## static friction co-efficient

1. The problem statement, all variables and given/known data
from the free body diagram of a block resting on a flat surface, I am trying to derive the coefficient of static friction.

2. Relevant equations

3. The attempt at a solution
I have the definition of the coefficient as:
us=Fn/Ffr

But in my free body diagram i am pulling on the block, which resists with the Ffr
I have
$$\Sigma$$Fx=0
F-Ffrus=0
us=F/Ffr

My confusion is because if the definition of the coefficient is above, why don't I derive it when I observe my free body diagram of the block. What am I missing?
The forces in the x direction are my pulling force and the friction force that resists it * us
How do i get the weight of the block into my derived equation?
 Mentor Blog Entries: 1 The weight of the block is Fn. (If that's not what you're looking for, please state the complete problem exactly as it was given.)
 It is not a problem as such. I am trying to determine the coefficient of static friction us experimentally. Then, I am trying to determine if it is a function of surface area or of mass. My experiment consists of a wooden block on a wooden flat surface. I am going to attach a spring scale and determine at what force the block overcomes the friction and moves. In trying to determine us from my free body diagram I am having a little difficulty as I tried to explain. From my free body diagram, if at equilibrium the forces in the x plane are zero, I get Fpull-Ffriction*us=0 but I don't see where I am getting the mass of the block to become part of my derived equation.

## static friction co-efficient

 Quote by yoleven It is not a problem as such. I am trying to determine the coefficient of static friction us experimentally. Then, I am trying to determine if it is a function of surface area or of mass. My experiment consists of a wooden block on a wooden flat surface. I am going to attach a spring scale and determine at what force the block overcomes the friction and moves. In trying to determine us from my free body diagram I am having a little difficulty as I tried to explain. From my free body diagram, if at equilibrium the forces in the x plane are zero, I get Fpull-Ffriction*us=0 but I don't see where I am getting the mass of the block to become part of my derived equation.
you also need to apply newton's second low in the Y-direction (vertical direction)

ma_y = F_normal - mg = 0

do you understand this equation ? Why is it equal to 0 ?

ps a_y is the component of the acceleration in the vertical direction
and F_normal is the normal force.

In this case, this is not gonna help you much because the table is horizontal.

marlon

Mentor
Blog Entries: 1
 Quote by yoleven From my free body diagram, if at equilibrium the forces in the x plane are zero, I get Fpull-Ffriction*us=0
What you should get is Fpull - Ffriction = Fpull - μFnormal = 0.
 Thank you.

 Quote by yoleven Is it correct to state that Fpull=Ffriction*us? or is it Fnormal=Ffriction*us?
Nope,

Fpull=Fnormal*us