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Question Number 10451 by ketto last updated on 09/Feb/17

Answered by amir last updated on 10/Feb/17

Σf_x =10cos45−10cos45+0=0  Σf_y =2×10sin45−9=10(√2)−9#5.41  R=(√((Σf_x )^2 +(Σf_y )^2 ))=(√(0^2 +5.4^2 1))=5.41■

$$\Sigma{f}_{{x}} =\mathrm{10}{cos}\mathrm{45}−\mathrm{10}{cos}\mathrm{45}+\mathrm{0}=\mathrm{0} \\ $$$$\Sigma{f}_{{y}} =\mathrm{2}×\mathrm{10}{sin}\mathrm{45}−\mathrm{9}=\mathrm{10}\sqrt{\mathrm{2}}−\mathrm{9}#\mathrm{5}.\mathrm{41} \\ $$$${R}=\sqrt{\left(\Sigma{f}_{{x}} \right)^{\mathrm{2}} +\left(\Sigma{f}_{{y}} \right)^{\mathrm{2}} }=\sqrt{\mathrm{0}^{\mathrm{2}} +\mathrm{5}.\mathrm{4}^{\mathrm{2}} \mathrm{1}}=\mathrm{5}.\mathrm{41}\blacksquare \\ $$

Answered by ridwan balatif last updated on 10/Feb/17

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