Question

alguém pode resolver a 8 pra mim?

Answer 1

$\begin{cases}P_A-T1=m_A.a\\P_B+T_1-T_2=m_B.a\\T_2-P_C=m_C.a\end{cases}$

Somando as três equações:
$(P_A-T1)+(P_B+T_1-T_2)+(T_2-P_C)=(m_A.a)+(m_B.a)+(m_C.a)\\\\P_A+P_B-P_C=m_A.a+m_B.a+m_C.a\\\\m_A.g+m_B.g-m_C.g=a.(m_A+m_B+m_C)\\\\g.(m_A+m_B-m_C)=a.(m_A+m_B+m_C)\\\\10.(3+1-2)=a.(3+1+2)\\\\a=\dfrac{1}{3}\,\text{m.s}^{-2}$

Faltava apenas a aceleração para descobrirmos diretamente as tensões 1 e 2.
$\begin{matrix}P_A-T_1=m_A.a\\\\m_A.g-T_1=m_A.a\\\\T_1=m_A.g-m_A.a\\\\T_1=3.10-3.\dfrac{1}{3}\\\\\boxed{T_1=29\,\text{N}}\end{matrix}\qquad\qquad\qquad\begin{matrix}T_2-P_C=m_C.a\\\\T_2-m_C.g=m_C.a\\\\T_2=m_C.g+m_C.a\\\\T_2=2.10+2.\dfrac{1}{3}\\\\\boxed{T_2\approx20{,}7\,\text{N}}\end{matrix}$