# Friction between 2 masses and maximum acceleration

1. Dec 7, 2008

### 12thString

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

2 masses, m1 and m2, m1 being on top of m2 and m2 is heavier than m1. $$\mu$$s is the coefficient of static friction anywhere and $$\mu$$k is the coefficient of kinetic friction between the floor and m2. What is the maximum acceleration m2 can have for m1 not to fall off?

2. Relevant equations

just FBDs.

3. The attempt at a solution

for m1

N1=m1g from $$\Sigma$$Fy

for m2

N2=(m1+m2)g

so for $$\Sigma$$Fx=Fapp-$$\mu$$k(m1+m2)g

I included a Fapp variable there. But i have the feeling that it doesn't really matter. Anyway, tinkering with m1 $$\Sigma$$Fx

$$\Sigma$$Fx=m1a2-$$\mu$$sm1g

which you can see that i equated m1 with the acceleration of m2.

which when manipulated will give (I assumed that m1's $$\Sigma$$Fx=0 for it not to move)

a2=$$\mu$$sg

but this is incredibly wrong. I'm quite lost now.

2. Dec 7, 2008

### alphysicist

Why do you say it is wrong? Unless I'm reading the question incorrectly it looks like the right answer to me.

I would interpret some of your work differently. For m1, I would say:

\begin{align} \sum F_x = m_1 a_x\nonumber\\ \mu_s m g = m_1 a\nonumber\ \end{align}

In other words, it's not that the forces go to zero, there is only one horizontal force and it equals m1 a. But it gives the same answer.

3. Dec 7, 2008

### 12thString

i thought it was wrong because it didn't used the $$\mu$$k variable. and it seemed too simple.

but upon thinking, does this mean that i don't even have to deal with m2 at all?

4. Dec 7, 2008

### alphysicist

That's right. They both have the same acceleration because they move together, and since there are fewer forces acting on m1 I think it's easier to work with.

5. Dec 7, 2008

### 12thString

oh, so i was just complicating things. i thought it was weird that the only component of m1's horizontal force was the frictional force. thank you very much for the help.

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