Respuesta :
Answer:
T1 = 1862 N
T2 = 1663 N
Explanation:
We can represent the situation with the following figure
The system is in equilibrium, so the sum of vertical and horizontal forces is 0. It means that we can write the following equations
[tex]\begin{gathered} \\ T_1\sin47-T_2\sin55=0 \\ T_1\cos47+T_2\cos55-mg=0 \end{gathered}[/tex]Replacing the values, we get:
[tex]\begin{gathered} T_1(0.731)-T_2(0.819)=0 \\ T_1(0.682)+T_2(0.574)-227(9.8)=0 \\ \\ 0.731T_1-0.819T_2=0 \\ 0.682T_1+0.574T_2-2224.6=0 \end{gathered}[/tex]Now, we can solve the first equation for T1
[tex]\begin{gathered} 0.731T_1=0.819T_2 \\ \\ T_1=\frac{0.819T_2}{0.731} \\ \\ T_1=1.12T_2 \end{gathered}[/tex]Then, replace this equation with the second equation and solve for T2
[tex]\begin{gathered} 0.682T_1+0.574T_2=2224.6 \\ 0.682(1.12T_2)+0.574T_2=2224.6 \\ 0.764T_2+0.574T_2=2224.6 \\ 1.338T_2=2224.6 \\ \\ T_2=\frac{2224.6}{1.338} \\ \\ T_2=1663 \end{gathered}[/tex]Finally, the value of T1 is
[tex]\begin{gathered} T_1=1.12T_2 \\ T_1=1.12(1662.505) \\ T_1=1862 \end{gathered}[/tex]Therefore, the tension of each rope is 1862 N and 1663 N.