Why does a string suspended from two supports become a parabola?

Tywin Lannister

A very interesting question which forced me to leave my recliner at office & delve back into the basic theories of Mechanics, Tension, etc. Ah ! The world of free-body diagrams, motion, energy and...teenage dreams...;)

Well, a string suspended from two supports take the form of what is called a "Catenary". It looks like a parabola, but it's not exactly a parabola.

Now, what's a Catenary?

Wiki says "In physics and geometry, a catenary is the curve that an idealized hanging chain or cable assumes under its own weight when supported only at its ends.

The catenary curve has a U-like shape, superficially similar in appearance to a parabola, but it is not a parabola.

It is also considered as scaled and rotated graph of the hyperbolic cosine function."

We see this most often in Electric power cables, suspension bridges & spider webs !!!

In fact, inverted Catenary arches have several applications in construction of kilns. To create the desired curve, the shape of a hanging chain of the desired dimensions is transferred to a form which is then used as a guide for the placement of bricks or other building material.

In free-hanging chains, the force exerted is uniform with respect to length of the chain, and so the chain follows the catenary curve. The same is true of a simple suspension bridge or "catenary bridge," where the roadway follows the cable.

A stressed ribbon bridge is a more sophisticated structure with the same catenary shape.

However, in a suspension bridge with a suspended roadway, the chains or cables support the weight of the bridge, and so do not hang freely. In most cases the roadway is flat, so when the weight of the cable is negligible compared with the weight being supported, the force exerted is uniform with respect to horizontal distance, and the result is a parabola, as discussed below (although the term "catenary" is often still used, in an informal sense). If the cable is heavy then the resulting curve is between a catenary and a parabola.

Now, why does this happen ?

In the mathematical model the chain (or cord, cable, rope, string, etc.) is idealized by assuming that it is so thin that it can be regarded as a curve and that it is so flexible any force of tension exerted by the chain is parallel to the chain.

So, for laymen like me, it is the resultant of the gravitational pull & the string tension.

For experts, of course, this is the mathematical equation (from wiki again !)

Let the path followed by the chain be given parametrically by r = (x, y) = (x(s), y(s)) where s represents arc length and r is the position vector. This is the natural parameterizationand has the property that

<span class="MathJax" id="MathJax-Element-1-Frame" tabindex="0" data-mathml="drds=u" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">drds=udrds=u

where u is a unit tangent vector.

Diagram of forces acting on a segment of a catenary from c to r. The forces are the tension T0 at c, the tension T at r, and the weight of the chain (0, −λgs). Since the chain is at rest the sum of these forces must be zero.

A differential equation for the curve may be derived as follows.

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Let c be the lowest point on the chain, called the vertex of the catenary.

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The slope dy/dx of the curve is zero at C since it is a minimum point. Assume r is to the right of c since the other case is implied by symmetry. The forces acting on the section of the chain from c to r are the tension of the chain at c, the tension of the chain at r, and the weight of the chain. The tension at c is tangent to the curve at c and is therefore horizontal without any vertical component and it pulls the section to the left so it may be written (−T0, 0) where T0 is the magnitude of the force. The tension at r is parallel to the curve at r and pulls the section to the right. The tension at r can be split into two components so it may be written Tu = (T cos φ, T sin φ), where T is the magnitude of the force and φ is the angle between the curve at r and the x-axis (see tangential angle). Finally, the weight of the chain is represented by (0, −λgs) where λ is the mass per unit length, g is the acceleration of gravity and s is the length of the segment of chain between c and r.

The chain is in equilibrium so the sum of three forces is 0, therefore

<span class="MathJax" id="MathJax-Element-2-Frame" tabindex="0" data-mathml="Tcos&#x2061;&#x03C6;=T0" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">Tcosφ=T0Tcos⁡φ=T0

and

<span class="MathJax" id="MathJax-Element-3-Frame" tabindex="0" data-mathml="Tsin&#x2061;&#x03C6;=&#x03BB;gs," role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">Tsinφ=λgs,Tsin⁡φ=λgs,

and dividing these gives

<span class="MathJax" id="MathJax-Element-4-Frame" tabindex="0" data-mathml="dydx=tan&#x2061;&#x03C6;=&#x03BB;gsT0." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">dydx=tanφ=λgsT0.dydx=tan⁡φ=λgsT0.

It is convenient to write

<span class="MathJax" id="MathJax-Element-5-Frame" tabindex="0" data-mathml="a=T0&#x03BB;g" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">a=T0λga=T0λg

which is the length of chain whose weight is equal in magnitude to the tension at c.

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Then

<span class="MathJax" id="MathJax-Element-6-Frame" tabindex="0" data-mathml="dydx=sa" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">dydx=sadydx=sa

is an equation defining the curve.

The horizontal component of the tension, T cos φ = T0 is constant and the vertical component of the tension, T sin φ = λgs is proportional to the length of chain between the r and the vertex.

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Derivation of equations for the curve[edit]

The differential equation given above can be solved to produce equations for the curve.

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From

<span class="MathJax" id="MathJax-Element-7-Frame" tabindex="0" data-mathml="dydx=sa," role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">dydx=sa,dydx=sa,

the formula for arc length gives

<span class="MathJax" id="MathJax-Element-8-Frame" tabindex="0" data-mathml="dsdx=1+(dydx)2=a2+s2a." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">dsdx=1+(dydx)2−−−−−−−−−√=a2+s2−−−−−−√a.dsdx=1+(dydx)2=a2+s2a.

Then

<span class="MathJax" id="MathJax-Element-9-Frame" tabindex="0" data-mathml="dxds=1dsdx=aa2+s2" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">dxds=1dsdx=aa2+s2−−−−−−√dxds=1dsdx=aa2+s2

and

<span class="MathJax" id="MathJax-Element-10-Frame" tabindex="0" data-mathml="dyds=dydxdsdx=sa2+s2." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">dyds=dydxdsdx=sa2+s2−−−−−−√.dyds=dydxdsdx=sa2+s2.

The second of these equations can be integrated to give

<span class="MathJax" id="MathJax-Element-11-Frame" tabindex="0" data-mathml="y=a2+s2+&#x03B2;" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">y=a2+s2−−−−−−√+βy=a2+s2+β

and by shifting the position of the x-axis, β can be taken to be 0. Then

<span class="MathJax" id="MathJax-Element-12-Frame" tabindex="0" data-mathml="y=a2+s2,y2=a2+s2." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">y=a2+s2−−−−−−√,y2=a2+s2.y=a2+s2,y2=a2+s2.

The x-axis thus chosen is called the directrix of the catenary.

It follows that the magnitude of the tension at a point (x, y) is T = λgy, which is proportional to the distance between the point and the directrix.

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The integral of the expression for dx/ds can be found using standard techniques, giving

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<span class="MathJax" id="MathJax-Element-13-Frame" tabindex="0" data-mathml="x=aarsinh&#x2061;(sa)+&#x03B1;." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">x=aarsinh(sa)+α.x=aarsinh⁡(sa)+α.

and, again, by shifting the position of the y-axis, α can be taken to be 0. Then

<span class="MathJax" id="MathJax-Element-14-Frame" tabindex="0" data-mathml="x=aarsinh&#x2061;(sa),s=asinh&#x2061;(xa)." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">x=aarsinh(sa),s=asinh(xa).x=aarsinh⁡(sa),s=asinh⁡(xa).

The y-axis thus chosen passes through the vertex and is called the axis of the catenary.

These results can be used to eliminate s giving

<span class="MathJax" id="MathJax-Element-15-Frame" tabindex="0" data-mathml="y=acosh&#x2061;(xa)." role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">y=acosh(xa).