled.

Equating The Magnitudes of Opposite Components

Now, we said earlier that all forces within the cable are balanced by equal and opposite forces, resulting in the cable's equilibrium state. This means that:

³Ù³ó±ðÌýhorizontalÌýcomponents ofÌýT_PÌý²¹²Ô»åÌýT_QÌýare equal and opposite.

But what about their vertical components? Well, obviously the vertical component ofÌýT_PÌýis zero. So what force could possibly be balancing the upward vertical component thatÌýT_QÌýclearly exhibits? Haven't we been forgetting another force that the cable must be experiencing, a force other than tension? Haven't we overlooked the force of gravity acting on the cable's mass, i.e. ³Ù³ó±ðÌýweightÌýof the cable?

Once we take the cable weight into consideration we have that:

³Ù³ó±ðÌýverticalÌýcomponent ofÌýT_QÌýand cable segmentÌýPQ's weight are equal and opposite.

So let's make a check list of the quantities we need in order to take advantage of the equality of components we mentioned above:

Magnitudes of Horizontal Components

• To the left:Ìýwe have a force of magnitude |T_P|.
• To the right:Ìýwe have a force of magnitude |T_Q| cosÌýθ.

Magnitudes of Vertical Components

• Upwards:Ìýwe have a force of magnitude |T_Q| sinÌýθ.
  Ìý
• Downwards:Ìýwe have a force equal to the weight of the cable. Unfortunately, "the weight of the cable" isn't specific enough to be very useful in setting up the equations we'll need, so let's do a little more analysis.

  Say that the cable is of constant thickness and density, (which is not a very strange assumption when you think about it.) If this is the case then we can talk about the cable'sÌýlinear weight densityÌýmeasured in, say, newtons/meter. Let's say that the cable in question has a linear weight density ofÌýwÌýn/m.

  Now, say that the length of the cable fromÌýPÌýtoÌýQÌýisÌýsÌýmeters. Putting this together with the fact that the cable weighsÌýwÌýnewtons/meter, we get that the entire segment of cable under consideration has a weight ofÌýw sÌýnewtons. Conclusion:

  Downwards:Ìýwe have a force of magnitudeÌýw s.

So let's get back to equating the magnitudes of the corresponding components.

Horizontally, we have:

magnitude of the force to the leftÌý=Ìýmagnitude of force to the right

or, in symbols:

|T_P| = |T_Q| cosÌýθ

Vertically, we have:

magnitude of the force upwardsÌý=Ìýmagnitude of force downwards

or, in symbols:

|T_Q| sinÌýθÌý=Ìýw s

which could also be written as:

w sÌý= |T_Q| sinÌýθ

Deriving ³Ù³ó±ðÌýs-θÌýRelationship

At last! Mathematical equations that we can manipulate! Now we're going to combine these two equations to our advantage. Notice how both:

w sÌý= |T_Q| sinÌýθ

and:

|T_P| = |T_Q| cosÌýθ

have |T_Q| on the right hand side. (That's why I traded sides a couple of steps back!)

This means that if we divide the first by the second we get:

(w s)Ìý/Ìý|T_P| = (|T_Q| sinÌýθ)Ìý/Ìý(|T_Q| cosÌýθ)

Canceling the |T_Q|'s on the right hand side, and combining the sine and cosine to give a tangent function we get the simpler equation:

(w s)Ìý/Ìý|T_P| = tanÌýθ

We must now ask ourselves whether this is the equation we were looking for. We started out to find an equation for the curve formed by a hanging cable. In our very first graph we superimposed the cable on anÌýxy-coordinate system, so it would only be natural to expect our final equation to express a relationship between the variablesÌýxÌý²¹²Ô»åÌýy. A glance at the equation we just derived reveals a relationship between the variablesÌýθÌý²¹²Ô»åÌýs. (Note thatÌýwÌýand |T_P| are constants.)

I suppose we could leave the equation like this, but it wouldn't be very useful. So, our quest for anÌýxy-equation continues.

Deriving ³Ù³ó±ðÌýx-yÌýRelationship

We now need to getÌýxÌýinto the picture, (not to mentionÌýy,) and get rid ofÌýθÌý²¹²Ô»åÌýs.

