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Question-56874




Question Number 56874 by Runzzy last updated on 25/Mar/19
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19
F_R =(√((0.2)^2 +(0.3)^2 +2×0.2×0.3cos95^o ))           =0.3457≈0.346  F_x ={0.2cos25^o −0.3cos(180^o −95^o −25^o )}i        =0.2cos25^o −0.3cos60^o         =0.1812615574−0.15        =0.0312615574i  F_y ={0.2sin25^o +0.3sin60^o }×−j        =−(0.3443312735)j  tanβ=(F_y /F_x )
$${F}_{{R}} =\sqrt{\left(\mathrm{0}.\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{0}.\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{3}{cos}\mathrm{95}^{{o}} }\: \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{3457}\approx\mathrm{0}.\mathrm{346} \\ $$$${F}_{{x}} =\left\{\mathrm{0}.\mathrm{2}{cos}\mathrm{25}^{{o}} −\mathrm{0}.\mathrm{3}{cos}\left(\mathrm{180}^{{o}} −\mathrm{95}^{{o}} −\mathrm{25}^{{o}} \right)\right\}{i} \\ $$$$\:\:\:\:\:\:=\mathrm{0}.\mathrm{2}{cos}\mathrm{25}^{{o}} −\mathrm{0}.\mathrm{3}{cos}\mathrm{60}^{{o}} \\ $$$$\:\:\:\:\:\:=\mathrm{0}.\mathrm{1812615574}−\mathrm{0}.\mathrm{15} \\ $$$$\:\:\:\:\:\:=\mathrm{0}.\mathrm{0312615574}{i} \\ $$$${F}_{{y}} =\left\{\mathrm{0}.\mathrm{2}{sin}\mathrm{25}^{{o}} +\mathrm{0}.\mathrm{3}{sin}\mathrm{60}^{{o}} \right\}×−{j} \\ $$$$\:\:\:\:\:\:=−\left(\mathrm{0}.\mathrm{3443312735}\right){j} \\ $$$${tan}\beta=\frac{{F}_{{y}} }{{F}_{{x}} } \\ $$$$\:\:\:\: \\ $$$$\:\:\:\: \\ $$$$\:\: \\ $$$$\:\:\:\: \\ $$

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