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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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