Question Number 190903 by Spillover last updated on 13/Apr/23
Answered by mr W last updated on 14/Apr/23
$${m}_{{P}} {a}={m}_{{P}} {g}\:\mathrm{sin}\:\alpha−{T}\:\:\:…\left({i}\right) \\ $$$${m}_{{Q}} {a}={T}−{m}_{{Q}} {g}\:\mathrm{sin}\:\beta\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right):\left({ii}\right): \\ $$$$\frac{{m}_{{P}} }{{m}_{{Q}} }=\frac{{m}_{{P}} {g}\:\mathrm{sin}\:\alpha−{T}}{{T}−{m}_{{Q}} {g}\:\mathrm{sin}\:\beta} \\ $$$${T}=\frac{{m}_{{P}} {m}_{{Q}} \left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right){g}}{{m}_{{P}} +{m}_{{Q}} } \\ $$$${T}=\frac{\mathrm{0}.\mathrm{65}×\mathrm{0}.\mathrm{65}\:\left(\frac{\mathrm{63}}{\mathrm{65}}+\frac{\mathrm{16}}{\mathrm{63}}\right)×\mathrm{10}}{\mathrm{0}.\mathrm{65}+\mathrm{0}.\mathrm{65}}=\mathrm{3}.\mathrm{95}\:{N}\checkmark \\ $$
Commented by Spillover last updated on 14/Apr/23
$${nice}\:{solution}.{thanks} \\ $$