On the Calculation of Flexible Threads

V. P. Shubin

1

In textbooks on the course in strength of materials, the question of calculating flexible threads is limited to consideration of the deformation and tension of a thread under the action of a uniformly distributed load (its own weight). Of great interest is the calculation of a flexible thread under the action of a concentrated force applied at some section along the length of the thread.

Let us imagine a flexible thread of finite length, fixed immovably at the points \(a\) and \(b\) in such a way that the thread is arranged horizontally. Naturally, in this case an initial tension of the thread arises, which we shall denote by \(H_0\). We assume that before the application of the force \(H_0\) the thread had length \(L_0\) and, correspondingly, \(l_0\).

Fig. 1

If now, at the section \(d\) of the thread, some force is applied that does not carry the material of the thread beyond the elastic limit, then the thread is arranged along the triangle \(acb\) (Fig. 1).

Denoting the increase in the length of the flexible thread by \(\Delta l\), we obtain

\[ \Delta l=\overline{ac}+\overline{cb}-\overline{ab}. \]

Drawing the segment \(\overline{cd}\perp \overline{ab}\) and taking into account that \(\overline{ab}=\overline{ad}+\overline{db}\), we have

\[ \Delta l=\overline{ac}+\overline{cb}-\overline{ad}-\overline{db} =(\overline{ac}-\overline{ad})+(\overline{cb}-\overline{db}) \]

or

\[ \Delta l=\overline{ac}\left(1-\frac{\overline{ad}}{\overline{ac}}\right) +\overline{cb}\left(1-\frac{\overline{db}}{\overline{cb}}\right) =\overline{ac}(1-\cos\alpha)+\overline{cb}(1-\cos\beta). \]

Let us denote the segment \(\overline{cd}=h\); then

\[ \overline{ac}=\frac{h}{\sin\alpha};\qquad \overline{bc}=\frac{h}{\sin\beta}. \]

Thus,

\[ \Delta l=h\cdot \frac{1-\cos\alpha}{\sin\alpha}+h\cdot \frac{1-\cos\beta}{\sin\beta}. \]

By simple trigonometric transformations one can show that

\[ \frac{1-\cos\alpha}{\sin\alpha}=\tg\frac{\alpha}{2};\qquad \frac{1-\cos\beta}{\sin\beta}=\tg\frac{\beta}{2}. \]

Consequently, the elongation of the flexible thread is expressed in the form

\[ \Delta l=h\,\tg\frac{\alpha}{2}+h\,\tg\frac{\beta}{2}. \tag{1} \]

The total elongation \(\Delta l\) may be divided into two components:

\[ \Delta l_1=h\cdot \tg\frac{\alpha}{2}\quad \text{and}\quad \Delta l_2=h\cdot \tg\frac{\beta}{2}, \]

\[ \Delta l=\Delta l_1+\Delta l_2. \tag{2} \]

2.

Let us show that, when the force applied to a flexible thread changes, and consequently the value \(h\) changes, the points \(c_1'\), \(c_2'\), \(c'\) (Fig. 2) are located on the hyperbolas \(\Gamma_1\) and \(\Gamma_2\).

Indeed, drawing from the point \(C\) (Fig. 2) a straight line parallel to \(\overline{ab}\), and from the points \(a\) and \(b\), with radii \(ac\) and \(bc\), the arcs \(CC_1\) and \(CC_2\), we obtain:

\[ \overline{dc_1}=\overline{ac}-\overline{ad}=\Delta l_1, \]

\[ \overline{dc_2}=\overline{cb}-\overline{db}=\Delta l_2. \]

Fig. 2

Obviously, the segment \(C_2'CC_1'=\Delta l\), i.e. the sum of the values

\[ \Delta l_1+\Delta l_2. \]

Applying this method for various values of the forces and the height \(h\), it is easy to show that the locus of points such as \(C_1'\), \(C_2'\), etc., is a hyperbola.

