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Question Number 99887 by mr W last updated on 23/Jun/20

Commented by mr W last updated on 23/Jun/20

$A d d i t i o n a l q u e s t i o n :$ $F i n d t h e m i n i m u m i n i t i a l v e l o c i t y v$ $s u c h t h a t t h e s m a l l b l o c k c a n r e a c h$ $t h e r i g h t e n d o f t h e b i g g e r b l o c k .$
Commented by Dwaipayan Shikari last updated on 23/Jun/20

$T i m e = (\sqrt{\frac{4 M l}{(M + m) μ g}})$ ${v e l o c i t y}_{m i n} = \frac{1}{2} \sqrt{\frac{(M + m) μ g l}{M}}$
Commented by Dwaipayan Shikari last updated on 23/Jun/20

$S i r i h a v e n o t d o n e i t p r o p e r l y . I a m t r y i n g .$
Commented by Dwaipayan Shikari last updated on 23/Jun/20
Sir can you show your prove?
Commented by mr W last updated on 24/Jun/20

$i g o t a s o l u t i o n w h i c h i t h i n k i s$ $c o r r e c t . s e e b e l o w .$
	Answered by ajfour last updated on 24/Jun/20

	Commented by ajfour last updated on 24/Jun/20
	$frame of reference: bigger block −μ(M+m)g+(μ/2)(mg)=−MA mA=((μmg)/(2M))(2M+m) f=((μmg)/2) mA−f=ma a=A−((μg)/2) = ((μg)/2){2+(m/M)−1} a=((μg(M+m))/(2M)) l=(1/2)at^2 ⇒ t=(√((4Ml)/(μ(M+m)g))) still thinking about v_(min) .....$
	$f r a m e o f r e f e r e n c e : b i g g e r b l o c k$ $- μ (M + m) g + \frac{μ}{2} (m g) = - M A$ $m A = \frac{μ m g}{2 M} (2 M + m)$ $f = \frac{μ m g}{2}$ $m A - f = m a$ $a = A - \frac{μ g}{2} = \frac{μ g}{2} {2 + \frac{m}{M} - 1}$ $a = \frac{μ g (M + m)}{2 M}$ $l = \frac{1}{2} {a t}^{2} \Rightarrow t = \sqrt{\frac{4 M l}{μ (M + m) g}}$ $s t i l l t h i n k i n g a b o u t v_{m i n} . . . . .$
	Answered by mr W last updated on 24/Jun/20

