Question-81804

Question Number 81804 by ajfour last updated on 15/Feb/20

Commented by ajfour last updated on 15/Feb/20

I f t e n s i o n i n s t r i n g, T = \frac{m g}{5}

a n d r o d r e s t s i n e q u i l i b r i u m,

f i n d l e n g t h o f s t r i n g .

Answered by mr W last updated on 15/Feb/20

Commented by mr W last updated on 16/Feb/20

β + θ = \frac{π}{2} - α

\Rightarrow β = \frac{π}{2} - (θ + α)

γ + θ = \frac{π}{2} + α

\Rightarrow γ = \frac{π}{2} - (θ - α)

\frac{B D}{\sin γ} = \frac{D C}{\sin β} = \frac{B C}{\sin 2 θ}

B D = \frac{L \sin γ}{2 \sin 2 θ} = \frac{L \cos (θ - α)}{2 \sin 2 θ}

D C = \frac{L \sin β}{2 \sin 2 θ} = \frac{L \cos (θ + α)}{2 \sin 2 θ}

l = B D + D C = \frac{L [\cos (θ - α) + \cos (θ + α)]}{2 \sin 2 θ}

\Rightarrow l = \frac{L \cos α}{2 \sin θ} \dots (i i i)

B D \sin θ = e \cos α

\frac{L \cos (θ - α) \sin θ}{2 \sin 2 θ} = e \cos α

e = \frac{L \cos (θ - α)}{4 \cos θ \cos α} = \frac{L}{4} (1 + \tan θ \tan α)

l e t T = \frac{m g}{τ} w i t h τ = 5 g i v e n

F = 2 T \cos θ

F (\frac{L}{2} + e) = m g \times \frac{L}{2}

\Rightarrow 3 \cos θ + \sin θ \tan α = τ \dots (i)

\frac{L \sin α}{2} + B D \times \cos θ = h

2 \sin α + \frac{\cos (θ - α)}{\sin θ} = \frac{4 h}{L}

w i t h η = \frac{h}{L}

\Rightarrow 3 \sin α + \frac{\cos α}{\tan θ} = 4 η \dots (i i)

f r o m (i) :

\frac{3}{\sqrt{9 + \tan^{2} α}} \cos θ + \sin θ \frac{\tan α}{\sqrt{9 + \tan^{2} α}} = \frac{τ}{\sqrt{9 + \tan^{2} α}}

\cos (θ - ϕ) = \frac{τ}{\sqrt{9 + \tan^{2} α}} w i t h \tan ϕ = \frac{\tan α}{3}

\Rightarrow θ = \tan^{- 1} (\frac{\tan α}{3}) + \cos^{- 1} (\frac{τ}{\sqrt{9 + \tan^{2} α}})

p u t t h i s i n t o (i i) w e c a n g e t α .

w i t h α a n d θ, w e g e t t h e l e n g t h o f

t h e s t r i n g a c c . t o (i i i) : l = \frac{L \cos α}{2 \sin θ} .

e x a m p l e : τ = \frac{m g}{T} = 3, η = \frac{h}{L} = \frac{4}{5} = 0.80

\Rightarrow α = 39.6988 °

\Rightarrow θ = 30.9364 °

\Rightarrow l = 3.7416 m

===================

i f s t r i n g l e n g t h l i s g i v e n, f i n d T = ?

f r o m (i i i) w e g e t

\frac{\cos α}{\sin θ} = \frac{2 l}{L} = λ

\Rightarrow \sin θ = \frac{\cos α}{λ}

p u t t h i s i n t o (i i) :

3 \sin α + \sqrt{λ^{2} - \cos^{2} α} = 4 η

\sqrt{λ^{2} - 1 + \sin^{2} α} = 4 η - 3 \sin α

8 \sin^{2} α - 24 η \sin α + 16 η^{2} + 1 - λ^{2} = 0

\Rightarrow \sin α = \frac{6 η - \sqrt{2 (2 η^{2} + λ^{2} - 1)}}{4}

w i t h α a n d θ w e c a n g e t f r o m (i) :

τ = 3 \cos θ + \sin θ \tan α

\Rightarrow τ = \frac{1}{λ} (3 \sqrt{λ^{2} - 1 + \sin^{2} α} + \sin α)

e x a m p l e : L = 4 m, h = 3 m, l = 3 m

η = \frac{h}{L} = \frac{3}{4} = 0.75, λ = \frac{2 l}{L} = \frac{6}{4} = 1.5

\sin α = \frac{6 η - \sqrt{2 (2 η^{2} + λ^{2} - 1)}}{4} = 0.580137

\Rightarrow τ = 2.905932

\Rightarrow T = \frac{m g}{τ} = 0.344 m g

Commented by ajfour last updated on 15/Feb/20

g r e a t w a y S i r, i t s r e a l l y n o t s o

c o m p l e x, t h a n k y o u, i t h i n k

m y s o l u t i o n a n s w e r s w e l l t o o .

Commented by mr W last updated on 15/Feb/20

{t h a t}^{'} s s u r e s i r! y o u r s o l u t i o n s o l v e s

t h e p r o b l e m a l s o v e r y w e l l . m a n y

w a y s l e a d t o t h e s a m e g o a l!

Commented by mr W last updated on 16/Feb/20

f o r t h e c a s e : l i s g i v e n, f i n d T = ?

a n e x a c t s o l u t i o n i s p o s s i b l e, s e e a b o v e .

Answered by ajfour last updated on 15/Feb/20

Commented by ajfour last updated on 15/Feb/20

I a m e n l i s t i n g j u s t t h e e q s .

t h a t i h a v e :

h - \frac{L}{2} \sin θ = p \cos θ \dots . (i)

h - L \sin θ = q \cos θ \dots (i i)

p \cos α = q \cos β

p \sin α + q \sin β = \frac{L}{2}

\cos α + 2 \cos β = 5 \cos θ

α - β = 2 θ

(o n e o f t h e m i s i s d e p e n d e n t)

\cos α + \frac{2 p \cos α}{q} = 5 \cos θ

\Rightarrow \cos α = \frac{5 \cos θ}{(1 + 2 p / q)}

\cos β = \frac{5 (p / q) \cos θ}{(1 + 2 p / q)}

p \sqrt{1 - {\frac{5 \cos θ}{(1 + 2 p / q)}}^{2}} +

q \sqrt{1 - {\frac{5 (p / q) \cos θ}{(1 + 2 p / q)}}^{2}} = \frac{L}{2}

w h i l e w e h a v e p (θ) a n d q (θ),

f r o m f i r s t t w o e q s .

________________________

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