What is the acceleration of the fixed triangular stand so that the object on it will not slide down.?

A body of mass m is on the fixed triangular stand with an angle theta. The body will slide because it is frictionless so we will move the stand on acceleration a. If we move the triangular stand on horizontal plane so that the object on it will not slide, what is the acceleration of the stand?What is the acceleration of the fixed triangular stand so that the object on it will not slide down.?
ok my mechanics is rusty but I beleive the answer goes like...





Resolve (down the slope) m.g.cos(90-theta) - F.cos(theta)=0





Where F is the force exerted on the stand.





F = m.a and cos(90-theta)=sin(theta) so...





m.g.sin(theta) - m.a.cos(theta) = 0 (divide by m)





g.sin(theta) - a.cos(theta)= 0 (rearrange to get)





a = g.sin(theta)/cos(theta)


(sin(theta)/Cos(theta)= tan(theta))





a = g.tan(theta)





Hope this helps although without the aid of a diagram I don't thinks its too great!!!