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Question Number 174758 by Kallu last updated on 10/Aug/22
where am i wrong ?
$${where}\:{am}\:{i}\:{wrong}\:?\: \\ $$$$ \\ $$
Commented by Kallu last updated on 10/Aug/22
  where am i wrong ?   Correct answer is 5/4
$$ \\ $$$${where}\:{am}\:{i}\:{wrong}\:?\: \\ $$$${Correct}\:{answer}\:{is}\:\mathrm{5}/\mathrm{4} \\ $$
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
ΣF_y : W_1 =W_2 cos37°+Tcos53°  ΣF_x : Tsin53°=W_2 sin37°⇒T=((sin37°)/(sin53°))W_2   ⇒W_1 =W_2 cos37°+((sin37°)/(sin53°))×cos53°W_2   ⇒(W_1 /W_2 )=cos37°+((sin37°)/(tan53°))≈1.2521≈1.25=(5/4)
$$\Sigma{F}_{\mathrm{y}} :\:{W}_{\mathrm{1}} ={W}_{\mathrm{2}} \mathrm{cos37}°+{T}\mathrm{cos53}° \\ $$$$\Sigma{F}_{{x}} :\:{T}\mathrm{sin53}°={W}_{\mathrm{2}} \mathrm{sin37}°\Rightarrow{T}=\frac{\mathrm{sin37}°}{\mathrm{sin53}°}{W}_{\mathrm{2}} \\ $$$$\Rightarrow{W}_{\mathrm{1}} ={W}_{\mathrm{2}} \mathrm{cos37}°+\frac{\mathrm{sin37}°}{\mathrm{sin53}°}×\mathrm{cos53}°{W}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{{W}_{\mathrm{1}} }{{W}_{\mathrm{2}} }=\mathrm{cos37}°+\frac{\mathrm{sin37}°}{\mathrm{tan53}°}\approx\mathrm{1}.\mathrm{2521}\approx\mathrm{1}.\mathrm{25}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by Kallu last updated on 11/Aug/22
Thanks I figured out my mistake.
$${Thanks}\:{I}\:{figured}\:{out}\:{my}\:{mistake}. \\ $$$$ \\ $$
Commented by Tawa11 last updated on 12/Aug/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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