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Question Number 110993 by ajfour last updated on 01/Sep/20

Commented by ajfour last updated on 01/Sep/20

$A s s u m e a l l s u r f a c e s f r i c t i o n l e s s .$ $F i n d a c c e l e r a t i o n o f c e n t e r o f m a s s$ $o f t h e s y s t e m .$
	Answered by mr W last updated on 01/Sep/20

	Commented by mr W last updated on 01/Sep/20
	$N sin θ=MA N+mA sin θ=mg cos θ mg sin θ+mA cos θ=ma ⇒g sin θ+A cos θ=a A((M/m)+sin^2 θ)=g sin θ cos θ ⇒A=((g sin θ cos θ)/((M/m)+sin^2 θ))=(dx_M ^2 /dt^2 ) ⇒a=((((M/m)+1)g sin θ)/((M/m)+sin^2 θ))=−(ds^2 /dt^2 ) x_S =(1/(M+m))[Mx_M +m(x_M +s cos θ)] =(1/(M+m))[(M+m)x_M +ms cos θ] y_S =(m/(M+m))s sin θ ⇒a_(S,x) =(d^2 x_M /dt^2 )=(1/(M+m))[(M+m)A−ma cos θ) =(1/(M+m)){(((M+m)g sin θ cos θ)/((M/m)+sin^2 θ))−((m((M/m)+1)g sin θ cos θ)/((M/m)+sin^2 θ))} =0 (as expected) ⇒a_(S,y) =−(d^2 y_S /dt^2 )=(m/(M+m))a sin θ=((g sin^2 θ)/((M/m)+sin^2 θ))$
	$N \sin θ = M A$ $N + m A \sin θ = m g \cos θ$ $m g \sin θ + m A \cos θ = m a$ $\Rightarrow g \sin θ + A \cos θ = a$ $A (\frac{M}{m} + \sin^{2} θ) = g \sin θ \cos θ$ $\Rightarrow A = \frac{g \sin θ \cos θ}{\frac{M}{m} + \sin^{2} θ} = \frac{{d x}_{M}^{2}}{{d t}^{2}}$ $\Rightarrow a = \frac{(\frac{M}{m} + 1) g \sin θ}{\frac{M}{m} + \sin^{2} θ} = - \frac{{d s}^{2}}{{d t}^{2}}$ $x_{S} = \frac{1}{M + m} [{M x}_{M} + m (x_{M} + s \cos θ)]$ $= \frac{1}{M + m} [(M + m) x_{M} + m s \cos θ]$ $y_{S} = \frac{m}{M + m} s \sin θ$ $\Rightarrow a_{S, x} = \frac{d^{2} x_{M}}{{d t}^{2}} = \frac{1}{M + m} [(M + m) A - m a \cos θ)$ $= \frac{1}{M + m} {\frac{(M + m) g \sin θ \cos θ}{\frac{M}{m} + \sin^{2} θ} - \frac{m (\frac{M}{m} + 1) g \sin θ \cos θ}{\frac{M}{m} + \sin^{2} θ}}$ $= 0 (a s e x p e c t e d)$ $\Rightarrow a_{S, y} = - \frac{d^{2} y_{S}}{{d t}^{2}} = \frac{m}{M + m} a \sin θ = \frac{g \sin^{2} θ}{\frac{M}{m} + \sin^{2} θ}$
	Commented by ajfour last updated on 01/Sep/20

	$B e a u t i f u l S i r, u n d e r s t o o d; TH a n KS!$
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