Dua buah benda A dan B yang massanya masing-masing 2 kg dan 6 kg digantungkan pada sebuah katrol berbentuk silinder pejal (M = 4 kg).
Tinjau benda A
\begin{equation*}
\begin{split}
\Sigma F & = m \:.\: a \\\\
T_1 - w_A & = m_A \:.\: a \\\\
T_1 - 20 & = 2 \:.\: a \\\\
T_1 & = 20 + 2a \quad {\color {red} \dotso (1)}
\end{split}
\end{equation*}
Tinjau benda B
\begin{equation*}
\begin{split}
\Sigma F & = m \:.\: a \\\\
w_B - T_2 & = m_B \:.\: a \\\\
60 - T_2 & = 6 \:.\: a \\\\
T_2 & = 60 - 6a \quad {\color {red} \dotso (2)}
\end{split}
\end{equation*}
Tinjau katrol
\begin{equation*}
\begin{split}
\Sigma tau & = I \:.\: \alpha \\\\
(T_2 - T_1) \:.\: R & = \frac 12 \:.\: M \:.\: R^2 \:.\: \frac {a}{R} \\\\
T_2 - T_1 & = \frac 12 \:.\: M \:.\: a \\\\
T_2 - T_1 & = \frac 12 \:.\: 4 \:.\: a \\\\
T_2 - T_1 & = 2a \quad {\color {red} \dotso (3)}
\end{split}
\end{equation*}
Substitusi persamaan (1) dan (2) ke dalam persamaan (3)
\begin{equation*}
\begin{split}
T_2 - T_1 & = 2a \\\\
60 - 6a - (20 + 2a) & = 2a \\\\
40 - 8a & = 2a \\\\
40 & = 10a \\\\
a & = 4 \text{ m/s}^2
\end{split}
\end{equation*}
Tegangan tali
\begin{equation*}
\begin{split}
T_1 & = 20 + 2a = 20 + 2 \:.\: 4 = 28 \text{ N} \\\\
T_2 & = 60 - 6a = 60 - 6 \:.\: 4 = 36 \text{ N}
\end{split}
\end{equation*}
