In the arrangement shown in the fig, the bloc | vedclass

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Question

From Constrain equation
$-b+0+0-b / 2+b \frac{\sqrt{3}}{2}+a-b \frac{\sqrt{3}}{2}=0$
$a=\frac{3 b}{2} \quad 	ext {..(1) }$
$F.B.D$ of $8\,kg$ bloc

Vedclass · Accepted Answer

$a =\frac{30 \sqrt{3}}{23}\,m / s ^2$

Vedclass · Answer

From Constrain equation
$-b+0+0-b / 2+b \frac{\sqrt{3}}{2}+a-b \frac{\sqrt{3}}{2}=0$
$a=\frac{3 b}{2} \quad 	ext {..(1) }$
$F.B.D$ of $8\,kg$ block
$T +\frac{ N \sqrt{3}}{2}=8 b$
$F.B.D$ of $2\,kg$ block
$mg \frac{\sqrt{3}}{2}+\frac{ mg }{2}- T = ma$
$N + mb \frac{\sqrt{3}}{2}=\frac{ mg }{2}$
On solving above four equation, we get
$b =\frac{30 \sqrt{3}}{23 b }$

In the arrangement shown in the fig, the block of mass $m=2\,kg$ lies on the wedge on mass $M=8\,kg$. Find the initial acceleration of the wedge if the surfaces are smooth and pulley and strings are massless.

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