Two blocks of masses 50 kg and 30 kg connected by a massless string pass over a light frictionless pulley and rest on two smooth planes inclined at angles 30° and 60° respectively with the horizontal. Determine the acceleration of the two blocks and the tension in the string. Take g = 10 ms^–2.

Suppose the mass of 50 kg slides down with an acceleration a. The forces acting on the two blocks are shown in figure. The components of the two weights perpendicular to the inclined planes are balanced by the normal reactions R₁ and R₂.

The tension T of each part of the string is same and also the acceleration a of each block is same.

∴ 50g sin 30° − T = 50a    … (i)

and  T − 30g sin 60° = 30a    … (ii)

Adding (i) and (ii), we get

(50 sin 30° − 30 sin 60°)g = (50 + 30)a

or  a = [(50 × 0.5 − 30 × 0.866) × 10] / 80 = −0.12 ms⁻²