# About the work of the friction force [closed]

This will surely be a stupid question, but it bugs me.

Let's consider an inclined plane with friction.

We all know that the friction force is given by

$$f = \mu N$$

Where $N = mg\cos\theta$, the normal force.

Now, when I have to deal with non conservation of energy, I set $\mathcal{L} = -\Delta E$, but here is my doubt.

$\mathcal{L}$ is the work done by the friction force, and by definition we have

$$\mathcal{L} = f\cdot s$$

Where $s$ is the displacement. Now the dot $\cdot$ means "time cosine", but the force $f$ itself is $\mu N$ which has a cosine within. Hence

$$\mathcal{L} = \mu m g s \cos^2\theta$$

But it's wrong, since we know it to be

$$\mathcal{L} = \mu m g s \cos\theta$$

Where is the other cosine?

Thank you and sorry for this stupid question.

• Where did the 2nd cosine come from? Dot product means times cosine of the angle between $f$ and $s$. But they are anti-parallel, so this cosine = -1. – sammy gerbil Dec 16 '17 at 20:08
• @sammygerbil I wrote it. $$\mathcal{L} = f\cdot s = (\mu N)\cdot s = \mu N s \cos \theta = \mu m g s \cos\theta \cos\theta$$ – Les Adieux Dec 16 '17 at 20:09
• Could you explain why the \cdot is replaced by cosine? Especially with the same angle? The way you frame the problem, the ‚thetas‘ are different angles. The first one is the slope on which the item experiences friction, the second angle is the one which describes the how much of the force is directed into the direction the item moves. – rul30 Dec 16 '17 at 20:16
• @rul30 which is the same angle since I'm on an inclined plane! – Les Adieux Dec 16 '17 at 20:20
• It is not necessarily the same angle. The first angle is relative to the gravitational pull, the second angle is relative to the movement. – rul30 Dec 16 '17 at 20:37

• Excuse me but I don't understand. If I am on an inclined plane, of angle $\theta$, becalm mechanics I know that $N$, the normal force or constraint reaction if you prefer, is just "the opposite" of the perpendicular componente of the weight, hence $N = mg \cos\theta$. The friction force we know it to be equal to $f = \mu N$, over an inclined plane with no other inclinations or forces acting on the system. Hence $$L = f\cdot s = fs\cos\theta = \mu N s \cos\theta = \mu m g s \cos\theta \cos\theta$$ Beccasse the angle is the same. The mass slides over the plane! – Les Adieux Dec 16 '17 at 20:57
• @HenryTuring The angle $\alpha$ in $f \cdot s = fs \cos \alpha$ is the angle between $f$ and $s$. In your case, this is just $\pi \neq \theta$ (the displacement vector is downhill and friction vector uphill). So, your formula should read $$L = f \cdot s = fs \cos \alpha = \mu N s \cos \alpha = \mu mg s \cos \theta \cos \alpha = -\mu mg s \cos \theta$$ – CAF Dec 16 '17 at 21:32