# Spring with Coulomb Friction

1. Oct 14, 2014

### azupol

1. The problem statement, all variables and given/known data
A mass m is attached to a spring with spring constant k. There is a coefficient of static friction,
us
The coefficient of kinetic friction is uk
Suppose you pull the mass to the right and release it from rest.
You find there is a limiting value of x = A0 > 0 below which the
mass just sticks and does not move. For x > A0 , it starts sliding
when you release it from rest. Find A0 .

2. Relevant equations
$x''(t) + \omega x(t) - ( \mu mg)/k =0$

3. The attempt at a solution
I solved the ODE for Simple Harmonic Motion, and I get that $x(t)=B sin ( \omega t) + C ( \omega t) -umg/k$, but I'm not sure where to go from there. The derivative at x = A0 must be zero, but how does that help me find A0 itself?

Last edited: Oct 14, 2014
2. Oct 14, 2014

### Dr.D

The place to go is back to Newton's 2nd law. Your DE is wrong, and your thinking about static friction is in error. Please try again.

3. Oct 14, 2014

### azupol

In light of your post, I figured this:
At x=A0, the block is at rest, so the forces acting on it must be balanced. Thus, -kx=umg, and at A0, -kA0=umg, so solving for A0 gives: A0=-umg/k

Am I on the right track now?

4. Oct 15, 2014

### NATURE.M

Wouldn't A0 be positive since the question indicates A0 > 0 ?