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F-1-4N-F-2-5N-R-8-32N-




Question Number 208069 by Davidtim last updated on 03/Jun/24
F_1 =4N,   F_2 =5N,   R=8.32N  θ=?
$${F}_{\mathrm{1}} =\mathrm{4}{N},\:\:\:{F}_{\mathrm{2}} =\mathrm{5}{N},\:\:\:{R}=\mathrm{8}.\mathrm{32}{N} \\ $$$$\theta=? \\ $$
Commented by mr W last updated on 03/Jun/24
θ=180°−cos^(−1) ((4^2 +5^2 −8.32^2 )/(2×4×5))=42.125°
$$\theta=\mathrm{180}°−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{8}.\mathrm{32}^{\mathrm{2}} }{\mathrm{2}×\mathrm{4}×\mathrm{5}}=\mathrm{42}.\mathrm{125}° \\ $$
Answered by efronzo1 last updated on 04/Jun/24
 R=(√(4^2 +5^2 +40 cos θ))   (8.32)^2 −41= 40 cos θ    θ= arccos ((((8.32)^2 −41)/(40)))=45.125°
$$\:\mathrm{R}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{40}\:\mathrm{cos}\:\theta} \\ $$$$\:\left(\mathrm{8}.\mathrm{32}\right)^{\mathrm{2}} −\mathrm{41}=\:\mathrm{40}\:\mathrm{cos}\:\theta \\ $$$$\:\:\theta=\:\mathrm{arccos}\:\left(\frac{\left(\mathrm{8}.\mathrm{32}\right)^{\mathrm{2}} −\mathrm{41}}{\mathrm{40}}\right)=\mathrm{45}.\mathrm{125}° \\ $$

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