A gondola lift is planned to be built on the Gellért Hill in Budapest. The peak of the hill is $84 \mathrm{~m}$ higher than its base. The entire length of the straight slope track is $140 \mathrm{~m}$. Two equally weighted cars travel up and down, hanging on a cable through a pulley. Each car can accommodate 40 people. The gondola carries the maximum load when the car going uphill is fully loaded, and the one going downhill is empty. Each car weighs 50 q. The average weight of a person is $70 \mathrm{~kg}$. The rolling friction coefficient on the steel cable is 0.005. Calculate the horsepower of the steam engine needed, if one trip takes one minute. (Neglect the stiffness and weight of the cable, the moment of inertia of the pulley, and air resistance in this approximate calculation.)
The weight of the upper car is $50 \mathrm{q}=5000 \mathrm{~kg}$, and the weight of the lower car with 40 people is $50 \mathrm{q}+70 \cdot 40 \mathrm{~kg}=7800 \mathrm{~kg}$; the component of the weight of the upper car along the inclined plane is:

$$
5000 \cdot \frac{84}{140}=3000 \mathrm{~kg} \text { weight }
$$

and that of the lower car is

$$
7800 \cdot \frac{84}{140}=4680 \mathrm{~kg} \text { weight. }
$$

The two forces, due to the worm gear transmission, act in opposite directions, so that they are in equilibrium at any point on the track:

$$
4680-3000 \mathrm{~kg}=1680 \mathrm{~kg}
$$

of force must be exerted by the engine. This force works over a distance of 140 m, during which

$1680 \cdot 140 \mathrm{~kg} \mathrm{~m}$ of work is done.

This work is completed in $60 \mathrm{~s}$, so the power is

$$
H_{p}=\frac{1680 \cdot 140}{60} \mathrm{kgm} \mathrm{sec}^{-1}
$$

or

$$
H_{p}=\frac{1680 \cdot 140}{60 \cdot 75} \text { horsepower }=52.3 \text { horsepower. }
$$

The engine must also overcome the resistance due to friction. The friction is proportional to the normal force, that is,

$$
S=G \cdot \cos \alpha \cdot \varepsilon
$$

where $S$ is the friction, $G$ is the total weight of the moving mass, $\alpha$ is the angle of the incline, and $\varepsilon$ is the coefficient of friction. Therefore,

$$
S=(5000+7800) \frac{\sqrt{140^{2}-84^{2}}}{140} \cdot 0.005 \mathrm{~kg}
$$

When the engine overcomes the friction $S$ over a distance of $140 \mathrm{~m}$,

$$
S \cdot 140=12800 \sqrt{140^{2}-84^{2}} \cdot 0.005 \mathrm{~kg} \mathrm{~m}
$$

of work is done; since this work is completed in $60 \mathrm{~s}$,

$$
H_{p}=\frac{S \cdot 140}{60 \cdot 75}=\frac{12800 \sqrt{140^{2}-84^{2}} \cdot 0.005}{60 \cdot 75} \text { horsepower }
$$

the engine must have a power of

$$
H_{p}=1.59 \text { horsepower. }
$$

Therefore, the total power required by the engine is

$$
53.9 \text { horsepower. }
$$

(Daniel Barna.)

The problem was also solved by D. Brandt, F. Bojedain, K. Goldziher, and T. Kármán.