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Question Number 217565 by mr W last updated on 16/Mar/25

Commented by mr W last updated on 16/Mar/25

$t h e s t r i n g s a r e m a s s l e s s a n d t h e r e$ $i s n o f r i c t i o n b e t w e e n s t r i n g s a n d$ $p u l l e y s .$ $f i n d t h e a c c e l e r a t i o n s o f a l l o b j e c t s .$
	Answered by ajfour last updated on 16/Mar/25
	$Acc. of 5kg block be A. relative to 4kg pulley let acc. of 3kg be a_1 ↓ and that acc. of 4kg block itself be a↓ relative to 5kg block. 3g−T_1 =3(A+a+a_1 ) ...(i) T_1 −2g=2(A+a−a_1 ) ...(ii) T_2 −g=A+a ....(iii) 2T_2 −T_1 =4(a−A) ...(iv) 200−2T_1 =5A ...(v) .................. 4(A+a+g)−2T_1 =8(a−A) 200−2T_1 =5A ⇒ 200−4(A+a+g)=13A−8a ..(1) addingr (i)&(ii) g=5A+5a+a_1 ....(2) uding (ii) in (v) 200−4g−4(A+a−a_1 )=5A ⇒ 200−4g=9A+4a−4a_1 now using (2) 200−4g=9A+4a−4(g−5A−5a) ⇒ 200=29A+24a from (1) {200−4g=17A−4a}×6 1200−24g=102A−24a 200=29A+24a adding above two 1400−24g=131A A=((1400−24g)/(131)) rest can be found now.$
	$A c c . o f 5 k g b l o c k b e A .$ $r e l a t i v e t o 4 k g p u l l e y l e t a c c . o f 3 k g b e a_{1} ↓$ $a n d t h a t a c c . o f 4 k g b l o c k i t s e l f b e a ↓ r e l a t i v e$ $t o 5 k g b l o c k .$ $3 g - T_{1} = 3 (A + a + a_{1}) . . . (i)$ $T_{1} - 2 g = 2 (A + a - a_{1}) . . . (i i)$ $T_{2} - g = A + a . . . . (i i i)$ $2 T_{2} - T_{1} = 4 (a - A) . . . (i v)$ $200 - 2 T_{1} = 5 A . . . (v)$ $. . . . . . . . . . . . . . . . . .$ $4 (A + a + g) - 2 T_{1} = 8 (a - A)$ $200 - 2 T_{1} = 5 A$ $\Rightarrow 200 - 4 (A + a + g) = 13 A - 8 a . . (1)$ $a d d i n g r (i) & (i i)$ $g = 5 A + 5 a + a_{1} . . . . (2)$ $u d i n g (i i) i n (v)$ $200 - 4 g - 4 (A + a - a_{1}) = 5 A$ $\Rightarrow 200 - 4 g = 9 A + 4 a - 4 a_{1}$ $n o w u s i n g (2)$ $200 - 4 g = 9 A + 4 a - 4 (g - 5 A - 5 a)$ $\Rightarrow 200 = 29 A + 24 a$ $f r o m (1)$ ${200 - 4 g = 17 A - 4 a} \times 6$ $1200 - 24 g = 102 A - 24 a$ $200 = 29 A + 24 a$ $a d d i n g a b o v e t w o$ $1400 - 24 g = 131 A$ $A = \frac{1400 - 24 g}{131}$ $r e s t c a n b e f o u n d n o w .$
	Commented by mr W last updated on 17/Mar/25

	$t h a n k s s i r!$ $y o u g o t A = \frac{1160}{131} \approx 8.85 m / s^{2}$ $i g o t a c c . o f 5 k g p u l l e y$ $A_{1} = \frac{5590}{421} \approx 13.28 m / s^{2}$
	Commented by ajfour last updated on 18/Mar/25

	$I l l c h e c k s i r . T h a n k s f o r s o l v i n g t h e$ $p r o j e c t i l e o n e t o o .$
	Commented by Tawa11 last updated on 21/Mar/25

	That first line of this solution shouldn't it be (a + a₁) - A instead of a + a₁ +A
	Commented by mr W last updated on 21/Mar/25

	$h e {d i d n}^{'} t s a y t h a t A i s i n ↑ d i r e c t i o n .$
	Commented by Tawa11 last updated on 22/Mar/25

	$Thanks sir .$
	Commented by Tawa11 last updated on 22/Mar/25

	$Sir, help me see to Q 217797$ $That differential equation .$
	Answered by mr W last updated on 17/Mar/25