	$a, u = a c c . & v e l o c i t y o f m a s s m$ $A, U = a c c . & v e l o c i t y o f m a s s M$ $m a = - \frac{μ}{2} m g$ $\Rightarrow a = - \frac{μ g}{2}$ $u = v - \frac{μ g}{2} t ⩾ 0$ $M A = \frac{μ m g}{2} - μ (M + m) g$ $\Rightarrow A = - μ g (1 + \frac{m}{2 M})$ $U = v - μ g (1 + \frac{m}{2 M}) t ⩾ 0$ $Δ a = a - A = \frac{μ g}{2} (1 + \frac{m}{M})$ $L = \frac{1}{2} (Δ a) t^{2}$ $\Rightarrow t = \sqrt{\frac{2 L}{Δ a}} = 2 \sqrt{\frac{L}{μ g (1 + \frac{m}{M})}} . . . (i)$ $(i) i s v a l i d o n l y i f$ $U = v - (2 + \frac{m}{M}) \sqrt{\frac{μ g L}{1 + \frac{m}{M}}} ⩾ 0$ $v ⩾ v_{2} = (2 + \frac{m}{M}) \sqrt{\frac{μ g L}{1 + \frac{m}{M}}}$ $t h a t m e a n s i f v ⩾ v_{2}, t h e s m a l l e r b l o c k$ $r e a c h e s t h e r i g h t e n d o f t h e b i g g e r$ $b l o c k b e f o r e t h e b i g g e r b l o c k s t o p s .$ $i f v < v_{2}, t h e b i g g e r b l o c k s s t o p s b e f o r e$ $t h e s m a l l b l o c k r e a c h e s t h e r i g h t e n d .$ $a s s u m e t h e s m a l l b l o c k m o v e s t h e$ $d i s t a n c e L_{1} o n t h e b i g b l o c k a t t i m e t_{1}$ $a s t h e b i g g e r b l o c k s t o p s . a f t e r t h i s$ $m o m e n t t h e b i g g e r b l o c k r e s t s o n t h e$ $t h e r o a d a n d t h e s m a l l b l o c k m o v e s$ $t h e r e m a i n i n g d i s t a n c e L - L_{1} .$ $U_{1} = v - (2 + \frac{m}{M}) \sqrt{\frac{μ {g L}_{1}}{1 + \frac{m}{M}}} = 0$ $t_{1} = 2 \sqrt{\frac{L_{1}}{μ g (1 + \frac{m}{M})}}$ $\Rightarrow L_{1} = \frac{(1 + \frac{m}{M}) v^{2}}{μ g {(2 + \frac{m}{M})}^{2}}$ $\Rightarrow t_{1} = \frac{2 v}{μ g (2 + \frac{m}{M})}$ $u_{1} = v - \frac{μ g}{2} t_{1} = v - \frac{v}{2 + \frac{m}{M}} = \frac{1 + \frac{m}{M}}{2 + \frac{m}{M}} v$ $s t a r t i n g w i t h t h i s v e l o c i t y t h e s m a l l e r$ $b l o c k s h o u l d b e a b l e t o m o v e t h e$ $r e m a i n i n g d i s t a n c e L - L_{1} .$ $\frac{1}{2} {m u}_{1}^{2} ⩾ \frac{μ m g (L - L_{1})}{2}$ $u_{1}^{2} ⩾ μ g (L - L_{1})$ ${(\frac{1 + \frac{m}{M}}{2 + \frac{m}{M}})}^{2} v^{2} ⩾ μ g [L - \frac{(1 + \frac{m}{M}) v^{2}}{μ g {(2 + \frac{m}{M})}^{2}}]$ $\frac{{(1 + \frac{m}{M})}^{2} + 1 + \frac{m}{M}}{{(2 + \frac{m}{M})}^{2}} v^{2} ⩾ μ g L$ $\frac{1 + \frac{m}{M}}{2 + \frac{m}{M}} v^{2} ⩾ μ g L$ $\Rightarrow v ⩾ \sqrt{(1 + \frac{1}{1 + \frac{m}{M}}) μ g L}$ $i . e . v_{m i n} = \sqrt{(1 + \frac{1}{1 + \frac{m}{M}}) μ g L} < v_{2}$ $s u m m a r y :$ $i f 0 < v < \sqrt{(1 + \frac{1}{1 + \frac{m}{M}}) μ g L},$ $s m a l l b l o c k {c a n}^{'} t r e a c h t h e r i g h t e n d,$ $b o t h b l o c k s s t o p e d m o v i n g b e f o r e .$ $i f \sqrt{(1 + \frac{1}{1 + \frac{m}{M}}) μ g L} ⩽ v < (2 + \frac{m}{M}) \sqrt{\frac{μ g L}{1 + \frac{m}{M}}}$ $s m a l l b l o c k r e a c h e s t h e r i g h t e n d,$ $b u t t h e b i g b l o c k s t o p e d m o v i n g b e f o r e .$ $i f v ⩾ (2 + \frac{m}{M}) \sqrt{\frac{μ g L}{1 + \frac{m}{M}}}$ $s m a l l b l o c k r e a c h e s t h e r i g h t e n d a n d$ $t h e b i g b l o c k i s a l s o i n m o t i o n a t t h i s$ $m o m e n t .$
	Commented by ajfour last updated on 24/Jun/20

	$g r e a t w o r k s i r, i s h a l l t r y t o$ $f o l l o w, a n d i k n e w b e f o r e h a n d$ $i t w o u l d g o a b i t l e n g t h y . .$
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