Denoting

\[ \overline{C_1'C}=\overline{cd}\ \text{by } y, \]

\[ \overline{ac_1}=\overline{ac}\ \text{by } x, \]

\(ad\) by \(l\); from triangle \(acd\) we have: \(\overline{ac}^{\,2}=\overline{ad}^{\,2}+\overline{cd}^{\,2}\), or \(x^2=y^2+l^2\), whence

\[ \frac{x^2}{l^2}-\frac{y^2}{l^2}=1 \tag{3} \]

We have obtained the equation of an equilateral hyperbola with vertex at the point \(d\) and with the origin of coordinates at the point \(a\). Similarly, the point \(C_2'\) lies on a hyperbola with equation

\[ \frac{x^2}{(L-l)^2}-\frac{y^2}{(L-l)^2}=1. \tag{4} \]

3

Let us investigate the influence of the height \(h\) on the elongation of a flexible thread. For this purpose, instead of hyperbolas, parabolas may be taken with a sufficient degree of accuracy.

For this, in equality (1), instead of \(\operatorname{tg}\frac{\alpha}{2}\) and \(\operatorname{tg}\frac{\beta}{2}\), one must take half the values of the functions of the whole angles. So long as the angles \(\alpha\) and \(\beta\) lie within the limits of small values, the difference between \(\operatorname{tg}\frac{\alpha}{2}\) and \(\frac{1}{2}\operatorname{tg}\alpha\) does not exceed \((1 \div 1.5)\%\).

Making this substitution, we obtain

\[ \Delta l=\frac{h}{2}\left(\operatorname{tg}\beta+\operatorname{tg}\alpha\right), \]

but

\[ \operatorname{tg}\alpha=\frac{h}{l},\qquad \operatorname{tg}\beta=\frac{h}{L-l}, \]

therefore

\[ \Delta l=\frac{h}{2}\left(\frac{h}{l}+\frac{h}{L-l}\right) =\frac{h^2}{2}\cdot\frac{L}{l(L-l)}. \tag{5} \]

Taking \(\Delta l\) as the abscissa and \(h\) as the ordinate of the curve \(\Delta l=f(h)\), we obtain a parabola of the form:

\[ y^2=2px. \]

Equality (5) in this case will be represented in the form

\[ h^2=2l\cdot\left(\frac{L-l}{L}\right)\cdot\Delta l, \tag{6} \]

where the parameter is

\[ p=\frac{l(L-l)}{L}. \]

4

Let us determine the influence of the lengths \(l\) and \((L-l)\) on the elongation of a flexible thread. Let us impose the condition: the point \(d\) of application of the force is variable; the height \(h\) and the length of the thread \(L\) are constant quantities,

\[ \Delta l=\overline{ac}+\overline{bc}-\overline{ab}. \]

It is clear that the locus of all points \(C\) that give different values of elongation is an ellipse with foci at the points \(a\) and \(b\). Let us consider Fig. 3.

The fundamental property of an ellipse: “the sum of the distances of any point of the ellipse from the foci is a constant.” From Fig. 3 it follows that

\[ \overline{ac'_1}+\overline{c'_1b} = \overline{ac'_2}+\overline{c'_2b} = \overline{ac}+\overline{cb}. \]

This same equality says that the elongation \(\Delta l=\mathrm{const}\), since we can always find it by subtracting one and the same quantity \(\overline{ab}\) from each sum.

Fig. 3

If we take \(\Delta l=\mathrm{const}\), and since \(\Delta l=\overline{ac}+\overline{cb}-L\), then

\[ \overline{ac}+\overline{cb}=\Delta l+L=\mathrm{const}. \]

Mark the midpoint of the length \(\overline{ab}\) (Fig. 4) by the point \(m\). In this case, since

\[ \overline{ap}+\overline{pb}=L+\Delta l,\qquad \text{but } \overline{ap}=\overline{pb}, \]

it follows that

\[ \overline{ap}=\frac{L+\Delta l}{2}, \tag{7} \]

\[ mp=\sqrt{(\overline{ap})^2-(\overline{am})^2} = \sqrt{\left(\frac{L+\Delta l}{2}\right)^2-\left(\frac{L}{2}\right)^2} = \frac{1}{2}\sqrt{2\Delta l\cdot L+\Delta l^2}. \tag{8} \]

