Question

No arranjo da figura (1), as massas m0, m1 e m2 dos corpos são iguais a m, as massas das polias e dos fios são desprezíveis. Ache a aceleração com a qual o corpo m0 cai, e a tensão do fio ligando os corpos de massa m1 e m2, se o coeficiente de atrito dinâmico entre esses corpos e a superfície horizontal são respectivamente μ1 e μ2

Answer 1

Façamos :

$\text T_2 \to \underline{\text{tra{\c c}{\~a}o entre m}_1 \ \text e \ \text m_2} \\\\ \text T_1 \to \underline{\text{tra{\c c}{\~a}o entre m}_0 \ \text e \ \text m_1}$

Fazendo as forças resultantes nos corpos :

$\underline{\text{Bloco 2}}:\\\\ \text{F}_\text r = \text T_2-\text {Fat}_2 \\\\ \text{m.a} = \text T_2 - \text{m.g.}\mu_2 \\\\ \underline{\text {Bloco 1}} : \\\\ \text{F}_\text r = \text T_1-\text T_2-\text {Fat}_1 \\\\ \text{m.a} = \text T_1-\text T_2-\text{m.g.}\mu_1 \\\\\ \underline{\text{sistema dos Blocos m1 e m2}}: \\\\\ \text{m.a} = \text T_2 - \text{m.g.}\mu_2 \\\\ \underline{\text{m.a} = \text T_1-\text T_2-\text{m.g.}\mu_1} \ \ + \\\\ \text{2.m.a} = \text T_1-\text{m.g}(\mu_1+\mu_2)$

Bloco m0 :

$\text{F}_\text r = \text P -\text T_1 \\\\ \text{m.a}=\text {m.g}-\text T_1 \\\\ \underline{\text{usando a equa{\c c}{\~a}o obtido por m1 e m2 }}: \\\\ \text{m.a}=\text {m.g}-\text T_1 \\\\ \underline{\text{2.m.a}=\text T_1-\text{m.g}(\mu_1+\mu_2)} \ \ + \\\\ \text{3.m.a} = \text {m.g}-\text{m.g}(\mu_1+\mu_2) \\\\\\ \displaystyle \huge\boxed{\text a = \frac{\text g.[ \ 1-(\mu_1+\mu_2) \ ]}{3}\ }\checkmark$

Usando a equação do fio ligando os corpos m1 e m2 :

$\displaystyle \text{m.a} = \text T_2-\text{m.g.}\mu_2 \\\\ \text T_2 = \text{m.a}+\text{m.g.}\mu_2 \\\\ \underline{\text{substituindo o valor de acelera{\c c}{\~a}o}}: \\\\\\ \text T_2=\frac{\text{m.g}.[\ 1-(\mu_1+\mu_2)\ ]}{3}+\text{m.g.}\mu_2 \\\\\\ \text T_2 = \frac{\text{m.g}-\text{m.g.}\mu_1-\text{m.g}.\mu_2+\text{3.m.g.}\mu_2}{3} \\\\\\ \text T_2=\frac{\text{m.g}-\text{m.g.}\mu_1+2.\text{m.g.}\mu_2}{3} \\\\\\ \huge\boxed{\text T_2=\frac{\text{m.g}.[\ 1 -\mu_1+2.\mu_2\ ]}{3}\ }\checkmark$