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Question Number 186026 by Michaelfaraday last updated on 31/Jan/23

Answered by mr W last updated on 31/Jan/23

$$\left(b\right)$$$$T_{A-B}=\mu m_{A}g=0.35\times 25=8.75N✓$$$$T_{B-C}=\mu m_{A}g+\mu m_{B}g\cos \theta +m_{B}g\sin \theta$$$$=(0.35+0.35\times \frac{4}{5}+\frac{3}{5})\times 25=30.75N$$$$\left(c\right)$$$$m_{C}g=T_{B-C}=30.75N✓$$$$\frac{m_C}{m_B}=\frac{30.75}{25}=1.23$$$$\left(d\right)$$$$(m_B+m_C)a=m_{C}g-m_{B}g\sin \theta -\mu m_{B}g\cos \theta +T_{A-B}$$$$(m_B+m_C)a=m_{C}g-m_{B}g\sin \theta -\mu m_{B}g\cos \theta +\mu m_{A}g$$$$a=\frac{(1.23-\sin \theta -\mu \cos \theta +\mu )g}{1+1.23}$$$$=\frac{1}{2.23}(1.23-\frac{3}{5}-0.35\times \frac{4}{5}+0.35)\times 10$$$$\approx 3.139m/s^2✓$$

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