Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
Question Number 56874 by Runzzy last updated on 25/Mar/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19
	$F_R =(√((0.2)^2 +(0.3)^2 +2×0.2×0.3cos95^o )) =0.3457≈0.346 F_x ={0.2cos25^o −0.3cos(180^o −95^o −25^o )}i =0.2cos25^o −0.3cos60^o =0.1812615574−0.15 =0.0312615574i F_y ={0.2sin25^o +0.3sin60^o }×−j =−(0.3443312735)j tanβ=(F_y /F_x )$
	$F_{R} = \sqrt{{(0.2)}^{2} + {(0.3)}^{2} + 2 \times 0.2 \times 0.3 c o s 95^{o}}$ $= 0.3457 \approx 0.346$ $F_{x} = {0.2 c o s 25^{o} - 0.3 c o s (180^{o} - 95^{o} - 25^{o})} i$ $= 0.2 c o s 25^{o} - 0.3 c o s 60^{o}$ $= 0.1812615574 - 0.15$ $= 0.0312615574 i$ $F_{y} = {0.2 s i n 25^{o} + 0.3 s i n 60^{o}} \times - j$ $= - (0.3443312735) j$ $t a n β = \frac{F_{y}}{F_{x}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum