A four-meter-long rod has a $20 \mathrm{~kg}$ weight attached to one end, and $30,40,50 \mathrm{~kg}$ weights are attached at one, two, and three meters from this end, respectively. At the other end of the rod, a $60 \mathrm{~kg}$ weight is attached. Where should the rod be supported so that it remains in balance? The weight of the rod itself can be neglected.
In equilibrium, the algebraic sum of static moments is $0\left(\sum P p=0\right)$. Let $x$ be the distance from the support point to the end of the rod where the $20 \mathrm{~kg}$ weight hangs, then

$$
\begin{gathered}
20 x+(x-1) 30+(x-2) 40+(x-3) 50+(x-4) 60=0 \\
20 x+30 x-30+40 x-80+50 x-150+60 x-240=0 \\
200 x=500 \\
x=2 \frac{1}{2}
\end{gathered}
$$

The rod must be supported $2 \frac{1}{2}$ meters from the $20 \mathrm{~kg}$ weight, and thus $1 \frac{1}{2}$ meters from the other end, to remain in equilibrium.

(Kántor Nándor, VII. class, Budapest Higher and Real Gymnasium)

The problem was also solved by: Friedmann Bernát, Fröhlich Károly, Geist Emil, Goldstein Zsigmond, Goldziher Károly, Grünhut Béla, Riesz Frigyes, Szabó István, Weisz Lipót.