Fig. 4

It is easy to show that the segment \(\overline{Am}=\overline{ap}=\dfrac{L+\Delta l}{2}\). Indeed, for the points \(C,p,A\), on the basis of the property of the ellipse we have

\[ \overline{ac}+\overline{bc} = \overline{ap}+\overline{pb} = \overline{Aa}+\overline{Ab}. \]

For the middle point of the ellipse,

\[ \overline{Aa}+\overline{Ab}=2\overline{ap}=L+\Delta l=2\overline{am}+\Delta l. \]

From Fig. 4 it follows that

\[ \overline{Ab}-\overline{Aa}=2\overline{am}, \]

Solving these two equalities jointly, we obtain

\[ \overline{Aa}=\frac{\Delta l}{2}. \]

Consequently,

\[ \overline{Am}=\overline{Aa}+\overline{am}=\frac{\Delta l}{2}+\frac{L}{2}=\frac{L+\Delta l}{2}. \]

Thus, for our ellipse the semiaxes are

\[ r_1=\frac{L+\Delta l}{2}, \]

\[ r_2=\frac{1}{2}\sqrt{2\Delta l L+\Delta l^2}. \]

The equation of the ellipse will be:

\[ \frac{x^2}{\left(\dfrac{L+\Delta l}{2}\right)^2} + \frac{y^2}{\left(\dfrac{1}{2}\sqrt{2\Delta l L+\Delta l^2}\right)^2} \tag{9} \]

On the basis of equation (5)

\[ \Delta l=\frac{h^2}{2}\cdot\frac{L}{l(L-l)}. \]

If \(h\) and \(L\) are constants, then \(\Delta l\) depends only on \(l\) and \(L-l\). Let us call \(l(L-l)=\eta\), and, differentiating, we obtain:

\[ \frac{d\eta}{dl}=L-2l \]

or, treating it as a differential,

\[ d\eta=dl(L-l)+l\,d(L-l)=dl(L-l)-l\,dl. \]

For \(L=2l\), or \(l=\dfrac{L}{2}\);

\[ \frac{d\eta}{dl}=0;\qquad \frac{d^2\eta}{dl^2}=-2. \]

Fig. 5

Thus the question is one of a minimum. In this case

\[ \Delta l_{\min} = \frac{h^2}{2}\cdot\frac{2l}{l(2l-l)} = \frac{h^2}{l} = \frac{2h^2}{L}. \]

For all other positions, when \(l \gtrless \dfrac{L}{2}\), the values of \(\Delta l\) will increase. Figure 5 shows the graph of the increase of \(\Delta l\) as a function of \(l\).

Let us determine the influence of the fixed length of a flexible thread on the elongation. For this purpose we assume that the height \(h\) does not change, and only the length of the flexible thread \(L\) changes, i.e., in correspondence with this the applied force and \(\Delta l\) increase or decrease.

If, in this case, the distance \(l\) is preserved, then \(\Delta l\) depends only on

\[ \frac{L}{L-l}, \]

which is always \(>1\) and the closer to unity, the larger the chosen value of \(L\).

Let us call

\[ \frac{L}{L-l}=y, \]

then

\[ \frac{1}{y}=\frac{L-l}{L}=1-\frac{l}{L}. \]

As \(L\) increases, the quantity \(\frac{l}{L}\) decreases, while \(1-\frac{l}{L}\) increases and \(y\) decreases; conversely, when \(1-\frac{l}{L}\) decreases with an increase of \(\frac{l}{L}\) (when \(L\) decreases), then \(y\) increases.

