What is the acceleration of the garden roller given certain parameters?

[PhDorBust]

A garden roller with external radius a is pulled along a rough horizontal path by force F which acts at a point on its axle and is inclined at an angle alpha with the horizontal.

Given that resultant force is at a right angle to F, what is the acceleration of the roller? Express in terms of radius a, alpha, and radius of gyration k.

My set-up (and failure):

Let H be resultant force, f_s is frictional force

Forces in x-dir:
$$Hsin( \alpha) = Fcos( \alpha) - f_s$$
Forces in y-dir:
$$-Hcos( \alpha) = Fsin( \alpha) - mg$$
Rotational equation of motion:
$$af_s = \frac{mk^2}{a} \ddot{x}$$
Assume no slipping:
$$\ddot{x} = a \ddot{ \theta}$$
Solve for x-acceleration:
$$\ddot{x} = \frac{a^2 F - a^2 mgsin( \alpha)}{mk^2 cos( \alpha)}$$
Correct answer:
$$\ddot{x} = \frac{ga^2 sin( \alpha) cos( \alpha)}{k^2 + a^2 sin^2 (\alpha)}$$

Where am I going wrong? I don't see why F isn't in the answer...

 

[BruceW]

I haven't tried to work out the answer myself yet, but you're right that the 'correct answer' looks ridiculous.
It would predict the same acceleration even when F approaches zero, which is ridiculous.