Question 12.30: The coefficients of friction between blocks A and C and the ...

We first check that static equilibrium is not maintained:

\begin{aligned}\left(F_{A}\right)_{m}+\left(F_{C}\right)_{m} & =\mu_{s}\left(m_{A}+m_{C}\right) g \\& =0.24(5+10) g \\& =3.6 g\end{aligned}

Since W_{B}=m_{B} g=10 \mathrm{~g}>3.6 \mathrm{~g}, equilibrium is not maintained.

Block A: \quad \Sigma F_{y}: \quad N_{A}=m_{A} g

F_{A}=\mu_{k} N_{A}=0.2 m_{A} g

\overset{+}{\longrightarrow } \Sigma F_{\lambda}=m_{A} a_{A}: \quad T-0.2 m_{A} g=m_{A} a_{A}   (1)

Block C: \quad \Sigma F_{y}: \quad N_{C}=m_{C} g

F_{C}=\mu_{k} N_{C}=0.2 m_{C} g

\overset{+}{\longleftarrow } \Sigma F_{x}=m_{C} a_{C}: T-0.2 m_{C} g=m_{C} a_{C}   (2)

Block B: \quad+\downarrow \Sigma F_{y}=m_{B} a_{B}

m_{B} g-2 T=m_{B} a_{B}    (3)

From kinematics: \quad a_{B}=\frac{1}{2}\left(a_{A}+a_{C}\right)    (4)

(a) Tension in cord. Given data: m_{A}=5 \mathrm{~kg}

m_{B}=m_{C}=10 \mathrm{~kg}

Eq. (1): T-0.2(5) g=5 a_{A} \quad a_{A}=0.2 T-0.2 g    (5)

Eq. (2): T-0.2(10) g=10 a_{C} \quad a_{C}=0.1 T-0.2 g    (6)

Eq. (3): 10 g-2 T=10 a_{B} \quad a_{B}=g-0.2 T    (7)

Substitute into (4):

(b) Substitute for T into (5), (7), and (6):