# Isn't friction on an incline coefficient of friction times the Reactional force?

Question - "As shown in figure a body of mass 1 kg is shifted from A to D on inclined planes by applying a force slowly such that the block is always in contact with the plane surfaces. Neglecting the jerks experienced at C and B, what is the total work done by the force?"

Given - $$\mu$$AB = 0.1 $$\mu$$BC = 0.2 $$\mu$$CD = 0.4

My approach was to simply calculate the frictional forces by using $$\mu mg\cos\theta$$ and multiplying them by their respective distances covered in each part. After that, I calculated the gain in potential energy.

But when I check the solutions to the problem, it is stated that the work done by friction is $$\mu mgl$$ in each case.

Shouldn't the Frictional force be $$\mu R$$ and then we substitute the Reactional force $$R$$ as $$R\cos\theta$$?

• What is $l$ and how does it compare to “distance covered”? – dmckee --- ex-moderator kitten Aug 7 '19 at 17:10
• L is the respective length of each portion AB, BC, CD. – Robin Singh Aug 7 '19 at 18:03
• I asked for a particular reason. For instance the distance traveled along segment BA is $2\sqrt{2} = 2/(\cos \pi/4)$. Notice the cosine in the denominator? If the solution key is using $l$ to mean "the horizontal length of each segment" then $l$ absorbs the cosine factor that you're looking for. This is an example of why clear communication is essential in technical subjects and of why the meaning of symbols should be made explicit by (or at least immediately after) their first use. – dmckee --- ex-moderator kitten Aug 7 '19 at 18:31

## 2 Answers

I believe @dmckee is correct.

The distance $$l$$ in each case has to be the horizontal component of each part of the path. You know that no friction work is done on vertical components of the path because there’s no force normal to the surface (cosine 90). Only on the horizontal components of the path is the force normal to the path (cosine 0). Thus the total friction work done is $$umgl$$ where $$l$$ is the horizontal component of each part of the path and u the coefficient of friction of each part.

Hope this helps

• By horizontal do you mean parallel to the path? – Aaron Stevens Aug 7 '19 at 19:24
• @Aaron Stevens by horizontal I mean the projections of AB, BC, and CD on the horizontal axis – Bob D Aug 7 '19 at 19:28
• I don't understand why only the horizontal components matter. Friction will act in the opposite direction of the displacement. So both friction and displacement have horizontal and vertical components. Why do we only care about the horizontal component? – Aaron Stevens Aug 7 '19 at 19:29
• I understand that you are saying that $l$ should be equal to $L \cos\theta$ where $L$ is the length from one point to another point, but your explanation of it confuses me. – Aaron Stevens Aug 7 '19 at 19:34
• @Aaron Stevens. Each section Of the path can be resolved into a vertical and horizontal component. What is the component of the weight of the mass normal to each vertical component? Zero. Therefore friction force for each vertical component is zero. That leaves only the horizontal components for friction work. – Bob D Aug 7 '19 at 19:35

Along each path (A to B, B to C, or C to D) the frictional force is indeed has magnitude of $$f=\mu mg\cos\theta$$. Since the frictional force is constant along each path, and because the displacement $$\Delta r$$ is opposite the force along each path, we know that the work done by friction is $$W_f=f\Delta r=-\mu mg\cos\theta\Delta r$$

Now, the only way to reconcile this with the answer you give is if $$l=\Delta r\cos\theta$$, i.e. $$l$$ is the horizontal component of the displacement ($$2$$, $$2$$, and $$1$$ for each path respectively).