# 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.

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 $g=9.8m/s^2, \cos\theta=1/2$.