Q:

The applied and friction forces of a particle along a straight wire in horizontal plane

A particle of mass $14~\text{kg}$, slides along a straight wire in a horizontal plane. The coefficient of dynamic friction $\mu_k = 0.6$ The equation of the line of the wire is $y = \sqrt{3}x$ so that the angle between the wire and the X-axis is $60^\circ$.
The particle accelerates with a constant acceleration whose magnitude is $a = 2$. At time $t = 3~\text{s}$ the particle is at $\mathtt{A}$. The acceleration is produced by an applied force $\mathbf{P}$ acting parallel to the X-axis.
Show that the magnitude of $\mathbf{P}$ is $220.8~\text{N}$.

Attempted solution:
$$\mathbf{P} - \mathbf{f} = ma \quad \& \quad \mathbf{f} = \mu_k \cdot R$$.
Now,
$$\mathbf{P} = ma + \mathbf{f}$$
$$\mathbf{f} = 0.6\cdot mg\cos\theta = 41.202$$
So,
$$\mathbf{P} = 14\times 2 + 41.202 =69.202$$
What am I missing here? The applied force is bigger then the friction force as indicated by the first equation and the rest is straight forward.

A:

The correct version:
$$P \cos\theta-F_{fr}=ma.$$
Here $P\cos\theta$ is a projection of $P$ on the direction of motion.
Next we use the expression for the friction force $F=\mu N=\mu mg$ (note that $N=mg$ in this case because the motion is happening on the horizontal plane). Now
$$P \cos\theta-\mu mg=ma.$$
From where we find
$$P=\frac{m(\mu g+a)}{\cos\theta}\approx 220.64 N. $$
Here I used $g=9.8m/s^2, \cos\theta=1/2$.

