Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 103: 34

Answer

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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) $a=\frac{96.0N}{10.0kg+10.0kg+10.0kg}=3.2\frac{m}{s^2}$ $\sum F_A=\frac{10.0kg\times 96.0N}{30.0 kg}=32.0N$ $\sum F_B=\frac{10.0kg\times 96.0N}{30.0kg}=32.0N$ $\sum F_C=\frac{10.0kg\times 96.0N}{30.0kg}=32.0N$ $F_{AB}=96.0N\Big(\frac{20.0kg}{30.0kg}\Big)=64N$ $F_{BC}=\frac{10.0kg\times 96.0N}{30.0kg}=32N$
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