Langmuir 1992,8, 1086-1089
1086
General Treatment of a Hanging String under the Effect of a Soap Film P. Mohazzabi, J. P. McCrickard, and F. Behroozi' Department of Physics, University of Wisconsin-Parkside,
Kenosha, Wisconsin 53141
Received September 17, 1991.I n Final Form: January 22, 1992 We report on the behavior of a uniform flexible string, suspended from a horizontal rod at two points, when a soap film covers the surface bounded by the string and the rod. We find that, for fixed support points, as the value of the film's surface tension is increased, the shape assumed by the string passes continuously from a concave shape (positive curvature) through a linear configuration (zero curvature) to a convex configuration (negative curvature). Furthermore, for a fixed surface tension, the shape assumed by the string can be made to evolve continuously through the above configurations by adjusting the distance between the support points.
Introduction
Y I
The mathematical function describing the curve assumed by a hanging string, known as the catenary, was first published without proof by Leibnitz in 1691, and later derived by Jacob Bernoulli. In this paper we report on the general behavior of the catenary under the combined action of gravity and surface tension; Le., we describe the shape assumed by a hanging string when asoap film covers the surface bounded by the string. We have already reported that the string assumes a convex shape (negative curvature) when the surface tension exceeds the gravitational force per unit length of the string.' Here we consider the general case when the surface tension may assume any value relative to the gravitational force per unit length. We find that, as the surface tension is increased from zero, the shape of the string evolves continuously from a regular catenary (positive curvature) through a linear configuration (zero curvature) into a convex curve (negative curvature) which smoothly merges with the case reported previously.
General Shape of the String Figure 1 shows a uniform flexible string of length 2L, which is suspended from a horizontal rod at two points separated by 2x0. A soap film covers the surface bounded by the string and the horizontal rod. The shape of the string is always symmetric with respect to the y-axis as shown, so we limit our solutions and discussions only to the right half of the string. In what follows, T stands for tension in the string, X is the linear mass density of the string, a designates the surface tension of the film, and g is the acceleration of gravity. The arc length s(8) is measured from the point where the string meets the y-axis to the point of interest on the string while 8 designates the angle between the tangent and the positive x-axis. This angle is called 80 a t the support point where x = X O , and Bi at the point where the string intercepts the y-axis. Note that the horizontal rod from which the string is hanging lies along the x-axis. Consider the equilibrium conditions for a segment ds of the string under the combined action of gravity and surface tension. In terms of force components tangent (1) Mohazzabi, P.; McCrickard, J. P.; Behroozi, F. Langmuir 1990,6, 1269.
0743-7463/9212408-1086$03.0010
X
Figure 1. A uniform flexiblestring of length 2L suspended from a horizontal rod. A soap film covers the surface bounded by the rod and the string. The forces acti.ng on a typical segment ds of the string are shown. Here u is the surface tension of the film, Tis the tension in the string, X is the mass per unit length, and g is the acceleration of gravity. The string assumes this concave shape when a = (2uIXg) < xo/L.
and normal to the string, we obtain d T = Ag ds sin 8
(1)
and Tdd + 2ads = Agds cos 8 (2) The force due to the surface tension is 2a ds because the film has two surfaces. Furthermore, though these equations are derived for the case when ds/dd > 0 (see Figure l ) , they also apply when ds/d8 < 0 which is the case shown in Figure 2. As an aid to discussion, eq 2 may be rearranged to give
T = Ag(ds/d8)(cos d - a) (3) where a (2a)/(Ag) is the ratio of surface tension to gravitational force per unit length. Note that since the string tension Tis always positive, eq 3 requires that when a > 1 we must have ds/d8 < 0, which implies that the string assumes a convex shape. This is the configuration (Figure 2) which has been treated in detail previously.' When a < 1,three possibilities arise. If cos 8 > a,then eq 3 requires dsId8 > 0 which yields a concave shape for the string (Figure 1). On the other hand, if cos 8 < a,we must have dsId8 < 0, which means the string assumes a convex shape (Figure 2). In the special case when cos 8 = a,we must have ds/d8 = a,which leads to two possible shapes for the string. The first possibility is that the string hands in a V-shape with 0 1992 American Chemical Society
Langmuir, Vol. 8, No. 4,1992 1087
Effect of a Soap Film on a Hanging String Y I
I
/
/
- x
jy
Equations of the String To obtain the parametric equations of the string x ( 0 ) and y(@,we begin by combining eqs 1 and 3 to get
Figure 2. Convex configuration for the string. L’ is the length
of the tangent line from the rod to the y-axis. The clump length is denoted by 1. The string assumes this configuration when a > xo/L. Y I -I
cos 8 which shows a perfect balance between the surface tension force and the component of the gravitational force normal to the string. Hence, the string assumes a linear configuration. In contrast, when a > cos 8, we have 2u > Ag cos 0; i.e., the surface tension wins and the string is pulled inward, giving rise to a convex configuration. Finally, when a < cos 6, the gravitational force component Ag cos 0 is greater than the surface tension; consequently, the string bows outward and a concave configuration results.