If now the quantity \(l\) changes with the change of \(L\), but in such a way that each time its ratio to the length \(L\) remains constant, then \(\Delta l\) is independent of

\[ \frac{L}{l(L-l)}, \]

where \(l=\frac{L}{h}\) for \(h=\mathrm{const}\), and

\[ y=\frac{L}{\dfrac{L}{h}\left(L-\dfrac{L}{h}\right)} =\frac{h}{L\left(1-\dfrac{1}{h}\right)} =\frac{h^2}{h-1}\cdot\frac{1}{L}. \]

We see that in this case as well, with an increase in the length of the flexible thread, the elongation decreases. It may be concluded that if the length of the flexible thread \(L\) increases or decreases, then for a constant value of \(l\), or for a proportional change of \(l\) and \(L\), the elongation decreases or increases.

Fig. 6

On the basis of formula (5) it may be concluded that, for constant \(h\), the elongation increases if the point of application of the force moves away toward the points of fastening of the flexible thread, i.e.,

\[ \Delta l_1>\Delta l_{\mathrm{avg}} \]

\[ \Delta l_2>\Delta l_{\mathrm{avg}} \]

where \(\Delta l_{\mathrm{avg}}\) is the elongation for the case when the force is applied at the middle of the length of the flexible thread; hence

\[ \frac{\Delta l_{\mathrm{avg}}}{\Delta l_1} = \frac{ \dfrac{h^2}{2}\cdot\dfrac{L}{l_1(L-l_1)} }{ \dfrac{h^2}{2}\cdot\dfrac{L}{l(L-l)} } = \frac{l_1(L-l_1)}{l(L-l)}. \]

If, with the same length \(l\), we move the fastening point \(v\) in the direction \(ab\) to the right or to the left, then the quantity \(L\) changes (Fig. 6).

Move point \(b\) by an amount \(s\).

For \(l=L+s\),

\[ \frac{\Delta l}{\Delta l_1}=y=\frac{l_1(L-l_1)}{l(L-l)} \]

will be replaced by

\[ y_1=\frac{l_1(L+s-l_1)}{l(L+s-l)} = \frac{l_1}{l}\cdot \frac{(L-l_1)+s}{(L-l)+s}. \]

The expressions for \(y\) and \(y_1\) contain identical quantities \(\frac{l_1}{l}\) (when comparing them, these may be discarded) and different quantities

\[ \frac{L-l_1}{L-l} \quad\text{and}\quad \frac{L-l_1+s}{L-l+s}, \]

which must be compared.

If \(s>0\), i.e., point \(b\) is shifted to the right and \(L\) increases, then the ratio will increase and, conversely, when \(m<0\), it will decrease.

6

Let us investigate the influence of raising or lowering one of the points of attachment of a flexible thread on its elongation.

If one of the points of attachment of the flexible thread, for example \(b\), is raised or lowered, then the perpendicularity of the lines \(\overline{ab}\) and \(\overline{cd}\) will no longer hold.

In order to draw any conclusions in this case, it is necessary to consider some properties of the hyperbola.

Let a hyperbola be constructed in a rectangular coordinate system,

\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1. \tag{10} \]

\(A\) and \(B\) are the foci of the two branches of the hyperbola; \(S\) and \(S_1\) are their vertices.

Fig. 7

It is known that \(SS_1=2a\), where \(a=\overline{OS}\);

\[ AB=2l; \]

\[ l^2=a^2+b^2; \]

\[ b^2=l^2-a^2. \]

It is also known that a property of the hyperbola is the following:

\[ 2a = Aq - Bq. \]

Take, somewhere on the ordinate of the point \(q\), a new point \(p\). The value of the ordinate of this point is \(\eta\), and the abscissa \(\xi = x\). Through the point \(p\) we draw a new hyperbola II, whose foci coincide with the foci \(A\) and \(B\) of hyperbola I.
The equation of hyperbola II will be:

\[ \frac{\xi^2}{a_1^2} - \frac{\eta^2}{b_1^2} = 1, \]

where \(a_1\) and \(b_1\) are the major and minor semiaxes of the hyperbola, and \(\xi\) and \(\eta\) are its current coordinates.

