What is the coefficient of static friction in this scenario?

[yoleven]

Homework Statement

from the free body diagram of a block resting on a flat surface, I am trying to derive the coefficient of static friction.

Homework Equations

The Attempt at a Solution

I have the definition of the coefficient as:
u_s=F_n/F_fr

But in my free body diagram i am pulling on the block, which resists with the F_fr
I have
\SigmaF_x=0
F-F_fru_s=0
u_s=F/F_fr

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 * u_s
How do i get the weight of the block into my derived equation?

 

[Doc Al]

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.)

 

[yoleven]

It is not a problem as such. I am trying to determine the coefficient of static friction u_s 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 u_s 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
F_pull-F_friction*u_s=0
but I don't see where I am getting the mass of the block to become part of my derived equation.

 

[marlon]

It is not a problem as such. I am trying to determine the coefficient of static friction u_s 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 u_s 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
F_pull-F_friction*u_s=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 going to help you much because the table is horizontal.

marlon

 

[Doc Al]

From my free body diagram, if at equilibrium the forces in the x plane are zero, I get
F_pull-F_friction*u_s=0

What you should get is F_pull - F_friction = F_pull - μF_normal = 0.

 

[yoleven]

Thank you.

 

[marlon]

Is it correct to state that F_pull=F_friction*u_s?
or is it F_normal=F_friction*u_s?

Nope,

F_pull=F_normal*u_s