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Question
Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.
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Solution
The free-body diagram of the system is shown below:
Let acceleration of the block of mass 2M be a.
So, acceleration of the block of mass M will be 2a.
M(2a) + Mgsinθ − T = 0
⇒ T = 2Ma + Mgsinθ ...(i)
2T + 2Ma − 2Mg = 0
From equation (i),
2(2Ma + Mgsinθ) + 2Ma − 2Mg = 0
4Ma + 2Mgsinθ + 2Ma − Mg = 0
6Ma + 2Mgsin30° + 2Mg = 0
6Ma = Mg
\[\Rightarrow a = \frac{g}{6}\]
Hence, the acceleration of mass
\[M = 2a = 2 \times \frac{g}{6} = \frac{g}{3} \left (\text{ up the plane }\right) .\]
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