Question Number 174758 by Kallu last updated on 10/Aug/22
$${where}\:{am}\:{i}\:{wrong}\:?\: \\ $$$$ \\ $$
Commented by Kallu last updated on 10/Aug/22
$$ \\ $$$${where}\:{am}\:{i}\:{wrong}\:?\: \\ $$$${Correct}\:{answer}\:{is}\:\mathrm{5}/\mathrm{4} \\ $$
Commented by Kallu last updated on 10/Aug/22
Commented by Kallu last updated on 10/Aug/22
Answered by Ar Brandon last updated on 10/Aug/22
$$\Sigma{F}_{\mathrm{y}} :\:{W}_{\mathrm{1}} ={W}_{\mathrm{2}} \mathrm{cos37}°+{T}\mathrm{cos53}° \\ $$$$\Sigma{F}_{{x}} :\:{T}\mathrm{sin53}°={W}_{\mathrm{2}} \mathrm{sin37}°\Rightarrow{T}=\frac{\mathrm{sin37}°}{\mathrm{sin53}°}{W}_{\mathrm{2}} \\ $$$$\Rightarrow{W}_{\mathrm{1}} ={W}_{\mathrm{2}} \mathrm{cos37}°+\frac{\mathrm{sin37}°}{\mathrm{sin53}°}×\mathrm{cos53}°{W}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{{W}_{\mathrm{1}} }{{W}_{\mathrm{2}} }=\mathrm{cos37}°+\frac{\mathrm{sin37}°}{\mathrm{tan53}°}\approx\mathrm{1}.\mathrm{2521}\approx\mathrm{1}.\mathrm{25}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by Kallu last updated on 11/Aug/22
$${Thanks}\:{I}\:{figured}\:{out}\:{my}\:{mistake}. \\ $$$$ \\ $$
Commented by Tawa11 last updated on 12/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$