EliminatingÌýs

Let's designate the coordinates of pointÌýQÌýas (x,y). This means that as the cable stretches fromÌýPÌýtoÌýQÌýit actually stretches between coordinates (0,0) and (x,y). A quick dig in your memory banks should help you to recall that there is a way of calculating the distance along a curve between two knownÌýx-values.

In your second calculus course you should have learned the following formula for the length,ÌýL, of a curve betweenÌýxÌý=ÌýaÌý²¹²Ô»åÌýxÌý=Ìýb:

In our particular case the curve length has been designated asÌýsÌýand the points we are measuring between are (0,0) and (x,y). Putting these quantities into the above equation we get:

So we've managed to get a way of transformingÌýsÌýintoÌýx's andÌýy's. But we still have a problem--the formula we just found involves anÌýintegral. Using it in this form will lead us to anÌýintegral equationÌý,ÌýnotÌýaÌýdifferential equation. (ThereÌýareÌýcourses devoted to integral equations, but this course isn't one of them.) How could we get rid of the integral?

The obvious answer is to use the fundamental theorem of calculus, (well, at least one of its flavors,) which states thatÌýdifferentiatingÌýan integral (under the right conditions) kills the integral. So applying the derivative with respect toÌýxÌýto both sides of the last equation gives us:

which condenses to:

This now contains no integral, but we unfortunately have the derivative ofÌýs, rather than just plainÌýs. We'll tackle this issue in a while. First we'll look at replacing the other unwanted variable in our equation—θ.

EliminatingÌýθ

Remember that the equation we're trying to "fix up" is:

(w s)Ìý/Ìý|T_P| = tanÌýθ

So far we have ³Ù³ó±ðÌýsÌý(kind of) taken care of. TheÌýθÌýstill remains. The trick here is toÌýnotÌýconcentrate onÌýθÌýalone, but to consider tanÌýθÌýas a whole. If you look back to the earlier diagrams, you should note that tanÌýθÌýis actually ³Ù³ó±ðÌýslopeÌýof the tension vectorÌýT_Q. (If can't see this think tanÌýθÌý=Ìýopp/adjÌý=Ìýrise/runÌý=Ìýslope.)

But another way of thinking of the slope of the tension vectorÌýT_QÌýis to remember that it lies tangent to the curve of the cable. You learned at the very beginning of your calculus course that the slope of a tangent line is given by ³Ù³ó±ðÌýderivative.

To summarize, then, we have:

slope ofÌýT_QÌý=ÌýtanÌýθ

and we also have:

slope ofÌýT_QÌý=Ìýdy/dx

Putting these two together gives us:

tanÌýθÌý=Ìýdy/dx

At last we have a way of replacing ³Ù³ó±ðÌýθÌýin the equation of the cable.

Putting it all Together

Once again, remember that the equation of the cable, as we have it so far is:

(w s)Ìý/Ìý|T_P| = tanÌýθ

We can immediately replace the tanÌýθÌýbyÌýdy/dx, according to the work we just completed. This makes the cable's equation become:

(w s)Ìý/Ìý|T_P| =Ìýdy/dx

or, flipping it around:

dy/dxÌý= (w s)Ìý/Ìý|T_P|

But ³Ù³ó±ðÌýsÌýstill needs to be dealt with! A little earlier we found that:

If we only hadÌýds/dxÌýin our cable equation, instead of justÌýs, we'd be all set. But wait! There's nothing to stop us from producing the derivative for ourselves. Let's just take the derivative of both sides of of the cable equation:

dy/dxÌý= (w s)Ìý/Ìý|T_P|

to get:

which can now be substituted into the curve length equation to get:

Whew! That's it! We got it! An equation for the hanging cable which uses only the variablesÌýxÌý²¹²Ô»åÌýy. As promised, it's aÌýdifferential equation. Worse yet, it's a second-orderÌýnon-linearÌýdifferential equation, which, you may have learned in class already, makes it much harder to solve. We'd really like to solve it since we'd like to see a final formula for the hanging cable in the formÌýyÌý=Ìýf(x).

Thankfully we'll be having the computer help us with this last part of the problem. It's now time toÌý, and get on with the job!

---

If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.