xe
-
-- x
sin6 de cos e - cy This relation is easily integrated to yield dT--
T
T = X g C cos e - a where C is a constant. Using this result in eq 3, we obtain
C -ds= (7) de (COS e - a ) 2 Since dx = ds cos 8, and dy = ds sin 0, we may immediately write down the governing equations of the string as -dx = de
I
Figure 3. Linear or V-shape configuration for the string. The string assumes this configuration when a = xo/L. two linear segments (see Figure 3). In this case cos 00 = a = x d L , and the equation describing the right half of the string is simply given by
Ccose (COS
e - a)2
4 Y = C sin e de (COS e - a ) 2
(9)
These two equations can be readily integrated to give -X =
asin8
+
C (1- a2)(cose - a)
The second possible configuration for the case cos e = a is that the half-string contains an inflection point which separates a concave section from a convex section. Later we will show that this mixed scenario is not admissible. However, the determination of the shape of the string based on a comparison of cos 0 and a is unsatisfactory. It is desirable to be able to determine the general shape of the string in terms of the string parameters X O , L , and a. In fact this can be done. Recall that when a = xo/L, the string assumes a linear shape given by eq 4. This linear shape forms a boundary between the concave and convex configurations. It is easy to show that when a < xo/L, one obtains a concave configuration, whereas when a > xo/L, the string assumes a convex form. To see this, consider the convex case depicted in Figure 2. Now, draw a tangent to the string at the point where the string meets the rod (e = eo). Let L’ denote the length of this tangent line between the rod and the y-axis. Since L’ < L for any convex configuration, we have cos 60 = xdL’ > xo/L. But a > cos 0 for all values of 8 for the convex case, so CY > XOIL.A similar argument can be used to show that concave solutions are characterized by a < XOIL. In summary, then, the shape of the string is convex when a > x d L , becomes linear when a = xo/L,and assumes a concave shape when cy < xo/L. These distinct configurations can be understood on an intuitive basis if we compare the forces acting normal to the string. Recall that a 2a/Ag. So, when a = cos 8, we have 2u = Ag X
(8)
+ +
+
(1 a)tan e12 (1- a2)1/2 1 1nl (1- a2)3’2 (1 a)tan 1912- (1- a2)1/2
J!=
C
1 COS^-^)
+
CY
(11)
Equations 10 and 11 are the parametric equations for x and y in terms of three undetermined constants C,, Cy, and C. The values of these constants are fixed by the choice of origin, and the ratio x0lL. One may obtain y ( x ) by eliminating B between the two equations, which for the case of a = 0 reduces to the well-known catenary equation (hyperbolic cosine). For a # 0, the function y ( x ) is too cumbersome to be useful. We mention in passing that eq 10 has been obtained under the assumption that a < 1. The corresponding equation for the case a > 1has been published previously.’ We are now in a position to ask the following question: Given a string of length 215, and support separation 2x0 < 2L, what exact form does the string assume for a given a? As discussed before, the value of a relative to XOIL determines whether the system is in the regime of Figure 1or that of Figure 2. When a < xo/L, the string assumes a concave configuration, whereas a > x d L leads to a convex shape, with cy = xo/L giving the linear configuration. We consider each case in greater detail. (a) Concave Configuration, This is the situation depicted in Figure 1,where a < xo/L which is equivalent to the condition cos 0 < a. In this case the boundary
Mohazzabi et al.
1088 Langmuir, Vol. 8, No. 4, 1992 conditions can easily be applied to eqs 7, 10, and 11 to determine the three constanLs C, C,, and Cy,which in turn determine the angle 00 that the string makes with the rod. Specifically, eq 7 gives
L = JoL ds = C S,""O/(cos
and 0 = X _
00,
eq 17 gives Cy. The results are
a sin 0 (1- a2)(cos0 - a ) (1 a) tan 012 1
+
+ + (1- C Y ~ ) ~ / ~ (1 + a ) tan 0/2 - (1- a')'/'
0 - a)2
which upon integration results in sin Bo Ii= c (1- a2)(cosBo - a ) +
1-
(1 a ( 1 - a2 )3/2 In (1
and
+ a ) tan 0,/2 + (1- a2)'l2 + tan 0,/2 - (1 -
L=
Further, since in the case under consideration, 0 = 0 at x = 0, and 0 = 00 at y = 0, one can easily determine that C, = 0, and C, = -l/(cos 00 - CY). Thus eqs 10 and 11 take the form -X =
c
a sin 0
(1 - a2)(cos 0 - a )
1 1 (cos 0 - a ) cos Bo - a
-a1
CY)
We can now determine the remaining parameters 1 and 00 in terms of L , xo, and a. Thus, eq 7 determines 00 in terms of 1 and L:
ds = -al d0
+
(19)
1 (cos 0 - a ) 2
and hence JoL-' ds = -a1 Jr:2dO/(cos
0 - a)2
or
Y=--
c
1
coso-
1 cy
(14)
cosoo- a
which gives
Using x o and 00 in eq 13, we have
"o= c
a sin Bo (1- a')(cos 0, - a )
(1- a2)332
In1
L ---
+
(1 + ix) tan 0,/2
I
+ (1-
