Minimum friction coefficient to prevent slipping between stacked boxes

[Andrew Mason]

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

 

[randomwinner]

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]

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?

 

[Doc Al]

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?

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?

 

[randomwinner]

Well, this is what it looks like.

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

So F_A = 2.5 m/s² x mass
And F_S = u_s x 9.8 m/s ² x mass

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

2.5 m/s² = u_s x 9.8 m/s²

Then I would divide and then have my answer. Correct?

 

[Doc Al]

Exactly correct.

Fs = ma = μmg
Thus, μ = a/g.

 

[randomwinner]

All right. Thanks for the help!