Three stacked blocks w/ friction between sliding on frictionless floor

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SUMMARY

The discussion revolves around a physics problem involving three stacked blocks on a frictionless floor, where the coefficient of static friction between the blocks is constant. Participants analyze the net static friction forces acting on each block, concluding that the net friction on Block A and Block B is equal, while Block C experiences no net friction. The confusion arises from the problem's wording, which leads to a misinterpretation of the static friction forces. The correct relationships established are: f_{A,Net} = f_{B,Net} = \mu mg and f_{C,Net} = 2\mu mg, contradicting the professor's assertion that f_{C,Net} equals zero.

PREREQUISITES
  • Understanding of Newton's Laws of Motion, particularly Newton's 2nd and 3rd Laws.
  • Familiarity with the concept of static friction and its mathematical representation (f = μN).
  • Ability to analyze Free Body Diagrams for multiple objects.
  • Knowledge of the relationship between mass, force, and acceleration (f = ma).
NEXT STEPS
  • Review the principles of static friction and its maximum value in different scenarios.
  • Study the application of Newton's Laws in multi-body systems.
  • Explore advanced topics in friction, including dynamic versus static friction.
  • Investigate common pitfalls in physics problem wording and how to clarify ambiguous problems.
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of friction in multi-body systems.

Saladsamurai
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So I was helping my girlfriend with this problem and we are convinced we are correct, but her professor disagrees. So I am taking it to PF court :smile:

Homework Statement



In the figure, all 3 blocks have the same mass. A force pulls the system along the frictionless floor and the 3 blocks all move together in unison.

If the coefficient of static friction between pairs of blocks is the same, compare the magnitudes of net static friction on each block.

Picture6-4.png






Homework Equations



Newton's 2nd and the definition of friction: f = mu*Normal

The Attempt at a Solution



Here is a Free Body Diagram of each block:

Picture7-4.png


Each pair of blocks has a Newton's 3rd Law reaction pair acting on each face whose magnitude is equal to mu*Normal force.

From Block A

f_{AB}=\mu*N_A = \mu mg

From Block B

f_{BC} = \mu*N_B = 2\mu mg

Therefore the net force from friction on A is given by:

f_{A,Net}=f_{AB} = \mu mg

The net force from friction on B is given by:

f_{B,Net} = f_{BC} - f_{AB} = \mu mg

The net force from friction on C is given by:

f_{C,Net}=f_{BC} = 2\mu mg

Therefore:

f_{A,Net} = f_{B,Net} = \frac{1}{2}f_{C,Net}

I am convinced this is correct, but the answer is supposedly supposed to be:

f_{A,Net} = \frac{1}{2}f_{B,Net} \text{ and } f_{C,Net} = 0

which to me makes no sense. If the NET friction on C was zero, then that would be equivalent to saying there is no friction between blocks C and B which clearly is not true, else block C would accelerate with respect to block B (they would not 'stick' together).

Thoughts from the jury?

:smile:
 
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Yes, your comparison of net static friction forces is correct, which apparently the problem has asked for, but the solution seems to give not the net friction forces acting on each block, but rather, f_ab, f_bc, and F_floor on C (the latter of which is 0, frictionless), giving a right answer to a wrong question. But also remember that the staic friction force is not usually uN, but often, less than uN. But the comparison of friction forces works out ok, regardless.
 


Yes, I thought that he might be giving the frictional forces in between surfaces only.

The problem is poorly worded in my opinion. You are correct that it is only mu*N when it is fMAX,static. How else could one make a comparison between fstatic? I am just curious.

I am just wondering if there is some other technique that I may have missed?
 


Since the acceleration of all blocks are the same, you could try
f_ab = ma
f_bc -f_ab =ma

Which shows that the net friction force on the top and middle blocks are equal, and that since from the 2nd equation f_bc = ma + f_ab , then plugging eq. 1 into eq. 2, f_bc = 2f_ab. I don't like the problem wording, either.
 

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