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 164256 by ajfour last updated on 15/Jan/22

	Answered by ajfour last updated on 16/Jan/22

	$A & B r e p r e s e n t p o i n t s a n d a l s o$ $f o r c e s a t t h o s e p o i n t s r e s p e c t i v e l y .$ $q = - (2 a \cos θ - p)$ $2 p = \tan α$ $2 q = - \tan β$ $T o r q u e a b o u t B$ $(A_{y} + B_{y}) a \cos θ = A_{y} (2 a \cos θ) + A_{x} (2 a \sin θ)$ $. . . . (i)$ $A_{x} = B_{x}, A_{x} = A_{y} \tan α = 2 {p A}_{y}$ $B_{x} = B_{y} \tan β = - 2 {q B}_{y}$ $\Rightarrow A_{x} = 2 (2 a \cos θ - p) B_{y}$ $2 a \sin θ = p^{2} - q^{2} \Rightarrow$ $n o w \frac{2 \sin θ}{a} = {(\frac{p}{a})}^{2} - {(2 \cos θ - \frac{p}{a})}^{2}$ $\Rightarrow \frac{2 \sin θ}{a} = 4 (\frac{p}{a}) \cos θ - 4 \cos^{2} θ$ $\Rightarrow \frac{p}{a} = t = \frac{\tan θ}{2 a} + \cos θ$ $f r o m . . . . (i)$ $1 + \frac{2}{2 (2 \cos θ - \frac{p}{a})} = 2 + 4 p \tan θ$ $\Rightarrow 1 + 4 a \tan θ (\frac{\tan θ}{2 a} + \cos θ)$ $= \frac{1}{\cos θ - \frac{\tan θ}{2 a}}$ $\Rightarrow \cos θ - \frac{\tan θ}{2 a} + 2 \tan^{2} θ \cos θ$ $- \frac{\tan^{2} θ}{a} + 4 a \sin θ \cos θ$ $- 2 \sin θ \tan θ = 1$ $\Rightarrow$ $2 a \cos^{3} θ - \sin θ \cos θ + 4 a \sin^{2} θ \cos θ$ $- 2 \sin^{2} θ + 8 a^{2} \sin θ \cos^{3} θ$ $- 4 a \sin^{2} θ \cos θ = 2 a \cos^{2} θ$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum