# What smallest value of kinetic friction

1. Oct 6, 2011

### fizzex

1. The problem statement, all variables and given/known data
(a) suppose the coefficient of μκ between m1 and the plane in Fig 4-57 is 0.15, and that m1=m2=2.7 kg. As m2 moves down, determine the magnitude of the acceleration of m1 and m2, given θ = 25°. (b) What smallest value of kinetic friction will keep this system from accelerating?

2. Relevant equations
There are many equations...
For part A, ƩF = Fg - Ft +Fn
Fg = mg
ƩF = Fg - Ft
Ff = μ(Fn)
Part B, I'm not too sure.
I was going to have Ft - Fg - Ff = ƩF, and have ƩF=0 because ƩF = ma, and acceleration would be 0, but then I was confused.

3. The attempt at a solution
I solved Part A... the acceleration is 2.16. If you want me to go through the whole process, I will, just for the sake of time, I was hoping to just start from part B, where i'm absolutely stuck. I know that I have to set acceleration equal to zero, but when I tried I got 0.14 instead of the answer, 0.64.

Any suggestions? ...and does that even make sense? Sorry, it's my first post.

2. Oct 6, 2011

### SammyS

Staff Emeritus
Re: Dynamics

Welcome to PH Forums.

It's hard to help without the figure!

3. Oct 6, 2011

### fizzex

Re: Dynamics

I can't copy it, but it's a pulley system. M2 is hanging over the side of the inclined plane, and M1 is sliding down the 25 degree angle. Does that help?

4. Oct 7, 2011

### Staff: Mentor

Re: Dynamics

I think you'll have to spell out your final equation for the acceleration (symbolic form, no numbers) so we can see where things stand.

5. Oct 7, 2011

### fizzex

Re: Dynamics

From Part A, there were two equations involving acceleration (ƩF = ma)... ƩF = Fg - Ft and ƩF = Ft - Fg - Ff. The first one was applied to the hanging box, box 2, because there was no friction. (Conversely, the second equation included Ff, as box 1 was sliding down the inclined plane.)

These formulas were then set equal to each other, which found the answer to part A, finding the acceleration.

I'm not sure how to put them into the second part of the problem.