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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 )
FR=(0.2)2+(0.3)2+2×0.2×0.3cos95o=0.34570.346Fx={0.2cos25o0.3cos(180o95o25o)}i=0.2cos25o0.3cos60o=0.18126155740.15=0.0312615574iFy={0.2sin25o+0.3sin60o}×j=(0.3443312735)jtanβ=FyFx

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