Question-56777

Question Number 56777 by ajfour last updated on 23/Mar/19

Commented by ajfour last updated on 23/Mar/19

Find x and T .

Answered by mr W last updated on 23/Mar/19

Commented by mr W last updated on 24/Mar/19

l e t λ = \frac{l}{L}, μ = \frac{m}{M}

\frac{x}{\sin α} = \frac{l - x}{\sin β} = \frac{L}{\sin (α + β)}

x = \frac{\sin α}{\sin (α + β)} \times L

l - x = \frac{\sin β}{\sin (α + β)} \times L

\Rightarrow \sin α + \sin β = λ \sin (α + β) \dots (i)

\frac{\frac{M}{M + m} \times \frac{L}{2}}{\sin γ} = \frac{h}{\sin α}

\frac{\frac{L}{2} + \frac{m}{M + m} \times \frac{L}{2}}{\sin γ} = \frac{h}{\sin β}

\Rightarrow \sin α = (1 + 2 μ) \sin β \dots (i i)

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

\sqrt{4 μ (1 + μ) + \cos^{2} α} + \cos α = \frac{2 (1 + μ)}{λ}

\sqrt{q^{2} + \cos^{2} α} = p - \cos α

\Rightarrow \cos α = \frac{p^{2} - q^{2}}{2 p} = \frac{1 + μ}{λ} - μ λ

\Rightarrow α = \cos^{- 1} (\frac{1 + μ}{λ} - μ λ)

s i m i l a r l y

\Rightarrow \cos β = \frac{1 + μ + μ λ^{2}}{λ (1 + 2 μ)}

\Rightarrow β = \cos^{- 1} \frac{1 + μ + μ λ^{2}}{λ (1 + 2 μ)}

\Rightarrow γ = \frac{π - (α + β)}{2}

\Rightarrow x = \frac{\sin α}{\sin (α + β)} \times L = \frac{l \sin α}{\sin α + \sin β} = \frac{(1 + 2 μ) l}{2 (1 + μ)}

\Rightarrow T = \frac{(M + m) g}{2 \cos γ} = \frac{(M + m) g}{2 \sin \frac{α + β}{2}}

\Rightarrow θ = \frac{π}{2} - γ = \frac{α + β}{2}

E x a m p l e :

L = 1 m, l = 2 m \Rightarrow λ = 2

M = 20 k g, m = 10 k g \Rightarrow μ = 0.5

\Rightarrow α = 104.477 °

\Rightarrow β = 28.955 °

\Rightarrow x = 1.333 m

\Rightarrow T = 163.3 N

Commented by ajfour last updated on 23/Mar/19

thank you Sir .

Commented by mr W last updated on 23/Mar/19

i t s h o u l d l o o k l i k e t h i s w h e n t h e s y s t e m

i s i n e q u i l i b r i u m .

Commented by mr W last updated on 24/Mar/19

Commented by mr W last updated on 24/Mar/19

y o u a r e r i g h t s i r! n o w c o r r e c t e d .

i n c a s e o f t h r e e f o r c e s i a l w a y s t r y t o

d e t e r m i n e t h e p o i n t o f t h e i r r e s u l t a n t

g e o m e t r i c a l l y .

Commented by ajfour last updated on 24/Mar/19

Torque about right end of rod,

TLsin β = \frac{MgLcos θ}{2}

\Rightarrow T = \frac{Mgcos θ}{2 \sin β} \dots (i)

Tcos (β - θ) = Tcos (α + θ)

\Rightarrow β - θ = α + θ \dots . (iia)

\Rightarrow θ = \frac{α + β}{2} \dots . . (iib)

xcos (β - θ) + (l - x) \cos (α + θ) = Lcos θ

\dots . (iii)

(l - x) \sin (α + θ) = xsin (β - θ) + Lsin θ

\dots . . (iv)

From Torque about pulley point,

\frac{\frac{L}{2} \cos θ - (l - x) \cos (α + θ)}{(l - x) \cos (α + θ)} = \frac{m}{M}

\Rightarrow \frac{Lcos θ}{2 (l - x) \cos (α + θ)} = \frac{m + M}{M} \dots (v)

from (iii), (iv) using (iia)

l \cos (α + θ) = Lcos θ \dots (1)

(l - 2 x) \sin (α + θ) = Lsin θ \dots (2)

Now (v) implies

\frac{l}{2 (l - x)} = \frac{m + M}{M}

\Rightarrow M l = 2 (m + M) (l - x)

____________________________

\Rightarrow x = (1 + \frac{m}{M + m}) \frac{l}{2}

____________________________

from (1) & (2)

{(\frac{Lcos θ}{l})}^{2} + {(\frac{Lsin θ}{l - 2 x})}^{2} = 1

\Rightarrow \frac{\cos^{2} θ}{l^{2}} + \frac{{(M + m)}^{2} \sin^{2} θ}{m^{2} l^{2}} = \frac{1}{L^{2}}

\Rightarrow \cos^{2} θ [{(\frac{M + m}{m})}^{2} - 1] = {(\frac{M + m}{m})}^{2} - \frac{l^{2}}{L^{2}}

\Rightarrow \cos^{2} θ = \frac{{(\frac{M + m}{m})}^{2} - {(\frac{l}{L})}^{2}}{{(\frac{M + m}{M})}^{2} - 1}

\sin β = \sin (β - θ + θ)

= \sin (α + θ + θ)

= (\frac{Lsin θ}{l - 2 x}) \cos θ + (\frac{Lcos θ}{l}) \sin θ

= (Lsin θ \cos θ) (\frac{1}{l} - \frac{M + m}{m l})

= - \frac{MLsin θ \cos θ}{m l}

And as T = \frac{Mgcos θ}{2 \sin β} \dots (i)

\Rightarrow T = - \frac{mg l}{2 Lsin θ} (θ < 0)

____________________________

T = \frac{mg l}{2 L \sqrt{1 - \frac{{(\frac{M + m}{m})}^{2} - {(\frac{l}{L})}^{2}}{{(\frac{M + m}{m})}^{2} - 1}}}

& x = (1 + \frac{m}{M + m}) \frac{l}{2}

____________________________.

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