Questão 4 10 Mecanica

F_{z} = 0 - T_{y} - T'_{y} - P_{m} T_{cxo} - T_{cxo} - mg = 0 (T - T') cos \varphi = mg T - T' = \frac{mg}{\cos \theta} Em (M) P/ Z/X0 x Tx se anula Em y 2T_{y} - Mg = 0 2 T_{cxo} = Mg cos \theta = \frac{Mg}{2T} \cos \theta = \frac{Mg}{2T} Substituicao T' em III \cos \theta = \frac{Ng}{\frac{Mmw l^{2}}{(2m+M)}} \cos \theta = \frac{g(c+m+M)}{mw^{2}l} (T+T') = m w^{2}r ? (T+T') = mwk T T = mw \sin \theta = \sin \theta T + T = m w^{2}l (T + T') = mw ^2 l m\frac{mg}{\mu g} Mg = \frac{Mg}{2T} 2Tm + T' \left(IV\right) IV m I \frac{g M}{n} T = \frac{Mm w l^{2}}{2(m+M)} (a) Substituicao T' em IV T = Hmw l^{3} 2(m+M) T = \frac{pmn w^{2}(2m+M)}{2cm+M} (b) T = m w^{2} (2m+M + M) T_{0} = \frac{nw l (2m+M)}{2(l+M)} T_{0} = \frac{m w^{2}l (2m+M + M)}{2 (l+M)}