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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

	$l e t λ = \frac{l}{L}, μ = \frac{m}{M}$ $\frac{x}{\sin α} = \frac{l - x}{\sin β} = \frac{L}{\sin (α + β)}$ $x = \frac{\sin α}{\sin (α + β)} \times L$ $l - x = \frac{\sin β}{\sin (α + β)} \times L$ $\Rightarrow \sin α + \sin β = λ \sin (α + β) . . . (i)$ $\frac{\frac{M}{M + m} \times \frac{L}{2}}{\sin γ} = \frac{h}{\sin α}$ $\frac{\frac{L}{2} + \frac{m}{M + m} \times \frac{L}{2}}{\sin γ} = \frac{h}{\sin β}$ $\Rightarrow \sin α = (1 + 2 μ) \sin β . . . (i i)$ $p u t (i i) i n t o (i) :$ $\sqrt{4 μ (1 + μ) + \cos^{2} α} + \cos α = \frac{2 (1 + μ)}{λ}$ $\sqrt{q^{2} + \cos^{2} α} = p - \cos α$ $\Rightarrow \cos α = \frac{p^{2} - q^{2}}{2 p} = \frac{1 + μ}{λ} - μ λ$ $\Rightarrow α = \cos^{- 1} (\frac{1 + μ}{λ} - μ λ)$ $s i m i l a r l y$ $\Rightarrow \cos β = \frac{1 + μ + μ λ^{2}}{λ (1 + 2 μ)}$ $\Rightarrow β = \cos^{- 1} \frac{1 + μ + μ λ^{2}}{λ (1 + 2 μ)}$ $\Rightarrow γ = \frac{π - (α + β)}{2}$ $\Rightarrow x = \frac{\sin α}{\sin (α + β)} \times L = \frac{l \sin α}{\sin α + \sin β} = \frac{(1 + 2 μ) l}{2 (1 + μ)}$ $\Rightarrow T = \frac{(M + m) g}{2 \cos γ} = \frac{(M + m) g}{2 \sin \frac{α + β}{2}}$ $\Rightarrow θ = \frac{π}{2} - γ = \frac{α + β}{2}$ $E x a m p l e :$ $L = 1 m, l = 2 m \Rightarrow λ = 2$ $M = 20 k g, m = 10 k g \Rightarrow μ = 0.5$ $\Rightarrow α = 104.477 °$ $\Rightarrow β = 28.955 °$ $\Rightarrow x = 1.333 m$ $\Rightarrow T = 163.3 N$
	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

	$Torque about right end of rod,$ $TLsin β = \frac{MgLcos θ}{2}$ $\Rightarrow T = \frac{Mgcos θ}{2 \sin β} . . . (i)$ $Tcos (β - θ) = Tcos (α + θ)$ $\Rightarrow β - θ = α + θ . . . . (iia)$ $\Rightarrow θ = \frac{α + β}{2} . . . . . (iib)$ $xcos (β - θ) + (l - x) \cos (α + θ) = Lcos θ$ $. . . . (iii)$ $(l - x) \sin (α + θ) = xsin (β - θ) + Lsin θ$ $. . . . . (iv)$ $From Torque about pulley point,$ $\frac{\frac{L}{2} \cos θ - (l - x) \cos (α + θ)}{(l - x) \cos (α + θ)} = \frac{m}{M}$ $\Rightarrow \frac{Lcos θ}{2 (l - x) \cos (α + θ)} = \frac{m + M}{M} . . . (v)$ $from (iii), (iv) using (iia)$ $l \cos (α + θ) = Lcos θ . . . (1)$ $(l - 2 x) \sin (α + θ) = Lsin θ . . . (2)$ $Now (v) implies$ $\frac{l}{2 (l - x)} = \frac{m + M}{M}$ $\Rightarrow M l = 2 (m + M) (l - x)$ $____________________________$ $\Rightarrow x = (1 + \frac{m}{M + m}) \frac{l}{2}$ $____________________________$ $from (1) & (2)$ ${(\frac{Lcos θ}{l})}^{2} + {(\frac{Lsin θ}{l - 2 x})}^{2} = 1$ $\Rightarrow \frac{\cos^{2} θ}{l^{2}} + \frac{{(M + m)}^{2} \sin^{2} θ}{m^{2} l^{2}} = \frac{1}{L^{2}}$ $\Rightarrow \cos^{2} θ [{(\frac{M + m}{m})}^{2} - 1] = {(\frac{M + m}{m})}^{2} - \frac{l^{2}}{L^{2}}$ $\Rightarrow \cos^{2} θ = \frac{{(\frac{M + m}{m})}^{2} - {(\frac{l}{L})}^{2}}{{(\frac{M + m}{M})}^{2} - 1}$ $\sin β = \sin (β - θ + θ)$ $= \sin (α + θ + θ)$ $= (\frac{Lsin θ}{l - 2 x}) \cos θ + (\frac{Lcos θ}{l}) \sin θ$ $= (Lsin θ \cos θ) (\frac{1}{l} - \frac{M + m}{m l})$ $= - \frac{MLsin θ \cos θ}{m l}$ $And as T = \frac{Mgcos θ}{2 \sin β} . . . (i)$ $\Rightarrow T = - \frac{mg l}{2 Lsin θ} (θ < 0)$ $____________________________$ $T = \frac{mg l}{2 L \sqrt{1 - \frac{{(\frac{M + m}{m})}^{2} - {(\frac{l}{L})}^{2}}{{(\frac{M + m}{m})}^{2} - 1}}}$ $& x = (1 + \frac{m}{M + m}) \frac{l}{2}$ $____________________________.$
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