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



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



Hence, the acceleration of mass


1