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Question Number 214395 by mr W last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

$f i n d a c c e l e r a t i o n o f w e d g e A = ?$
	Answered by mr W last updated on 07/Dec/24

	$A = a c c e l e r a t i o n o f M (\leftarrow)$ $a_{x} = a c c e l e r a t i o n o f m (\to)$ $a_{y} = a c c e l e r a t i o n o f m (↓)$ $N = c o n t a c t f o r c e b e t w e e n M a n d m$ $N \sin α = M A = {m a}_{x}$ $\Rightarrow a_{x} = \frac{M A}{m}$ $\Rightarrow N = \frac{M A}{\sin α}$ $m g - N \cos α = {m a}_{y}$ $a_{y} = g - \frac{M A \cos α}{m \sin α}$ $\tan α = \frac{a_{y}}{a_{x} + A} = \frac{g - \frac{M A \cos α}{m \sin α}}{\frac{M A}{m} + A}$ $\Rightarrow A = \frac{g \sin α \cos α}{\frac{M}{m} + \sin^{2} α} = \frac{g \sin α \cos α}{(\frac{M}{m} + 1) - \cos^{2} α}$
	Commented by ajfour last updated on 07/Dec/24

	$y e s r i g h t, i r e m e m b e r t h e r e s u l t t o o .$
	Commented by mr W last updated on 07/Dec/24

	$i n c a s e o f a r o l l i n g o b j e c t, t h e$ $a c c e l e r a t i o n o f w e d g e s h o u l d b e$ $s m a l l e r t h a n t h i s . s o t h e r e s u l t i n$ $y o u r v i d e o$ $A = \frac{g \sin α \cos α}{\frac{3}{2} (\frac{M}{m} + 1) - \cos^{2} α} w i t h r o l l i n g$ $c y c l i n d e r i s b y t r e n d r i g h t .$
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