Sum

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

\[M = 2a = 2 \times \frac{g}{6} = \frac{g}{3} \left (\text{ up the plane }\right) .\]

Concept: Newton’s Second Law of Motion

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