How Is the Limiting Displacement A0 Determined in a Spring System with Friction?

[azupol]

Homework Statement

A mass m is attached to a spring with spring constant k. There is a coefficient of static friction,
u_s
The coefficient of kinetic friction is u_k
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 .

Homework Equations

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

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?

 

[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.

 

[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?

 

[NATURE.M]

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

 

[Nickie]

I would need more information, such as the initial conditions (position and velocity) or the value of the coefficient of static friction (us) in order to fully solve for A0. However, based on the given information, we can make some observations.

First, we know that the mass will not move when x < A0, meaning that the force from the spring (kx) is less than or equal to the maximum static friction force (usmg). This can be represented mathematically as:

kx ≤ usmg

Second, when x > A0, the mass will start to slide and the force from the spring (kx) will be greater than the kinetic friction force (ukmg). This can be represented as:

kx > ukmg

Combining these two equations and using the fact that the mass is released from rest (meaning x' = 0 at t = 0), we can solve for A0:

kA0 = ukmg

A0 = ukmg/k

So, the limiting value of A0 is ukmg/k. This value can also be interpreted as the maximum displacement of the mass before it starts to slide.

In summary, to fully solve for A0, we would need more information. However, based on the given information, we can make some observations and find a general expression for A0 in terms of the given parameters.