Find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light.

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Solution

$2 M a + M g sin θ - T = 0$

$\Rightarrow T = 2 M a + M g s i n θ$ ....(i)

Now 2T +2Ma -2Mg = 0

From equation (i),

$2 (2 M a + M g s i n θ) + 2 M a - 2 M g = 0$

$\Rightarrow 4 M a + 2 M g s i n θ + 2 M a - 2 M g = 0$

$\Rightarrow 6 M a + 2 M g s i n 30^{\circ} + 2 M g = 0$

$\Rightarrow 6 M a = M g$

$\Rightarrow a = \frac{g}{6}$

Hence acceleration of mass,

M = 2a = 2 $\times \frac{g}{6}$

= $\frac{g}{3}$ up the plane.

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Find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light.

2Ma+Mg sin θ−T=0 ⇒T=2Ma+Mg sin θ ....(i) Now 2T +2Ma -2Mg = 0 From equation (i), 2(2Ma+Mg sinθ)+2Ma−2Mg=0 ⇒4Ma+2Mg sinθ+2Ma−2Mg=0 ⇒6Ma+2Mg sin30∘+2Mg=0 ⇒6Ma=Mg ⇒a=g6 Hence acceleration of mass, M = 2a = 2 ×g6 = g3 up the plane.