Comparing the two hyperbolas, we obtain

\[ a_1^2 + b_1^2 = a^2 + b^2 = l^2,\quad \text{and also } \xi = x \]

and

\[ \frac{\xi^2}{a_1^2} - \frac{\eta^2}{b_1^2} = \frac{x^2}{a_1^2} - \frac{\eta^2}{b_1^2} = \frac{x^2}{a_1^2} - \frac{\eta^2}{l^2 - a_1^2} = 1, \]

whence

\[ \eta^2 = \left(\frac{x^2}{a_1^2} - 1\right)\left(l^2 - a_1^2\right), \tag{11} \]

similarly, from equality (10), we obtain

\[ y^2 = \left(\frac{x^2}{a^2} - 1\right)\left(l^2 - a^2\right). \tag{12} \]

Comparing the right-hand sides of expressions 11 and 12, we can conclude

\[ a > a_1,\quad y < \eta \]

and conversely, when

\[ \eta \gtrless y,\quad \text{then } a_1 \lessgtr a. \]

If \(a_1 \lessgtr a\), then likewise \(2a_1 \lessgtr 2a\), but

\[ 2a_1 = Ap - Bp, \]

\[ 2a = Aq - Bq. \]

Thus, for \(\eta \gtrless y\), we shall have:

\[ Ap - Bp \lessgtr Aq - Bq \tag{13} \]

From this one can make, in general, the following conclusion: if we have two fixed points \(A\) and \(B\) in a plane and take, in this same plane, a third point \(q\), which we connect by rays with \(A\) and \(B\), then, when it is displaced in the direction perpendicular to \(AB\), the difference \(Ap - Bq\) will be the smaller, the greater the distance of \(q\) from the straight line \(AB\); this difference will be a maximum when \(q\) lies on \(AB\), and will be equal to zero when \(q\) is infinitely distant from \(AB\), for in that case \(Aq \parallel Bq\).

7

Let us consider the question of determining the tension of a flexible thread when some concentrated load is applied to it. For this it is necessary to recall that, before the application of the load, the thread, being in the horizontal position \(ab\), was already stressed by the tension of its ends, and when a new force is applied, an additional tension of the thread is obtained.

Let, before application of the load, the tension of the thread have been \(H_0\), and let the additional tensions at the ends of the thread be \(P_1\) and \(P_2\); then the total tension will be:

\[ S_1=P_1+H_0;\qquad S_2=P_2+H_0 \]

\[ \Delta l_1=h\operatorname{tg}\frac{\alpha}{2}\simeq \frac{h^2}{2l};\qquad \Delta l_2=h\operatorname{tg}\frac{\beta}{2}\simeq \frac{h^2}{2(L-l)}. \]

Fig. 8

Considering the problem in the region of elastic deformations, we have

\[ S_1=H_0+P_1=\frac{EF}{l_0}\,\Delta l_{\text{total}}, \]

where \(l_0\) is the length of the thread before application of the load \(H_0\); \(\Delta l_{\text{total}}\) is the total elongation under the action of the force \(S_1\).

Under the influence of only one force \(H_0\), the absolute elongation of the thread will be:

\[ \frac{L-L_0}{L_0}, \]

therefore

\[ \Delta l_{\text{total}}=\frac{L-L_0}{L_0}\cdot l_0+\Delta l_1 =\frac{L-L_0}{L}\,l_0+\frac{h^2}{2l}, \]

Then, respectively,

\[ S_1=\frac{EF}{l_0}\left(L-L_0\right)+\frac{EF}{l_0}\cdot\frac{h^2}{2l}. \tag{14} \]

With some approximation it may be assumed that

\[ \frac{h}{l_0}=\frac{h}{l}=\operatorname{tg}\alpha;\qquad \text{then } P_1=\frac{EF}{2}\cdot\frac{h^2}{l^2} \]

and, analogously,

\[ P_2=\frac{EF}{2}\cdot\frac{h^2}{(L-l)^2}. \]

Hence we obtain

\[ P_1:P_2=(L-l)^2:l^2 \]

The additional tensions in the thread are directly proportional to the square of the deflection and inversely proportional to the square of the corresponding lengths of the thread.

