How to find the coefficient of friction?

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Homework Help Overview

The discussion revolves around determining the coefficient of friction in a physics problem involving forces, mass, and acceleration. The original poster provides a free body diagram and equations related to the forces acting on an object.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants examine the equations presented by the original poster, questioning the algebraic steps taken and the clarity of the setup. There are attempts to clarify the relationships between the forces involved.

Discussion Status

Some participants have raised concerns about the clarity of the original post and the accuracy of the algebraic manipulations. There is an ongoing examination of the equations and their implications, with no clear consensus reached yet.

Contextual Notes

Participants note potential confusion in the original poster's presentation and the need for clearer definitions or assumptions regarding the forces involved.

trASHf
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Homework Statement
Unknown object has a mass of 3.2kg on steel plank.

Add force:
|FA|=22.8N
|a|=3.02m/s^2

What is the material?
Relevant Equations
Newton's Second Law - sum of all forces = mass * acceleration = ...
GR (bold = mathematical convention)
m = 3.2kg
|applied force| = + 22.8 N
|acceleration| = + 3.02 m/s^2

FBD-hypothetical
^ normal force
|
friction force <-----*-----------> applied force
|
gravitational force

A
ma = FA + FfK
ma = FA -Mmg
ma/mg = FA - M
ma/mg - FA = -M
 
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Hi,

Can you imagine it is hard to make sense of this ? Read what you posted
 
Last edited by a moderator:
BvU said:
Hi,

Can you imagine it is hard to make sense of this ? Read what you posted

other than the attempted solution I put there, no I don't think it's hard to make sense of this.
 
trASHf said:
ma = FA + FfK
ma = FA -Mmg
ma/mg = FA - M
ma/mg - FA = -M
Your algebra is faulty when you go from the second equation to the third. You need to divide all the terms on the right side by mg.
 
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