Coefficient of friction Problem

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Homework Help Overview

The problem involves a small box resting on a larger box, both of which are accelerating together when a horizontal force is applied to the larger box. The task is to determine the smallest coefficient of friction that will prevent the small box from slipping off the larger box, given an acceleration of 2.5 m/s².

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the forces acting on the boxes, including normal force and friction. Some mention using Newton's second law to relate acceleration to frictional forces. Others question the necessity of knowing the mass to determine the coefficient of friction.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided hints and guidance without revealing complete solutions, while others express confusion about the role of mass and the application of formulas related to friction.

Contextual Notes

There is a noted lack of consensus on whether the mass is necessary to calculate the coefficient of friction, with some participants suggesting that it may cancel out in the equations. The original poster's understanding of the problem is also being clarified through the discussion.

  • #31
randomwinner said:
Now that is impossible. It all comes back to mass. A certain force value will not make all things accelerate at the same rate. It depends on the mass.
Doc Al is trying to get you to see that while mass is certainly a part of the analysis, you don't need to know what it actually is. To see this, you have to do the analysis.

You know what the horizontal accelerating force must be applied to the top block by the bottom one if it accelerates at 2.5 m/sec^2. So write it out.

You know what determines the maximum force that the lower block can apply to the top block. Write that out.

You are asked to assume that this maximum force is applied to the top block when the acceleration is 2.5 m/sec^2. So what kind of relationship exists between the forces? Put that relationship between the expressions for the forces and you will solve the problem.

AM
 
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  • #32
So are you saying that it wouldn't make a difference if the box was 20 or 50 kg, as long as it was accelerating at 2.5 m/s squared? That the certain value for the coefficient of friction would be enough to hold both boxes of different weight to the larger box if they were accelerating at 2.5 m/s squared?
 
Last edited:
  • #33
Doc Al said:
Luckily, we don't need to know the actual mass to solve this problem.

Wouldn't you need a higher coefficient of friction to hold a smaller box in place while accelerating?
 
  • #34
randomwinner said:
So are you saying that it wouldn't make a difference if the box was 20 or 50 kg, as long as it was accelerating at 2.5 m/s squared? That the certain value for the coefficient of friction would be enough to hold both boxes of different weight to the larger box if they were accelerating at 2.5 m/s squared?

randomwinner said:
Wouldn't you need a higher coefficient of friction to hold a smaller box in place while accelerating?
Why don't you do the analysis and find out?
 
  • #35
Well, this is what it looks like.

The applied force on the box would have to equal the force of static friction.

So FA = 2.5 m/s2 x mass
And FS = us x 9.8 m/s 2 x mass

Therefore I can cancel out the mass which leaves me with:

2.5 m/s2 = us x 9.8 m/s2

Then I would divide and then have my answer. Correct?
 
  • #36
Exactly correct.

Fs = ma = μmg
Thus, μ = a/g.
 
  • #37
All right. Thanks for the help!
 

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