Another Equilibrium Dilemma: Air resistance holding a truck in equillibrium

[Senjai]

**Edit: Solved the problem, need help with some of the concepts though..[/color]

Homework Statement

The diagram shows a cart on a ramp which is fixed to the floor in the back of a truck. Friction in the wheels is negligible. At what rate must the truck accelerate forward for the cart to remain stationary with respect to the ramp. The angle of the incline is 23 degrees from the floor of the back of the truck.

The Attempt at a Solution

Another tricky question which doesn't give me much data, sorry i couldn't post a picture.

If the truck was allowed to move free, the horizontal component of the gravity along the plane would accelerate the truck down the incline.

When i ask myself "what force could hold it in equilibrium" i believe the only answer would be air resistance. So how could i generate enough Air Resistance to hold the cart in place.I stated that:

Fgx = Fairx
i know that Fgx = mg \cdot sin 23

But i know that Air resistance would be exerted directly horizontal (not on the incline) so i know that the horizontal component would have to equal Fgx, but i don't know theta, or how to determine how to retrieve acceleration out of this..

I guessed and figured it was 67 degrees because Fair makes a right angle with Fg, and the theta between Fg and Fgy is 23 degrees.

So i tried:

mg \cdot sin 23 = ma sin 67

Cancelled out M, and solved a to be .. the right answer... I just realized this after i figured out everything haha. What i don't understand is why Fair = ma? The object will be put into equillibrium why are we talking about net force? and why isn't a negative if were talking about the trucks motion which is opposite of Fair?

 

[rock.freak667]

The resultant force is ma, since the cart is moving down the plane,

ma=F_air - mg (I didn't write in the components parallel to the plane)

at constant velocity a = 0 so

F_aircosθ=mgsinθ

 

[Senjai]

haha thank you rocky! And thanks for constantly bailing me out :)