What is the coefficient of friction for a given surface and two weights?

  • Thread starter Thread starter SimonJ
  • Start date Start date
  • Tags Tags
    Friction Physics
Click For Summary

Homework Help Overview

The discussion revolves around determining the coefficient of friction for a surface given two weights, specifically 0.5672 kg and 0.050 kg. The original poster presents an equation related to the forces acting on the masses but expresses uncertainty about how to proceed with the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants inquire about the context of the weights' actions, with one suggesting the need to analyze the forces involved. There is a focus on understanding the relationship between frictional force and normal force, as well as the definition of the coefficient of friction.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the forces acting on the masses and the role of friction. Some guidance has been offered regarding the relationship between frictional force and normal force, but no consensus or resolution has been reached yet.

Contextual Notes

There is a lack of detailed information about the setup of the problem, including the specific actions of the weights and any assumptions regarding the system, such as frictionless pulleys.

SimonJ
Messages
4
Reaction score
0

Homework Statement



I need to find the coefficient of friction on a surface. Two weights were given, 0.5672kg and 0.050kg

Homework Equations



m1a + Mm1g = m2g - m2a

The Attempt at a Solution



(0.5672kg)a + M (0.5672)(9.81) = (0.050kg)(9.81) - (0.050kg) a


M represents the coefficient of friction. I need to solve for that, I just don't know where to take it from here.
 
Physics news on Phys.org
That's not enough of a question - what were the two weights doing?
 
246t8h4.jpg


That is what is happening
 
That helps.
Now try drawing in the forces.
The frictional force is = coeff_friction * normal force (ie weight)
 
Can you explain that a little more please?
 
There is a force down on the 0.05kg mass = m g
This same force pulls the larger mass to the right, so an equal friction force acts to the left (assuming no loss in the pulleys).
The definition of coeeficient of friction is
friction force = coeef * normal force

The normal force is the force acting down on the suface = M g
 
Last edited:
Ok thanks!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 4 ·
Replies
4
Views
3K