Question Number 120413 by aurpeyz last updated on 31/Oct/20
Commented by JDamian last updated on 31/Oct/20
I cannot understand how the solution is unique when the first force is not clearly defined
Answered by TANMAY PANACEA last updated on 31/Oct/20
$${F}_{\mathrm{4}} =−\left({F}_{\mathrm{1}} +{F}_{\mathrm{2}} +{F}_{\mathrm{3}} \right) \\ $$$$=−\left(\mathrm{20}{i}+\mathrm{50}{cos}\mathrm{100}\:{i}+\mathrm{50}{sin}\mathrm{100}\:{j}+\mathrm{150}{cos}\mathrm{240}\:{i}+\mathrm{150}{sin}\mathrm{240}{j}\right) \\ $$$$=−{i}\left(\mathrm{20}+\mathrm{50}{cos}\mathrm{100}+\mathrm{150}{cos}\mathrm{240}\right)−{j}\left(\mathrm{50}{sin}\mathrm{100}+\mathrm{150}{sin}\mathrm{240}\right) \\ $$$$\mid{F}_{\mathrm{4}} \mid=\sqrt{\left(\mathrm{20}+\mathrm{50}{cos}\mathrm{100}+\mathrm{150}{cos}\mathrm{240}\right)^{\mathrm{2}} +\left(\mathrm{50}{sin}\mathrm{100}+\mathrm{150}{sin}\mathrm{240}\right)^{\mathrm{2}} } \\ $$$${tan}\theta=\frac{\mathrm{50}{sin}\mathrm{100}+\mathrm{150}{sin}\mathrm{240}}{\mathrm{20}+\mathrm{50}{cos}\mathrm{100}+\mathrm{150}{cos}\mathrm{240}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{calculate}} \\ $$