(15) (1 + a ) tan 0,/2 - (1- a2)1/2 Finally, eqs 12 and 15 suffice to determine the constants 80 and C in terms of the parameters L , X O , and a. With these constants determined, eqs 13 and 14 give x , y pairs for any value of 0 I 0 I 00, hence providing a complete description of the shape of the string. (b) Convex Configuration. This is the case for which xolL < a < 1, which is equivalent to the condition cos 0 < a. Under this condition the string hangs in a convex configuration with the lower portion clumped in a vertical line as shown in Figure 2. As reported elsewhere' we have observed this behavior experimentally. Since any physical string is not infinitely flexible, the angle 0 at the clump . point (point C in Figure 2) assumes a value of ~ 1 2 Hence, besides C,, C y , and C, a fourth undetermined parameter 1 appears, which is the clump length extending along the -y-axis. The clump length 1 serves to specify the tension Ti a t the clump point. With Oi = ir/2, we have Ti = lAg, which when used in eq 6 gives C = -al. Now the parametric equations of the string can easily be written as 1
L - 1 = a1 S,y2dO/(cos 0 - a)' -a (1+ a) + (1- a 2 p 2 + ( 1 - 2 ) ( 1 -aa2 )3/21n (1 + a) - (1- a2)1'2 sin Bo
I+
(1- a2)(a- cos Bo) a
(1- a3312 In
I
(1 + a ) tan 0,/2
+ (1 - a2)1/2
(1 + a ) tan 0,/2 - (1- a2)l/'
Finally, since x ( 0 0 ) = X O , eqs 18 and 20 may be used to determine 00 and 1 in terms of X O , L , and a. With all the constants thus determined, eqs 18 and 19 give x,y pairs for the range of 80 I 0 I ~ / 2 . It is important to note that in the case under consideration there exists an essential indeterminacy which in principle results in an infinitely degenerate solution. This interesting situation arises because a knowledge of L and x o suffices to determine only two of the three parameters 00, 1, and Oi (note that C, and Cy are always determined by the choice of origin). In our analysis above, we assume Oi = ~ 1 2since , this is the choice forced on the problem due to limited flexibility of any physical string.2 If we consider a perfectly flexible string, then Oi can assume any value in the range 1r/2 IOi I0, hence giving rise to an infinite number of solutions that satisfy the remaining boundary conditions.
(16)
Discussion In Figure 4, we have used the results of a numerical calculation to obtain the shape of the right half of the string for the particular choice of xo/L = 112. For a < 112 we obtain concave solutions, while for a > 112 the solutions are convex. Of special interest is the case of a = 1/2, where the shape of the string consists of two straight sections given by eq 4 (see Figure 3) with a slope of * ( ( L z X O ~ ) / X O ) ' / ~ In . this case 0 = 00 = c0s-l xo/L = c0s-l a.
When the boundary condition is applied at the clump point where x = 0 and 0 = n/2, eq 16 yields C,. Similarly, using the boundary condition at the support point where y = 0
(2) In any real string a discontinuity in the slope is not permitted because of the enormous cost in elastic energy. In contrast, a perfectly flexible string can conform to any shape with no cost in elastic energy.
_ --a1 X
a sin 0 (1- a2)(cos0 - a ) (1 a ) tan 012 1 +
+
1
+ (1 - a2)'I2 + C,
Effect of a Soap Film on a Hanging String
Langmuir, Vol. 8, No. 4, 1992 1089
X
X
Erl ,
I
1
-2.0 -1.8
1
1
I
I
I
Figure 4. Shape of the right half of the string for the particular choice of xdL = 1/2 for several values of a.
In the discussion following eq 4, we pointed out that the half-string cannot contain an inflection point. As was observed earlier at an inflection point dsld0 = w. Since the tension T must be finite, then eq 3 implies that cos d - a = 0 a t an inflection point. However, eqs 13 and 14 or eqs 18 and 19 show that this condition can only be satisfied at x: = 01 and y = 03. Consequently,as long as the string has a finite length, it cannot show an inflection point and the string cannot assume a shape with both concave and convex sections present. In an earlier paper,' we reported on the shape of the string for a > 1 which is seen to be a subset of the a > xo/L case. In fact, when a > 1, though x ( 0 ) has a somewhat different functional form compared to eq 18,the qualitative
Figure 5. Shape of the right half of the string for a fixed value of a and L but for several values of ro.
behavior of the string is similar. The functional form of x ( 0 ) for the a = 1 case is also different compared to the general case of a > xo/L; nevertheless, the shape of the string evolves continuously through the a = 1boundary (see Figure 4). When a >> 1,the string assumes a shape very close to the arc of a circle' of radius lla. It is interesting to note that, for a given L and for a < 1, the shape of the string can be made to evolve continuously from a concave to a convex configuration as xo is reduced (see Figure 5 ) . Specifically, one must begin with x o < L,but with xo > d, to ensure a concave shape. As xo is reduced, 80 increases until at zo = d, 00 = c o d a, and the string becomes linear. As 3c0 is reduced further, the string assumes a convex configuration. Finally as x : ~ approaches zero, the clump length 1 increases to L and the entire string hangs in a vertical line.
