How Do You Calculate the Coefficient of Friction on an Inclined Plane?

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Homework Help Overview

The problem involves calculating the coefficient of friction for a block sliding down an inclined plane, where the first half of the plane is frictionless and the second half has friction. The block comes to a stop at the bottom of the slope, and participants are discussing the reasoning behind the calculations and methods used to arrive at the coefficient of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss a method involving balancing forces and constant acceleration to derive the coefficient of friction. Questions arise about the correctness of the method and the assumptions made regarding the symmetry of the problem.

Discussion Status

Some participants express confidence in the explanation provided, while others seek clarification on the proof of the method. There is an ongoing exploration of different ways to express the reasoning behind the calculations, indicating a productive dialogue without a definitive consensus.

Contextual Notes

Participants note the importance of constant acceleration in the problem and question the assumptions made in the original explanation. There is also a mention of needing to prove the reasoning behind the calculations, highlighting the need for thorough understanding rather than just accepting the answer.

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Homework Statement


A block starts sliding on an inclined plane inclined at an angle θ with the horizontal. The first half of the plane is frictionless and the second half has a co-efficient of friction μ. When the block reaches the bottom of the slope, it has velocity zero. Calculate the co-efficient of friction.

Homework Equations


W = mg, N = mgcosθ, F = μN

The Attempt at a Solution


Actually, my friend has solved the problem and has got the correct answer but I don't know if his method is correct. He says, if we apply a force F to an object continuously(gravitational force in this case). Then, if we apply an equal and opposite force -F, then the body will acquire a constant velocity but if we apply twice that force, the body will come to rest. So, in this case,
2 μ mg cosθ = mg sinθ and μ = 2 tanθ which is the correct value for the co-efficient of friction. As the answer is correct, the method seems correct. But I don't know for sure. Is it correct?
 
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I think his explanation is excellent.



Edit:
The reason it works is only because of constant accelerations.

Since there is a constant acceleration, his method results in an equal average velocity for both halfs (and therefore equal times for which the force is being applied)

It's a sort of "symmetry" about the problem, and I think that's probably the way the problem was intended to be solved
 
Last edited:
Okay. Thanks.
 
Yashbhatt said:
in this case,
2 μ mg cosθ = mg sinθ and μ = 2 tanθ which is the correct value
You meant μ mg cosθ = 2 mg sinθ, right?
 
Oh yes. Was in a hurry. So . . .
Can you show how does one go about proving it because just saying so might not be enough.
 
Yashbhatt said:
Oh yes. Was in a hurry. So . . .
Can you show how does one go about proving it because just saying so might not be enough.
There are other ways of expressing what is essentially the same argument.
E.g. in terms of work, the downslope component of the gravity acts over the full length of the slope: W = L mg sinθ. The same work is done against friction, since no KE left at the end: (L/2) mg cos θ.
 

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