Question

Find the acceleration of the block of mass M in the situation shown in figure (5−E15). All the surfaces are frictionless and the pulleys and the string are light.

Answer 1

Given :

We can calculate by the FBD

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

⇒a=g/6

Hence, the acceleration of mass

M=2a=2×g/6=g/3