# Friction problem - Relation between coefficient of friction

1. Sep 27, 2013

### Saitama

1. The problem statement, all variables and given/known data
A man of mass m is applying a horizontal force to slide a box of mass m' on a rough horizontal surface. It is known that the man does not slide. The coefficient of friction between the shoes of the man and the floor is $\mu$ and between the box and the floor is $\mu'$. In which of the following cases it is certainly not possible to slide the box?
A) μ>μ', m<m'
B) μ<μ', m<m'
C) μ<μ', m>m'
D) μ>μ', m>m'

2. Relevant equations

3. The attempt at a solution
If man applies a force F on box, due to Newton's third law, he also experiences a force F in the opposite direction. According to the question, F<=μmg and F<=μ'm'g. Also, since it is given that man does not slide, F has to be less than or equal to μmg. The box can slide if μ'm'g<F<μmg. The block never slides if μmg<μ'm'g. But I don't see how to find relation between the coefficient of friction and masses by considering the above inequalities.

Any help is appreciated. Thanks!

2. Sep 27, 2013

### Staff: Mentor

OK.

Go through the set of choices and see which ones make that second inequality true.

3. Sep 27, 2013

### Saitama

Call f as the friction force on man and f' on box. For the first case,
Let μ=0.6, μ'=0.5, m=3 kg and m'=5kg. Then f=18N and f'=25N. Here f<f' so A is the answer but the given answer is B.

4. Sep 27, 2013

### Staff: Mentor

Don't plug in specific numbers. You want an answer that will make it impossible to slide for any number in the given range.

What's an easy way to ensure that μmg<μ'm'g? One of the given choices ensures that will inequality will be met.

5. Sep 27, 2013

### Saitama

I am still confused. The inequality can be rearranged as m'>(μ/μ')m. For the first case, μ/μ'>1 so I guess (μ/μ')m may become greater than m' so first is false. Is this a correct way to put it?

6. Sep 27, 2013

### Staff: Mentor

Yes. That shows that choice A won't work.

Now look at choice B.

7. Sep 27, 2013

### Saitama

In this case μ<μ' and m<m'. Rearranging as before, m'>(μ/μ')m. Already, m < m' and since (μ/μ')<1, the product on the RHS becomes more smaller and hence it is always true. Correct?

8. Sep 27, 2013

### Staff: Mentor

Yes.

9. Sep 27, 2013

### Saitama

Thanks a lot Doc Al!