Question-52938

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 \dots (i)

x \sin γ = s \cos α

\Rightarrow 5 x \sin γ = 3 s \dots (i i)

s \sin α = L \sin θ

\Rightarrow 4 s = 25 a \sin θ \dots (i i i)

x \cos γ + s \sin α = 4 a

\Rightarrow 5 x \cos γ + 4 s = 20 a \dots (i v)

b \tan β = 4 a \dots . (v)

(7 a - x) \cos β = b \dots . (v i)

N \cos α + T \cos γ + R + T \sin β = M g

\dots . . (v i i)

N \sin α + T \sin γ = T \cos β \dots (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) \dots (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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