Question Number 211279 by Davidtim last updated on 04/Sep/24
Commented by Davidtim last updated on 04/Sep/24
$${T}_{\mathrm{1}} =? \\ $$$${T}_{\mathrm{2}} =? \\ $$$${T}_{\mathrm{3}} =? \\ $$
Answered by mr W last updated on 04/Sep/24
$${T}_{\mathrm{1}} \mathrm{sin}\:\theta=\mathrm{20}\:\Rightarrow{T}_{\mathrm{1}} =\frac{\mathrm{20}}{\mathrm{sin}\:\theta} \\ $$$${T}_{\mathrm{3}} \mathrm{sin}\:\mathrm{37}°=\mathrm{30}\:\Rightarrow{T}_{\mathrm{3}} =\frac{\mathrm{30}}{\mathrm{sin}\:\mathrm{37}°}=\frac{\mathrm{30}}{\mathrm{0}.\mathrm{6}}=\mathrm{50}\:{N} \\ $$$${T}_{\mathrm{2}} ={T}_{\mathrm{1}} \mathrm{cos}\:\theta={T}_{\mathrm{3}} \mathrm{cos}\:\mathrm{37}° \\ $$$$\Rightarrow\frac{{T}_{\mathrm{1}} }{{T}_{\mathrm{3}} }=\frac{\mathrm{cos}\:\mathrm{37}°}{\mathrm{cos}\:\theta}=\frac{\mathrm{20}×\mathrm{sin}\:\mathrm{37}°}{\mathrm{sin}\:\theta×\mathrm{30}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{20}×\mathrm{tan}\:\mathrm{37}°}{\mathrm{30}}=\frac{\mathrm{20}×\mathrm{0}.\mathrm{6}}{\mathrm{30}×\mathrm{0}.\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}},\:\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{\mathrm{20}×\sqrt{\mathrm{5}}}{\mathrm{1}}=\mathrm{20}\sqrt{\mathrm{5}}\:{N} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\mathrm{50}×\mathrm{0}.\mathrm{8}=\mathrm{40}\:{N} \\ $$
Commented by Davidtim last updated on 04/Sep/24
$${thanks}\:{dear} \\ $$