	Commented by mr W last updated on 17/Mar/25

	$δ_{1} r e l a t i v e a c c . o f s t r i n g t o p u l l e y 1$ $δ_{2} r e l a t i v e a c c . o f s t r i n g t o p u l l e y 2$ $a_{1} = A_{1} + δ_{1}$ $A_{2} = A_{1} - δ_{1}$ $a_{2} = A_{2} + δ_{2} = A_{1} - δ_{1} + δ_{2}$ $a_{3} = A_{2} - δ_{2} = A_{1} - δ_{1} - δ_{2}$ $F_{2} = m_{2} (g + a_{2}) = m_{2} (g + A_{1} - δ_{1} + δ_{2})$ $F_{2} = m_{3} (g + a_{3}) = m_{3} (g + A_{1} - δ_{1} - δ_{2})$ $F_{1} = m_{1} (g + a_{1}) = m_{1} (g + A_{1} + δ_{1})$ $F_{1} - 2 F_{2} = M_{2} (g + A_{2}) = M_{2} (g + A_{1} - δ_{1})$ $F - 2 F_{1} = M_{1} (g + A_{1})$ $F - 2 m_{1} (g + A_{1} + δ_{1}) = M_{1} (g + A_{1})$ $(M_{1} + 2 m_{1}) A_{1} + 2 m_{1} δ_{1} = F - (M_{1} + 2 m_{1}) g$ $7 A_{1} + 2 δ_{1} = 130 . . . (i)$ $m_{1} (g + A_{1} + δ_{1}) - 2 m_{2} (g + A_{1} - δ_{1} + δ_{2}) = M_{2} (g + A_{1} - δ_{1})$ $(m_{1} - 2 m_{2} - M_{2}) A_{1} + (M_{2} + m_{1} + 2 m_{2}) δ_{1} - 2 m_{2} δ_{2} = (M_{2} - m_{1} + 2 m_{2}) g$ $- 7 A_{1} + 9 δ_{1} - 4 δ_{2} = 70 . . . (i i)$ $m_{2} (g + A_{1} - δ_{1} + δ_{2}) = m_{3} (g + A_{1} - δ_{1} - δ_{2})$ $(m_{2} - m_{3}) A_{1} + (- m_{2} + m_{3}) δ_{1} + (m_{2} + m_{3}) δ_{2} = (m_{3} - m_{2}) g$ $- A_{1} + δ_{1} + 5 δ_{2} = 10 . . . (i i i)$ $δ_{1} = \frac{7800}{421}$ $δ_{2} = \frac{400}{421}$ $A_{1} = \frac{7800}{421} + 5 \times \frac{400}{421} - 10 = \frac{5590}{421}$ $A_{2} = \frac{5590}{421} - \frac{7800}{421} = - \frac{2210}{421}$ $a_{1} = \frac{5590}{421} + \frac{7800}{421} = \frac{13390}{421}$ $a_{2} = - \frac{2210}{421} + \frac{400}{421} = - \frac{1810}{421}$ $a_{3} = - \frac{2210}{421} - \frac{400}{421} = - \frac{2610}{421}$
	Commented by Tawa11 last updated on 18/Mar/25

	$Sir, you assumed the cylinder are rotating here .$ $But Ajfour did not assume the cylinder is rotating .$
	Commented by mr W last updated on 18/Mar/25

	$T h e p u l l e y s {d o n}^{'} t r o t a t e a n d i a l s o$ ${d i d n}^{'} t a s s u m e t h a t t h e y r o t a t e .$
	Commented by Tawa11 last updated on 18/Mar/25

	$Thanks sir .$
	Commented by mr W last updated on 18/Mar/25

	$i f t h e r e i s n o f r i c t i o n b e t w e e n s t r i n g$ $a n d p u l l e y, t h e s t r i n g c a n a l s o$ $m o v e r e l a t i v e l y t o t h e p u l l e y w i t h o u t$ $t h a t t h e p u l l e y r o t a t e s .$ $i n t h e w o r k i n g s a b o v e δ_{1} a n d δ_{2} a r e$ $t h e r e l a t i v e m o t i o n s o f t h e s t r i n g s$ $a l o n g t h e p u l l e y s . t h e r e a r e n o t t h e$ $r o t a t i o n o f t h e p u l l e y s!$
	Commented by mr W last updated on 18/Mar/25

	Commented by Tawa11 last updated on 18/Mar/25

	$Ohh, thanks sir .$ $I appreciate your time .$
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