Before the load is applied, the tension in the parts of the flexible thread was the same; therefore the ratio of the tensions will be:

\[ \frac{S_1}{S_2}=\frac{H_0}{H_0}=1 \]

After the load is applied,

\[ \frac{S_1}{S_2} = \frac{H_0+P_1}{H_0+P_2} = \frac{H_0+\dfrac{EFh^2}{2l^2}} {H_0+\dfrac{EFh^2}{2(L-l)^2}} . \]

The greater the value of \(h\), the greater are the values of \(P_1\) and \(P_2\), and the closer the ratio \(\dfrac{S_1}{S_2}\) approaches the value \(\dfrac{(L-l)^2}{l^2}\).

8

For the bending of a flexible thread fixed between points \(a\) and \(b\), mechanical work must be performed. Consider the case when the points of the thread \(a\) and \(b\) are fixed and the thread is under a tension \(H_0\):

\[ S_1=\frac{EF}{L_0}(L-L_0)+\frac{EF}{l_0}\cdot\frac{h^2}{2l}; \]

\[ S_2=\frac{EF}{L_0}(L-L_0)+\frac{EF}{L_0-l_0}\cdot\frac{h^2}{2(L-l)}. \]

According to Fig. 8, the vertical force \(V_1\), equal to the sum of the projections of the tensions \(S_1\) and \(S_2\) on the direction \(\overline{cd}\), will be

\[ V_1=S_1\sin\alpha+S_2\sin\beta . \tag{15} \]

Replacing the sines of the angles by tangents, we have:

\[ V_1=S_1\operatorname{tg}\alpha+S_2\operatorname{tg}\beta \tag{16} \]

or

\[ V_1=S_1\frac{h}{l}+S_2\frac{h}{L-l}. \]

Substituting here the values of \(S_1\) and \(S_2\), we obtain

\[ V_1= \left(H_0+\frac{EF}{2}\cdot\frac{h^2}{l^2}\right)\cdot\frac{h}{l} + \left(H_0+\frac{EF}{2}\cdot\frac{h^2}{(L-l)^2}\right)\frac{h}{L-l} \]

or

\[ V_1 = H_0\left(\frac{h}{l}+\frac{h}{L-l}\right) + \frac{EF}{2} \left( \frac{h^3}{l^3} + \frac{h^3}{(L-l)^3} \right), \]

or, finally,

\[ V_1 = H_0\frac{L}{l(L-l)}\cdot h + \frac{EF}{2}\cdot \frac{(L-l)^3+l^3}{l^3(L-l)^3}\cdot h^3 . \tag{17} \]

When the magnitude of the thread lift is increased by \(dh\), we obtain the increment of work

\[ V_1\,dh=S_1\,d\Delta l_1+S_2\,d\Delta l_2, \tag{18} \]

strictly speaking, one should have written

\[ S_1\,d(\overline{ac})+S_2\,d(\overline{bc}), \]

but

\[ d(\overline{ac})=d(l+\Delta l_1)=d\Delta l_1 \]

\[ d(\overline{bc})=d(L-l+\Delta l_2)=d\Delta l_2 \]

Dividing equation (18) by \(dh\), we obtain

\[ V_1=S_1\frac{d\Delta l_1}{dh}+S_2\frac{d\Delta l_2}{dh}. \]

From formula 5 it follows that:

\[ \Delta l_1=\frac{h^2}{2l}; \qquad \Delta l_2=\frac{h^2}{2(L-l)}; \]

therefore,

\[ \frac{d\Delta l_1}{dh}=\frac{h}{l}; \qquad \frac{d\Delta l_2}{dh}=\frac{h}{L-l}; \]

