Bagi yang menggunakan smartphone, jika persamaan terpotong, bisa dilihat dengan cara swap ke kiri. Terima kasih :)
Diagram benda bebas sistem,
Gambar 1
Tinjau gaya pada silinder dalam arah sumbu \(\displaystyle x\),
\begin{align*}
\Sigma F_{x} & =0\\
f_{1} -N_{2} & =0\\
f_{1} & =N_{2}\\
\mu N_{1} & =N_{2} \tag{1}
\end{align*}
Tinjau gaya pada silinder dalam arah sumbu \(\displaystyle y\),
\begin{align*}
\Sigma F_{y} & =0\\
T+N_{1} +f_{2} -Mg & =0\\
T+N_{1} +f_{2} & =Mg\\
T+N_{1} +\mu N_{2} & =Mg \tag{2}
\end{align*}
Tinjau torsi pada silinder,
\begin{align*}
\Sigma \tau & =0\\
f_{1} R+f_{2} R-TR & =0\\
\mu N_{1} +\mu N_{2} & =T \tag{3}
\end{align*}
Tinjau gaya pada balok dalam arah sumbu \(\displaystyle y\),
\begin{align*}
\Sigma F_{y} & =0\\
T-mg & =0\\
T & =mg \tag{4}
\end{align*}
Substitusi (1) ke (3),
\begin{align*}
\mu N_{1} +\mu ( \mu N_{1}) & =T\\
N_{1}\left( \mu ^{2} +\mu \right) & =T\\
N_{1} & =\frac{T}{\mu ^{2} +\mu } \tag{5}
\end{align*}
Substitusi (1), (4). dan (5) ke (2),
\begin{align*}
T+N_{1} +\mu ( \mu N_{1}) & =Mg\\
T+N_{1}\left( 1+\mu ^{2}\right) & =Mg\\
T+\frac{T}{\mu ^{2} +\mu }\left( 1+\mu ^{2}\right) & =Mg\\
T\left( 1+\frac{1+\mu ^{2}}{\mu ^{2} +\mu }\right) & =Mg\\
mg\left(\frac{\mu ^{2} +\mu +1+\mu ^{2}}{\mu ^{2} +\mu }\right) & =Mg\\
\frac{m}{M} & =\textcolor[rgb]{0.82,0.01,0.11}{\frac{\mu ^{2} +\mu }{2\mu ^{2} +\mu +1}} \tag{Jawaban: c}
\end{align*}
