First of all, we can divided that system into two dependent systems. See picture below
System 1
From system 1, we got :
$\begin{eqnarray*} \Sigma F &=& m_{A} . a \\
T - F_{k} &=& m_{A} . a \\
T - W_{A} . \mu _{k} &=& m_{A} . a \\
T - m_{A} . g . \mu _{k} &=& m_{A} . a \cdots (1) \end{eqnarray*}$
System 2
From system 2, we got :
$\begin{eqnarray*} \Sigma F &=& m_{B} . a \\
W_{B} - T &=& m_{B} . a \\
m_{B} . g - T &=& m_{B} . a \cdots (2) \end{eqnarray*}$
Then, eliminate $(1)$ and $(2)$.
If we add them, we will get :
$\begin{eqnarray*} T - m_{A} . g . \mu _{s} &=& m_{A} . a \\
m_{B} . g - T &=& m_{B} . a \\
m_{B} . g - m_{A} . g. \mu_{k} &=& (m_{A} + m_{B}) . a \\
a &=& \frac {(m_{B} - m_{A} . \mu_{k})}{(m_{A} + m_{B})}g\end{eqnarray*} $
Then, we can determine the tension. See it : ...
$\begin{eqnarray*}T - m_{A} . g . \mu _{k} &=& m_{A} . a \\
T &=& m_{A} . g . \mu_{k} + m_{A} . \frac {(m_{B} - m_{A} . \mu_{k})}{(m_{A} + m_{B})}g \\
T &=& m_{A} . g . \frac {\mu_{k}.(m_{A} + m_{B}) + (m_{B} - m_{A} . \mu_{k})} {(m_{A} + m_{B})}\\
T &=& m_{A} . g . \frac {\mu_{k}.m_{A} + \mu_{k}.m_{B} + m_{B} - m_{A} . \mu_{k}} {(m_{A} + m_{B})}\\
T &=& m_{A} . g . \frac {\mu_{k}.m_{B} + m_{B}} {(m_{A} + m_{B})}\\
T &=& \frac {m_{A} . m_{B} . (\mu_{k} + 1)} {(m_{A} + m_{B})}g\\
\end{eqnarray*}$
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ReplyDeleteOke :D
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