The ratio of acceleration of blocks A placed | vedclass

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Question

(a)

$\sum F_y=0$

$\Rightarrow N =10 g \cos 	heta / m _{ A }=10\,kg$

$\sum F_x=m_A \cdot a_A$

$\Rightarrow 10 g \cdot \sin 	heta=10 . a _

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(a)

$\sum F_y=0$

$\Rightarrow N =10 g \cos 	heta / m _{ A }=10\,kg$

$\sum F_x=m_A \cdot a_A$

$\Rightarrow 10 g \cdot \sin 	heta=10 . a _{ A }$

$\Rightarrow a_A=g \sin 	heta---$ (i)

$\sum F_y=0$

$N =20 g \cos 	heta----$ (ii) $/ m _{ B }=20\,kg$

$\sum F _{ x }= m _{ B } a _{ B }$

$\Rightarrow 20\, g \sin 	heta- fk =20 a _{ B }$

$\Rightarrow 20\, g \sin 	heta-\mu_{ k } \cdot 20 g \cos 	heta=20 a _{ B } g$

$f _{ k }=\mu_{ k } \cdot N$

$=\mu_{ k } \cdot 20\, g \cos 	heta$

from $(ii)$

$\Rightarrow \sin 	heta-\mu_{ k } \cdot \cos 	heta=\frac{a_{ B }}{g}$

$\Rightarrow a _{ B }= g \left(\sin 	heta- mu _{ k } \cos 	hetaight)---$ (iii)

$(i) \div (iii):$

$\frac{a_A}{a_B}=\frac{g \sin 	heta}{g\left(\sin 	heta-\mu_k \cos 	hetaight)}=\frac{\sin 	heta}{\sin 	heta\left(1-\mu_{ K } \cdot \cot 	hetaight)}$

$\Rightarrow \frac{2}{1}=\frac{1}{1-\mu_{ k }} \Rightarrow\left(1-\mu_{ k }ight) 2=1$

$	heta=45^{\circ}$

$\cot 45^{\circ}=1$

$\Rightarrow 2-2 \mu_{ k }=1 \Rightarrow \mu_{ k }=0.5$

The ratio of acceleration of blocks $A$ placed on smooth incline with block $B$ placed on rough incline is $2: 1$. The coefficient of kinetic friction between block $B$ and incline is .........

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