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Question Number 85606 by Rio Michael last updated on 23/Mar/20

Commented by Rio Michael last updated on 23/Mar/20

$$\mathrm{given}\mathrm{that}\mathrm{three}\mathrm{weights}{\mathrm{W}}_1,{\mathrm{W}}_{2}and5N\mathrm{are}\mathrm{suspended}\mathrm{as}\mathrm{shown}\mathrm{above}.$$$$\mathrm{A}\mathrm{light}\mathrm{inextensible}\mathrm{string}\mathrm{passing}\mathrm{over}\mathrm{smooth}\mathrm{fixed}\mathrm{pulleys}\mathrm{makes}\mathrm{angles}$$$$40°\mathrm{and}60°\mathrm{with}\mathrm{the}\mathrm{vertical}\mathrm{and}\mathrm{all}\mathrm{the}\mathrm{stings}\mathrm{are}\mathrm{taunt}\mathrm{at}\mathrm{point}\mathrm{A}.$$$$\mathrm{calculate}\mathrm{the}\mathrm{weights}{\mathrm{W}}_1\mathrm{and}{\mathrm{W}}_2.$$

Commented by mr W last updated on 23/Mar/20

$$\frac{W_1}{\sin 40°}=\frac{W_2}{\sin 60°}=\frac{G}{\sin 100°}$$$$G=5N$$$$W_1=\frac{5\sin 40°}{\sin 100°}=3.26N$$$$W_2=\frac{5\sin 60°}{\sin 100°}=4.40N$$

Commented by Rio Michael last updated on 23/Mar/20

$$\mathrm{thanks}\mathrm{sir},\mathrm{but}\mathrm{how}\mathrm{did}\mathrm{you}$$$$\mathrm{get}\mathrm{to}\mathrm{using}\mathrm{the}\mathrm{sine}\mathrm{rule}{\mathrm{that}}^{\prime }\mathrm{s}$$$$\mathrm{where}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}.$$

Commented by mr W last updated on 23/Mar/20

$$\overrightarrow{{\mathit{W}}_2}+\overrightarrow{{\mathit{W}}_1}+\overrightarrow{\mathit{G}}=0$$

Commented by mr W last updated on 23/Mar/20

Commented by Rio Michael last updated on 23/Mar/20

$$\mathrm{thanks}\mathrm{sir}$$$$$$

Commented by john santu last updated on 25/Mar/20

$$\frac{w_1}{\sin {140}^o}=\frac{5}{\sin {100}^o}\Rightarrow w_1=\frac{5\sin {140}^o}{\sin {100}^o}$$$$\frac{w_2}{\sin {120}^o}=\frac{5}{\sin {100}^o}\Rightarrow w_2=\frac{5\sin {120}^o}{\sin {100}^o}$$

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