Odpowiedź:
Minimalna wartość siły naciągu liny to ≈ 37905,72 N
Wyjaśnienie:
[tex]\text{Dane:}\\m = 2000 kg\\\alpha = 15^{\circ}\\\\\overrightarrow{F_{g}} = m*g\\\overrightarrow{F_{g}} = 2000 kg * 9,81 \frac{m}{s^{2}} = 19620 N[/tex]
[tex]\text{Korzystamy z rownowagi momentow sil:}\\M = F * r * \sin (\overrightarrow{r} ; \overrightarrow{v}) \text{, gdzie r = odleglosc od podstawy slupa}\\\\M_{g} = 19620 N * \frac{1}{2} l * \sin (90^{\circ}) = 9810 N * l\\M_{n} = F_{n} * l * \sin (15^{\circ})\\\\M_{g} = M_{n}\\9810 N * l = F_{n} * l * \sin (15^{\circ})\\F_{n} = \frac{9810 N}{\sin (15^{\circ})} = \frac{9810 N}{0,2588} \approx 37905,72 N[/tex]
