Newton's law problem on homogeneous flexible rope

[Titan97]

Homework Statement

A homogeneous flexible rope rests on a wedge whose sides make angles α and β with horizontal. The centre of rope lies on C. With what acceleration should the wedge be moved for the rope to stay stationary with respect to wedge? (all surfaces are smooth).

Homework Equations

$F_{ext}=Ma$

The Attempt at a Solution

My first doubt is the direction of tension. I think its horizontal.
If α>β, then the wedge should be accelerated in left direction. Let that acceleration be $a$.
Using Newton's laws and concept of pseudo forces,
$Tcos\alpha+\frac{m}{2}acos\alpha=\frac{m}{2}gsin\alpha$
Similarly,
$Tcos\beta-\frac{m}{2}acos\beta=\frac{m}{2}gsin\beta$
I can easily find $a$ now by eliminating $T$. But is my method correct?

 

[Orodruin]

Tension will always be in the direction of the rope, which makes your method a bit difficult. You cannot assume the tension to change direction at the edge without the edge providing a non negligible force. If you want to do it with tension, the tension shoulf be the same on both sides and in the direction of the rope. However, the easier way is to search for an acceleration where the angles differ from the gravitational acceleration + acceleration vector by the same amount.

 

[Titan97]

I did not understand that. I will try solving using tension.