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