where-am-i-wrong-

Question Number 174758 by Kallu last updated on 10/Aug/22

w h e r e a m i w r o n g ?

Commented by Kallu last updated on 10/Aug/22

w h e r e a m i w r o n g ?

C o r r e c t a n s w e r i s 5 / 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

Σ F_{y} : W_{1} = W_{2} \cos 37 ° + T \cos 53 °

Σ F_{x} : T \sin 53 ° = W_{2} \sin 37 ° \Rightarrow T = \frac{\sin 37 °}{\sin 53 °} W_{2}

\Rightarrow W_{1} = W_{2} \cos 37 ° + \frac{\sin 37 °}{\sin 53 °} \times \cos 53 ° W_{2}

\Rightarrow \frac{W_{1}}{W_{2}} = \cos 37 ° + \frac{\sin 37 °}{\tan 53 °} \approx 1.2521 \approx 1.25 = \frac{5}{4}

Commented by Kallu last updated on 11/Aug/22

T h a n k s I f i g u r e d o u t m y m i s t a k e .

Commented by Tawa11 last updated on 12/Aug/22

Great sir

Facebook Tweet Pin