P3 - F128 - Discursiva Diurno

QUESTÃO DISCURSIVA P3 DINAMO I = \frac{BM L^3}{2} b) bloco 1. N_1 - M_1g = 0 T_1 - F_at = M_1a T_1 - \mu M_1g = M_1 a (1) bloco 2. N_2 - M_1g\cos \theta = 0 M_2g\sen \theta - T_2 = M_2 a (2) T_2 - T_R = I \alpha T_R - \frac{B M R^2 \alpha}{K} = 0 T_2 - T_1 = \beta M a T_2 - T_1 = \beta M a T_1 = \mu M_1g + M_1a \vert \ M_2g \sen \theta = M_2\alpha - \mu M_1g - M_1a = \beta M a (I) \ T_1 + \mu M_1g + (3\sen \theta - \mu)T_1 \alpha - = \frac{BMR^2 \alpha}{K} v) - T_1 - \mu M_1g = T_1 + \mu M_1g = \frac{M_2 g\theta - \mu M_1g}{\beta M + H_2 + M_1} VERSÃO 1: \beta = \frac{1}{3} M2=3M1 M=4M1 (d) \ a = \frac{3\nu\sen\theta - \mu M_1g}{2M_1 + 3M_1 + M_1} = \frac{3\sen \theta - \mu graph}{6} (e) \ T_1 = \mu M_1g + (3\sen\theta - \mu)\frac{M_1g}{6} = (5\mu + 3sen-\theta\theta)M1\ g T_2 = (5\mu + 3sen\theta)\frac{M_1g}{6} = M_1g \left [3\theta + \mu 3\sen_g\theta\right ] - M_1g[\frac{\mu + 3sen\theta}{2}] VERSÃO 2: p=\frac{1}{2} M2=4M1 M=2M1 (d) \ a = \frac{4\mu_1\cos\theta - \mu M_1g}{M_1+4M_1+M1} = \frac{(4\sen \theta - \mu)g}{6} (e) \ T_1 = \mu M_1g + (\sen \theta - \mu)\frac{M_1g}{6} = (5\mu + \sen \theta) \frac{M_1g}{6} T_2 = \frac{(5\mu + 4\sen \theta)M_1g + (15\mu - \mu)M_1g}{36} = \frac{(3\mu + 9\nu)M_1g}{36} = \frac{(49\sen \theta + 2\mu)M}{36} VERSÃO 3: p=\frac{3}{5} M2=5M1 M=3M1 (d) \ a = \frac{5\mu\sen\theta - \mu M_1g}{2M_1+8M_1 + M_1} = \frac{(5\sen \theta - \mu)g}{8} (e) \ T_1 = \mu M_1g + (5\sen \theta - \mu)\frac{M_1g}{8} = (7 \mu + 5\sen \theta) T_2 = T_1 + \frac{(3\mu\theta)\frac{M_1}{8}}{(1-x + 5\sen\theta) + M_1g} = \frac{M_1g}{8} (5\mu + 15\mu sen \theta) f) Mon. angular. entrando na página ⊗