In the arrangement shown in the figure, a horizontal beam of length $d$ and weight $G$ is placed in a rotating joint built into a vertical wall. At the end of the beam, there is a load $G_{1}$, and it is supported by a wire with a diameter of $2r$, whose end is firmly embedded in the wall at a height $h$ from the joint. The modulus of elasticity of the wire is $E$. Calculate at what value of $h$ the elongation of the wire will be minimal, what this value is, and what the tensile force will be. (Given data: $d=0.5 \mathrm{~m}, G=5 \mathrm{kp}, G_{1}=15 \mathrm{kp}, r=0.1 \mathrm{~cm}$, $\left.E=2 \cdot 10^{6} \mathrm{kpcm}^{-2}.\right)$

![](https://cdn.mathpix.com/cropped/2024_05_02_222ce839f0f914e0279fg-1.jpg?height=387&width=377&top_left_y=365&top_left_x=858)
We assume the following: the rod is connected to the wall with a horizontal axis hinge, so the tensile force in the rod does not act due to the elongation and sagging of the rope, and the angular change due to sagging can be neglected.

![](https://cdn.mathpix.com/cropped/2024_05_02_6b7d17aa4ac057343236g-1.jpg?height=442&width=526&top_left_y=278&top_left_x=794)

Figure 1

The equilibrium equations of forces (Figure 1)

$$
\begin{aligned}
& A_{x}-F \cos \alpha=0 \\
& A_{y}+F \sin \alpha-G-G_{1}^{\prime}=0
\end{aligned}
$$

The equilibrium equation of moments (about point O):

$$
G \cdot \frac{d}{2}+G_{1} \cdot d-F \cdot \sin \alpha \cdot d=0
$$

and

$$
\sin \alpha=\frac{h}{\sqrt{h^{2}+d^{2}}}
$$

Solving the equations:

$$
F=\left(\frac{G}{2}+G_{1}\right) \sqrt{1+\frac{d^{2}}{h^{2}}}
$$

The elongation

$$
\lambda=\frac{F \cdot l}{E \cdot q}, \text { where }
$$

$l=\sqrt{d^{2}+h^{2}}$ is the length of the rope and $q=r^{2} \pi$ is the cross-sectional area. Thus,

$$
\lambda=\frac{\frac{G}{2}+G_{1}}{E \cdot q} \cdot \frac{h^{2}+d^{2}}{h}
$$

At the minimum, the first derivative of the function vanishes. $\lambda=\min$ if

$$
\left(\frac{h^{2}+d^{2}}{h}\right)^{\prime}=1-\frac{d^{2}}{h^{2}}=0, \text { i.e., } h=d
$$

(The value $h=-d$ is physically meaningless.)

$$
\lambda_{\min }=\frac{\frac{G}{2}+G_{1}}{E \cdot q} \cdot 2 d=0.278 \text{ mm}
$$

and the force in the rope corresponding to the minimum elongation

$$
F=\left(\frac{G}{2}+G_{1}\right) \sqrt{2}=24.74 \text{ kp}
$$

Katalin Fittler (Mosonmagyaróvár, Kossuth Gymnasium, 2nd year) László Szabó (Győr, Révai M. Gymnasium, 2nd year)

Remark. The minimum can also be calculated using another method.

![](https://cdn.mathpix.com/cropped/2024_05_02_6b7d17aa4ac057343236g-2.jpg?height=453&width=244&top_left_y=137&top_left_x=927)

Figure 1

From Figure 2, we can write

$$
\frac{C A}{C D}=\frac{C B}{C A}, \quad C D=y=\frac{C A^{2}}{C B}=\frac{h^{2}+d^{2}}{h}
$$

The segment $y$ is cut out from the line at a distance $d$ by the two sides of the right angle rotating around point $A$. It can be shown that $y$ is larger the larger of $C B$ and $D B$ is. $y$ is the smallest when the larger of $C B$ and $D B$ is the smallest, i.e., when they are equal.

Ágnes Fischer (Budapest, Móricz Zs. Gymnasium, 3rd year)

Let $h=d+x$, then it can be shown that the expression $y=\frac{d^{2}+(d+x)^{2}}{d+x}$ has a minimum at $x=0$. Indeed, starting from the fact that $x^{2} \geq 0$ (equality holds when $x=0$).

Add $2 d^{2}+2 d x$ to both sides:

$$
d^{2}+d^{2}+2 d x+x^{2} \geq 2 d^{2}+2 d x
$$

Therefore,

$$
d^{2}+(d+x)^{2} \geq 2 d(d+x)
$$

Since $h=d+x>0$, we can write:

$$
\frac{d^{2}+(d+x)^{2}}{d+x} \geq 2 d
$$

Equality holds when $x=0$, i.e., when $h=d$.