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Question Number 190903 by Spillover last updated on 13/Apr/23

Answered by mr W last updated on 14/Apr/23

m_P a=m_P g sin α−T   ...(i)  m_Q a=T−m_Q g sin β   ...(ii)  (i):(ii):  (m_P /m_Q )=((m_P g sin α−T)/(T−m_Q g sin β))  T=((m_P m_Q (sin α+sin β)g)/(m_P +m_Q ))  T=((0.65×0.65 (((63)/(65))+((16)/(63)))×10)/(0.65+0.65))=3.95 N✓

$${m}_{{P}} {a}={m}_{{P}} {g}\:\mathrm{sin}\:\alpha−{T}\:\:\:...\left({i}\right) \\ $$$${m}_{{Q}} {a}={T}−{m}_{{Q}} {g}\:\mathrm{sin}\:\beta\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right):\left({ii}\right): \\ $$$$\frac{{m}_{{P}} }{{m}_{{Q}} }=\frac{{m}_{{P}} {g}\:\mathrm{sin}\:\alpha−{T}}{{T}−{m}_{{Q}} {g}\:\mathrm{sin}\:\beta} \\ $$$${T}=\frac{{m}_{{P}} {m}_{{Q}} \left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right){g}}{{m}_{{P}} +{m}_{{Q}} } \\ $$$${T}=\frac{\mathrm{0}.\mathrm{65}×\mathrm{0}.\mathrm{65}\:\left(\frac{\mathrm{63}}{\mathrm{65}}+\frac{\mathrm{16}}{\mathrm{63}}\right)×\mathrm{10}}{\mathrm{0}.\mathrm{65}+\mathrm{0}.\mathrm{65}}=\mathrm{3}.\mathrm{95}\:{N}\checkmark \\ $$

Commented by Spillover last updated on 14/Apr/23

nice solution.thanks

$${nice}\:{solution}.{thanks} \\ $$

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