# Three stacked blocks w/ friction between sliding on frictionless floor

1. Oct 27, 2009

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

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

Here is a Free Body Diagram of each block:

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?

2. Oct 27, 2009

### PhanthomJay

Re: Friction

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.

3. Oct 28, 2009

Re: Friction

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?

4. Oct 28, 2009

### PhanthomJay

Re: Friction

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.