Answer to Question #212360 in Physics for wplll

Question

https://docs.google.com/drawings/d/1QNR2JD0TWfX6jTZ0HCIZR2GEXD_mDnLqPwEVQgAh4Bo/edit?usp=sharing

find the tension with the steps

Answer 1

From Newton's second law:

$$\begin{cases} \overrightarrow{T_1}+\overrightarrow{w_1}=m\overrightarrow{a_1} \\ \overrightarrow{T_1}+\overrightarrow{w_2} =2m\overrightarrow{a_1} \end{cases} \rightarrow \begin{cases} T_1-mg=ma_1 \\ 2mg-T_1 =2ma_1 \end{cases}$$ $$2mg-T_1 =2(T_1 -mg) \rightarrow T_1=mg$$

and

$$\begin{cases} \overrightarrow{T_2}+\overrightarrow{w_4}=2m\overrightarrow{a_2} \\ \overrightarrow{T_2}+\overrightarrow{w_3} =(m+m+2m)\overrightarrow{a_2} \end{cases} \rightarrow$$ $$\begin{cases} T_2-2mg=2ma_2 \\ (m+m+2m)g-T_2 =(m+m+2m)a_2 \end{cases}$$ $$4mg-T_2=2(T_2-2mg) \rightarrow T_2=2mg$$

In the first system of equations, we do not take into account the acceleration a₂, since this acceleration determines the motion of the entire system and not of individual parts. We can choose an appropriate coordinate system in which a₂=0.

Answer: $T_1=mg, \space T_2=2mg$