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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.

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

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