[tex]Dane:\\h = 19 \ cm = 0,19 \ m\\m = 75 \ kg\\d_{l} = 900\frac{kg}{m^{3}}\\d_{w} = 1000\frac{kg}{m^{3}}\\Szukane:\\S=?[/tex]
Rozwiązanie
Z warunków pływania
[tex]F_{w} = F_{c}+F_{l}\\\\F_{w} = d_{w}\cdot g\cdot V_{l}\\F_{c} = m\cdot g\\F_{l} = m_{l}\cdot g\\\\d_{w}\cdot g\cdot V_{l} = m\cdot g + m_{l}\cdot g \ \ /:g\\\\d_{w}\cdot V_{l} = m + m_{l}\\\\V_{l} = S\cdot h\\m_{l} = V\cdot d_{l} = S\cdot h\cdot d_{l}\\\\d_{w}\cdot S\cdot h = m+d_{l}\cdot S\cdot h\\\\d_{w}\cdot S\cdot h - d_{l}\cdot S\cdot h = m\\\\S\cdot h(d_{w}-d_{l}) = m \ \ /:h\\\\S = \frac{m}{h(d_{w}-d_{l})}[/tex]
[tex]S = \frac{75 \ kg}{0,19 \ m(1000\frac{kg}{m^{3}}-900\frac{kg}{m^{3}})}\\\\\boxed{S_{min} \approx 3,95 \ m^{2}}[/tex]