A homogeneous flexible rope rests on a wedge whose side edges make angles `alpha` and `beta` with the horizontal. The central part of the rope lies on the upper rib C of the wedge. With what acceleration should the wedge be pulled to the left along the horizontal plane in order to prevent the displacement of the rope with respect to the wedge? [Consider all surfaces to be smooth]

Explanation

Consider the parts of rope on either size of incline as two different elements. Draw FBD. Apply Newton's law on these parts separately

Taking components of acceleration along incline surface,
For left side part of rope: `(mg)/(2) sin alpha -T=m/2 a cos alpha`
For right side part of rope: `T-(mg)/(2) sin beta = m/2 a cos beta`
On solving, we get `a=(g[sin alpha-sin beta])/(cos beta+ cos alpha)`.