$R$ = Normal reaction force exerted on the particle by the ground
$Fr$ = Frictional contact force between the particle and the ground
All forces are measured in Newtons ($N$)
Using the angles we can calculate the forces acting on the particle.
$$m=1, \ \ \ mg = g = 9.8N$$
Netwton's Third Law of motion states that if A exerts a force magnitude N on B, then B exerts a force of maginitude N on A, but in the opposite direction. Hence:
$$R = 9.8 + 30 \ sin(30)$$ $$R = 24.8N$$
Now, if we also go by Newton's Third Law, then if the particle is pushing forwards on the ground with a force magnitude N, then the ground is pushing backwards on the particle with a force magnitude N, which should be friction. However:
$$30 \ cos(30) = 26.0N$$ $$Fr_{(max)} = R \mu = 2.48N$$ $$Fr_{(max)} < 26.0$$
Due to the limit on friction set by the coefficient of friction between the surface and the particle, the particle accelerates with acceleration $a$. However, this appears to contradict Newton's Third Law, which clearly states that $Fr$ should equal $30 \ cos(30)$.
This is especially strange since $R$ follows Newton's Third Law perfectly without a limit, so why does friction appear to disobey it?
What am I missing here?