Friction between 2 masses and maximum acceleration

[12thString]

Homework Statement

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?

Homework Equations

just FBDs.

The Attempt at a Solution

for m1

N₁=m₁g from \SigmaF_y


for m2

N₂=(m₁+m₂)g



so for \SigmaF_x=F_app-\mu_k(m₁+m₂)g

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

\SigmaF_x=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 \SigmaF_x=0 for it not to move)

a₂=\mu_sg

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

 

[alphysicist]

Homework Statement

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?

Homework Equations

just FBDs.

The Attempt at a Solution

for m1

N₁=m₁g from \SigmaF_y


for m2

N₂=(m₁+m₂)g



so for \SigmaF_x=F_app-\mu_k(m₁+m₂)g

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

\SigmaF_x=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 \SigmaF_x=0 for it not to move)

a₂=\mu_sg

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

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:

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

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.

 

[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?

 

[alphysicist]

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?

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.

 

[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.