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Question Number 52938 by ajfour last updated on 15/Jan/19

Commented by ajfour last updated on 16/Jan/19

$F i n d x, t e n s i o n T, a n d θ .$ $A s s u m e f r i c t i o n i s n o t p r e s e n t$ $a n d r o d i s i n e q u i l i b r i u m .$
Commented by ajfour last updated on 16/Jan/19

Commented by ajfour last updated on 16/Jan/19

$\tan α = \frac{4}{3}$ $s \cos α + b = L \cos θ$ $\Rightarrow \frac{3 s}{5} + b = 5 a \cos θ . . . (i)$ $x \sin γ = s \cos α$ $\Rightarrow 5 x \sin γ = 3 s . . . (i i)$ $s \sin α = L \sin θ$ $\Rightarrow 4 s = 25 a \sin θ . . . (i i i)$ $x \cos γ + s \sin α = 4 a$ $\Rightarrow 5 x \cos γ + 4 s = 20 a . . . (i v)$ $b \tan β = 4 a . . . . (v)$ $(7 a - x) \cos β = b . . . . (v i)$ $N \cos α + T \cos γ + R + T \sin β = M g$ $. . . . . (v i i)$ $N \sin α + T \sin γ = T \cos β . . . (v i i i)$ $Σ T o r q u e a b o u t B = 0$ $\Rightarrow M g \cos θ = N \sin (θ + \frac{π}{2} - α)$ $+ T \sin (θ + \frac{π}{2} - γ) . . . (i x)$ $u n k n o w n s b e i n g$ $θ, x b, s, β, γ, R, N, T .$ $S o l v i n g a l l t h e s e e q s . i n m y l i t t l e$ $n o t e b o o k i o b t a i n e d,$ $\frac{x}{a} = \sqrt{\frac{625}{16} \sin^{2} θ - 40 \sin θ + 16}$ $T = \frac{4 M g \cos θ}{4 \cos (γ - θ) + (3 \cos θ + 4 \sin θ) (\cos β - \cos γ)}$ $\tan β = \frac{16}{5 (4 \cos θ - 3 \sin θ)}$ $\tan γ = \frac{15 \sin θ}{16 - 20 \sin θ}$ $w h i l e θ i s f o u n d f r o m$ $\frac{25}{4} \sqrt{\sin^{2} (\sin^{- 1} \frac{4}{5} - θ) + \frac{256}{625}}$ $+ \sqrt{\frac{625}{16} \sin^{2} θ - 40 \sin θ + 16} = 7$ $\Rightarrow θ \approx 31.86 °, x \approx 2.402 a$
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