Friction Help

1. Feb 13, 2010

ƒ(x)

Problem: Ive been assigned a problem that involves a block on a table that is attached by a string to a block hanging over the edge. They are of different masses, and I am given coefficients for both static and kinetic friction. The former is .50, and the latter is .30. I have to find the acceleration of the system if it is released from rest.

My problem: ok, so do I need to use the coefficient of kinetic friction at all? I do not think so...that's pretty much my question.

2. Feb 13, 2010

ƒ(x)

3. Feb 13, 2010

c7ng23

Can you show what equations you used? Kinetic friction should be used for surfaces moving relative to each other.

4. Feb 13, 2010

tiny-tim

Hi ƒ(x)!

Do good ol' Newton's second law on each block separately, plus the fact that their accelerations must be the same (because the string length is constant).

What do you get?

5. Feb 13, 2010

ƒ(x)

m1 (on table) = 10 kg
m2 = 4 kg

.5*Fn = Fs
.5*10*9.8 = 49 N = Fs

Fnet = Fx - Fs = 4*9.8 - 49 = -9.8 N

ma = -9.8
a = -9.8/m = -9.8/(4+10) = -.7 m/s/s

6. Feb 13, 2010

tiny-tim

ohh! you didn't give the masses before

no wonder you got a negative answer
That doesn't make sense … how can the mass be accelerating upward?

What does it mean if the weight is less than the µsN ?

7. Feb 14, 2010

ƒ(x)

I'm guessing it means that the system isn't moving.

8. Feb 14, 2010

tiny-tim

(why guessing? )

That's right! …

if the system is released from rest, it will never move (even though if it was given a little nudge, the low µk would enable it to keep accelerating).