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.

Figure

Answer 1

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