Answer
See Answer
Work Step by Step
a) See Image
b) $a=\frac{F}{m}$
$a=\frac{F}{m_A+m_B+m_C}$
c) Each block has the same acceleration
$\sum F_A=m_A\times a_A=\frac{m_A\times F}{m_A+m_B+m_C}$
$\sum F_B=m_B\times a_B=\frac{m_B\times F}{m_A+m_B+m_C}$
$\sum F_C=m_C\times a_C=\frac{m_C\times F}{m_A+m_B+m_C}$
d) $F_{AB}=F-F_A=F\Big(\frac{m_B+m_C}{m_A+m_B+m_C}\Big)$
$F_{BC}=F_{AB}-F_B=F\Big(\frac{m_B+m_C}{m_A+m_B+m_C}\Big)-\frac{m_B\times F}{m_A+m_B+m_C}=\frac{m_C\times F}{m_A+m_B+m_C}$
e) $a=\frac{96.0N}{10.0kg+10.0kg+10.0kg}=3.2\frac{m}{s^2}$
$\sum F_A=\frac{10.0kg\times 96.0N}{30.0 kg}=32.0N$
$\sum F_B=\frac{10.0kg\times 96.0N}{30.0kg}=32.0N$
$\sum F_C=\frac{10.0kg\times 96.0N}{30.0kg}=32.0N$
$F_{AB}=96.0N\Big(\frac{20.0kg}{30.0kg}\Big)=64N$
$F_{BC}=\frac{10.0kg\times 96.0N}{30.0kg}=32N$
