\section*{Task 2 - 021022}

In a coal mine, a hoisting installation is to be built for a shaft that is \(800 \mathrm{~m}\) deep. Loads of up to \(12 \mathrm{Mp}\) are to be hoisted.

The hoisting rope consists of steel wires and can withstand a load of \(20 \mathrm{kp}\) per \(\mathrm{mm}^{2}\) cross-sectional area, taking into account the necessary safety.

How large must the metallic cross-sectional area of the rope be so that it can carry both its own weight and the load to be hoisted? (Weight of steel \(\gamma=7.8 \frac{p}{c m^{3}}\) )
}

The rope must carry the weight of \(12 \mathrm{Mp}\), as well as its own weight. The self-weight is given by \(V \cdot \gamma = x \cdot 624 \frac{\mathrm{kp}}{\mathrm{cm}^{2}}\). Here, \(\mathrm{x}\) is the cross-sectional area of the rope in \(\mathrm{cm}^{3}\).

The rope carries \(2000 \frac{\mathrm{kp}}{\mathrm{cm}^{2}}\) per cross-sectional area. Therefore, the rope requires a cross-sectional area of

\[
x=\frac{12000 \mathrm{kp}}{2000 \frac{\mathrm{kp}}{\mathrm{cm}^{2}}-624 \frac{\mathrm{kp}}{\mathrm{cm}^{2}}} \approx 8.721 \mathrm{~cm}^{2}
\]