It's a good thought to try to take out the components that are complicating the problem and then reexamine it, but in this case it fundamentally changes the problem. The point of wheels on landing gear is to significantly reduce the frictional force between the airplane and the ground.
In order to bring this entire discussion to a point, we must start with the equation:
F=ma (1)
with F being force, m being mass, and a being acceleration. The reason this equation is so important is that the fundamental question is whether or not there is acceleration. With that known, let's breakdown the forces on the airplane. If I were face-to-face with you right now, I would simply draw you a free body diagram, but due to the medium, my explanation will have to due.
A few assumptions: up is up, +j is up, down is down, -j is down, right is forward, +i is forward, left is backwards, -i is backwards, gravity goes down.
Now, we all can agree that there is gravity acting on the airplane and that the acceleration due to gravity is constant at g = -9.81 m/s^2. Using equation 1, the force due to gravity is the mass multiplied by g and is constant, W. The ground exerts a force on the plane equal to the force the plane exerts on ground. This is called the normal force, N. Lift, L, works in the same direction as the weight and normal forces. The only important part of the lift calculation in this problem is that there has to be some sort of freestream velocity, therefore, a positive velocity experienced by the plane will translate into lift. That is everything in the j-direction. Adding all the forces together and using equation 1 gives us equation 2:
maj = Wj + Nj + Lj (2)
In the i-direction there is thrust, T, friction, Ff, and drag, D. Since we are looking the airplane from rest to take-off, drag will be considered negligible since it is dominated by the velocity term which will be relatively small when compared to the friction and thrust terms. Friction is determined by using equation 3:
Ff= -Cr*N (3)
where Cr is the coefficient of roll friction. Cr is determined by the materials involved. It would be very unlikely for it to be anything greater than 0.05, therefore, we can safely say that Ff=-0.05N. The i-direction equation is:
mai = Ffi + Ti = -0.05Ni + Ti (4)
In order for the plane to take off, maj > 0 and therefore, mai must be > 0. While I do not know what the thrust of the hypothetical plane is, I am certain that it is greater than 1/20th of the weight of the plane. If the plane does indeed generate thrust greater than 1/20th of the weight, that will result in acceleration which will steadily increase the velocity creating lift and thus allowing the plane to take off.
I hope that makes sense to everyone. Again, tell me if there is anything that is unclear or you might feel is dubious. Thank you.