6. To lift a load, it is attached to the hook of a crane using slings made of steel cable. The calculated mass of the load is $M=20$ t, the number of slings $n=3$. Each sling forms an angle $\alpha=30^{\circ}$ with the vertical. All slings carry the same load during the lifting of the cargo. According to safety requirements, the tensile strength of the cable should exceed the calculated load by a factor of $k=6$. The cable consists of a very large number of steel threads, each of which can withstand a maximum load of $q=10^{3} \mathrm{H} / \mathrm{mm}^{2}$. Indicate the smallest diameter of the cable that can be used to manufacture the slings, if the available cables have any integer diameter in millimeters. The acceleration due to gravity $g$ is considered to be $10 \mathrm{~m} / \mathrm{s}^{2}$.
Answer: 26 mm

Solution. For each of the $n$ lower tie-downs, the force of the cargo weight is $\frac{P}{n}$. Then the tension force in the tie will be $N=\frac{P}{n \cdot \cos \alpha}$. Therefore, the strength of the rope must be $Q \geq k T=\frac{k P}{n \cdot \cos \alpha}$.

Since the strength of the rope $Q$ is determined by the cross-sectional area of all the threads $S$ and the strength of each thread $q(Q=S \cdot q)$, then $S q \geq \frac{k P}{n q \cdot \cos \alpha} \Rightarrow S \geq \frac{k P}{n q \cdot \cos \alpha}$.

Let's establish the relationship between the cross-sectional area of all the threads $S$ and the cross-sectional area of the rope $A$, assuming the number of threads is very large.

Tile the cross-section of the rope with hexagons of area $\mathrm{S}_{6}$, in which we inscribe circles of area $\mathrm{S}_{0}$. Then $\frac{A}{S}=\frac{S_{6}}{S_{0}}=\frac{6 \cdot 0.5 R \cdot 2 R}{\pi R^{2} \sqrt{3}}=\frac{2 \sqrt{3}}{\pi}$.

For the diameter of the rope $D$, we get

$$
A=\frac{\pi D^{2}}{4}=\frac{2 \sqrt{3}}{\pi} \cdot \frac{k P}{n q \cdot \cos \alpha} \Rightarrow D=\frac{2}{\pi} \cdot \sqrt{\frac{2 \sqrt{3} \cdot k M g}{n q \cdot \cos \alpha}} .
$$

Substituting the numerical values gives $D=\frac{80}{\pi}$. Since this number is between 25 and 26, the smallest diameter of the rope is 26 mm.

Answer to variant 172: 13 mm

Answer to variant 173: 26 mm

Answer to variant 174: 13 mm