# Homework Help: Maximal static friction problem

1. Jun 2, 2007

### Black Riven

This is mostly a theoretical question, I'm studying mechanics by myself so I have no teacher to ask this.

1. The problem statement, all variables and given/known data

The minimum force needed to move a stationary object is equal to the max static friction. However, I just encountered a question that included a sloped surface and an object moving upwards across it.

Eventually, mgcos(a)+Fk bring it to a halt, and the question is what should the minimal static friction constant be in order for the object to remain stationary.

3. The attempt at a solution

Fs must be strong enough to hold out against mgcos(a). The equation we are supposed to have is Fs-mgcos(a)=0
However, this confuses me. If when a force applied on the object equals to Fs(max) it causes it to move, Shouldn't the force needed to keep the object from moving be Fs>mgcos(a)?

In other words, is Fs(max) the breaking point or the last point before movement occurs?

Last edited: Jun 3, 2007
2. Jun 2, 2007

### turdferguson

It sounds like the question should read "minimal static friction constant"

3. Jun 3, 2007

### PhanthomJay

Assuming a level surface, why do you say "equal to"? If an object is at rest and the max static friction force available is 20N and you push on it with a 20N force, will it move?
As noted above, change the word 'kinetic' to 'static'.
does it?
What does Newton's first law tell you?

4. Jun 3, 2007

### Black Riven

This is the problem then. My intuition tells me that no, it will not move because the force must be bigger than Fs(max). However, the book I'm using says
Based on your answer I assume my intuition is correct? If Fs(max)=F the object will not move, and the minute F>Fs(max) it will start moving.
If that's the case any idea why the book says what it does?
Done.

5. Jun 3, 2007

### PhanthomJay

I think it's just saying that if the max static friction force is 20.000000000N, a force of 20.000000001N will, in theory, cause it to move. Close enough to 20, I guess.