Thus,

\[ V_1=S_1\frac{h}{l}+S_2\frac{h}{L-l}. \]

Introduce the following notation in formula (17):

\[ \frac{H_0L}{l(L-l)}=K_1; \qquad \frac{EF}{2}\cdot \frac{(L-l)^3+l^3}{l^3(L-l)^3}=K_2, \]

then

\[ V_1=K_1h+K_2h^3. \]

The magnitude of the work is

\[ A_1=\int_0^h V_1\,dh=\int_0^h K_1h\,dh+\int_0^h K_2h^3\,dh, \]

\[ A_1=\frac{K_1}{2}h^2+\frac{K_2}{4}h^4. \tag{19} \]

Finally,

\[ A_1=H_0\frac{L}{2l(L-l)}h^2+\frac{EF}{8}\frac{(L-l)^3+l^3}{l^3(L-l)^3}\cdot h^4. \tag{20} \]

When, in the values \(S_1\) and \(S_2\), the quantities \(P_1\) and \(P_2\) predominate over \(H_0\), as is usually the case, the resultant has the direction indicated in Fig. 8. If, however, both partial tensions are equal in magnitude and equal to \(S_0\), then the resultant is inclined to the other side.

Under this assumption, it is evident that \(P_1=P_2=P\) and

\[ S_0=H_0+P=\frac{EF}{L_0}(L-L_0+\Delta l_0) \]

or

\[ S_0=\frac{EF}{L}(L-L_0)+\frac{EF}{L_0}\frac{h^2L}{2l(L-l)} \tag{21} \]

Taking \(L_0=L\) in the second term of this equality, we obtain:

\[ S_0=\frac{EF}{L_0}(L-L_0)+\frac{EF}{2l}\cdot\frac{h^2}{(L-l)} =H_0+\frac{EF}{2}\frac{h^2}{l(L-l)} \tag{22} \]

In this case

\[ V_2dh-S_0(d\Delta l_1+d\Delta l_2)=S_0\,d\Delta l \]

\[ V_2=S_0\frac{d\Delta l}{dh}= \left[H_0+\frac{EF}{2}\cdot\frac{h^2}{l(L-l)}\right]\cdot\frac{d\Delta l}{dh} \]

Accordingly, equation (21) for

\[ \Delta l=\frac{h^{2}}{2}\,\frac{L}{l(L-l)};\qquad \frac{d\Delta l}{dh}=\frac{L}{l(L-l)}\cdot h \]

\[ V_{2}=H_{0}\frac{L}{l(L-l)}\cdot h+\frac{EF}{2}\cdot \frac{L}{l^{2}(L-l)^{2}}\,h^{3} \tag{23} \]

The work in this case will be:

\[ A_{2}=\int_{0}^{h} V_{2}\,dh =\int_{0}^{h} H_{0}\frac{L}{l(L-l)}\cdot h\,dh +\int_{0}^{h}\frac{EF}{2}\cdot \frac{L}{l^{2}(L-l)^{2}}\,h^{3}\,dh \]

or

\[ A_{2}=H_{0}\frac{L}{2l(L-l)}\cdot h^{2} +\frac{EF}{8}\cdot \frac{L}{l^{2}(L-l)^{2}}\cdot h^{4} \tag{24} \]

Comparing 17 and 23, 20 and 24, we see that the forces will be functions of the 1st and 3rd powers of the height \(h\), while the quantities of work depend on the 2nd and 4th powers of the same height.

Earlier we introduced the notations:

\[ K_{1}=\frac{H_{0}L}{l(L-l)} \]

\[ K_{2}=\frac{EF}{2}\cdot \frac{(L-l)^{3}+l^{3}}{l^{3}(L-l)^{3}} \]

\[ K_{3}=\frac{EF}{2}\,\frac{L}{l^{2}(L-l)^{2}} \]

Let us examine the influence of the quantity \(l\), for constant length \(L\), on the coefficients \(K_{1}; K_{2}; K_{3}\).

