[tex]Dane:\\F = 50 \ N\\S_1 = 5 \ cm^{2}\\S = 4S_1 = 4\cdot5 \ cm^{2} = 20 \ cm^{2} =0,002 \ m^{2}\\Szukane:\\p = ?\\\\p = \frac{F}{S}\\\\p = \frac{50 \ N}{0,002 \ m^{2}}\\\\\boxed{p = 25 \ 000 \ Pa = 25 \ kPa}\\\\(1 \ kPa = 1000 \ Pa)[/tex]
Odp. Ten kot wywiera ciśnienie na podłogę p = 25 kPa.