Sebuah benda dengan massa 10 kg tergantung pada sistem tali di bawah ini.
Tentukan besar tegangan tali.
Diagram gaya
Tinjau benda
\begin{equation*} \begin{split} \Sigma F & = 0 \\\\ T_1 - w & = 0 \\\\ T_1 & = w \\\\ T_1 & = mg \\\\ T_1 & =100 \text{ N} \end{split} \end{equation*}
Tinjau titik pertemuan tali
Kesetimbangan dalam arah X
\begin{equation*} \begin{split} \Sigma F_x & = 0 \\\\ T_2 \cos 45 - T_3 \cos 30 & = 0 \\\\ T_2 \cos 45 & = T_3 \cos 30 \\\\ \frac 12 \sqrt{2} \: T_2 & = \frac 12 \sqrt{3} \: T_3 \\\\ T_2 & = \frac {\sqrt{3}}{\sqrt{2}} \: T_3 \quad {\color {red} \dotso \: (1)} \end{split} \end{equation*}
Kesetimbangan dalam arah Y
\begin{equation*} \begin{split} \Sigma F_y & = 0 \\\\ T_2 \sin 45 + T_3 \sin 30 - T_1 & = 0 \\\\ T_2 \sin 45 + T_3 \sin 30 & = T_1 \\\\ T_2 \:.\: \frac 12 \sqrt{2} + T_3 \:.\: \frac 12 & = 100 \\\\ T_2 \:.\: \sqrt{2} + T_3 & = 200 \quad {\color {red} \dotso \: (2)} \end{split} \end{equation*}
Substitusi persamaan (1) ke persamaan (2)
\begin{equation*} \begin{split} \Sigma F_y & = 0 \\\\ T_2 \:.\: \sqrt{2} + T_3 & = 200 \\\\ \frac {\sqrt{3}}{\sqrt{2}} \: T_3 \:.\: \sqrt{2} + T_3 & = 200 \\\\ T_3 \:.\: \sqrt{3} + T_3 & = 200 \\\\ T_3 \:.\: (\sqrt{3} + 1) & = 200 \\\\ T_3 & = \frac {200}{\sqrt{3} + 1} \quad {\color {blue} \times \frac {\sqrt{3} - 1}{\sqrt{3} - 1}}\\\\ T_3 & = \frac {200 (\sqrt{3} - 1)}{3 - 1} \\\\ T_3 & = \frac {200 (\sqrt{3} - 1)}{2} \\\\ T_3 & = 100 (\sqrt{3} - 1) \text{ N} \end{split} \end{equation*}
Menentukan T2
\begin{equation*} \begin{split} T_2 & = \frac {\sqrt{3}}{\sqrt{2}} \:.\: 100 (\sqrt{3} - 1) \\\\ T_2 & = \frac {100(3 - \sqrt{3})}{\sqrt{2}} \quad {\color {blue} \times \frac {\sqrt{2}}{\sqrt{2}}} \\\\ T_2 & = \frac {100(3 \sqrt{2} - \sqrt{6})}{2} \\\\ T_2 & = 50(3 \sqrt{2} - \sqrt{6}) \text{ N} \end{split} \end{equation*}