For the coefficient \(K_{1}\), the fraction \(\frac{1}{l(L-l)}\) will be smaller the smaller \(l\) is; therefore \(K_{1}\) will be larger the smaller \(l\) is, but this will hold only as long as \(l<\frac{L}{2}\), with a maximum at \(l=\frac{L}{2}\).

For the coefficient \(K_{2}\)

\[ K_{2}=\frac{EF}{2}\cdot \frac{(L-l)^{3}+l^{3}}{l^{3}(L-l)^{3}} =\frac{EF}{2}\left[\frac{1}{l^{3}}+\frac{1}{(L-l)^{3}}\right] \]

Taking the first derivative with respect to \(l\), we obtain

\[ \frac{dK_{2}}{dl} =\frac{EF}{2}\left[-\frac{3l^{2}}{l^{6}}+\frac{-3(L-l)^{2}\cdot(-1)}{(L-l)^{6}}\right] =\frac{3}{2}\,EF\left[-\frac{1}{l^{4}}+\frac{1}{(L-l)^{2}}\right], \]

for \(l<L-l\) we have \(l^{4}<(L-l)^{4}\), and

\[ \frac{1}{l^{4}}>\frac{1}{(L-l)^{4}} \quad\text{or}\quad \frac{1}{(L-l)^{4}}-\frac{1}{l^{4}}<0, \]

reaching, at \(l=\frac{L}{2}\), a value equal to zero.

Hence we conclude that the quantity \(K_2\) decreases as \(l\) increases, reaching a minimum at \(l=\dfrac{L}{2}\), since

\[ \frac{dK_2}{dl}<0;\quad a\frac{d^2K_2}{dl^2}=6EF\left[\frac{1}{l^5}+\frac{1}{(L-l)^5}\right]>0 \]

For the coefficient \(K_3\), its value varies in the same way as in the case of \(K_1\), for it depends on \(l(L-l)\).

Let us determine the tension of a flexible thread when there is a deflection in it equal to \(h\). For this, consider, for example,

\[ S_1=\frac{EF}{L_0}(L-L_0)+\frac{EF}{l_0}\cdot\frac{h^2}{2l}. \]

The stress per unit area of the cross section of the flexible thread is

\[ \sigma_{z1}=\frac{S_1}{F}=E\cdot\frac{L-L_0}{L_0}=E\frac{h^2}{2ll_0}. \]

Discarding the initial stress in the flexible thread, we obtain

\[ \sigma_{z1}=\frac{E}{2}\frac{h^2}{l^2}=\frac{E}{2}\operatorname{tg}^2\alpha \quad (\text{if we put } l_0=l) \]

Thus the stress depends on the angle \(\alpha\). Obviously, by analogy,

\[ \sigma_{z2}=\frac{E}{2}\operatorname{tg}^2\beta. \]

Since \(\alpha>\beta\), then \(\operatorname{tg}\alpha>\operatorname{tg}\beta\); \(\sigma_{z1}>\sigma_{z2}\), therefore, when calculating the stress, it is advisable to take the larger angle.

As the value of \(l\) decreases, the value of \(\sigma_z\) will increase rapidly.

For the case when the tension in both parts of the flexible thread is the same, obviously, the stress will be proportional to the total elongation \(\Delta l\), and we would have

\[ \sigma_z=E\frac{\Delta l}{L}=\frac{E}{L}\cdot\frac{h^2L}{2l(L-l)} \quad \text{or} \quad \sigma_z=\frac{E}{2}\operatorname{tg}\alpha\cdot\operatorname{tg}\beta. \]

Conclusion

We have considered a range of questions connected with the theory of flexible threads. The assumption of a horizontal position of the thread before application of the force causing the thread deflection, as well as the constancy of the sag \(h\), when considering the phenomenon within the limits of elastic deformations, gives results close to the actual ones calculated under the assumption of a preliminary sag of the thread. In the problem, the length of the flexible thread is finite. This made it possible to investigate a number of special cases of interest in the practice of calculating flexible threads.