Finding the force of kinetic friction without a coefficient

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Homework Help Overview

The discussion revolves around a lab assignment involving a Fletcher's trolley system, where participants are tasked with calculating the normal force and the force of friction, ultimately aiming to find the coefficient of kinetic friction. The original poster provides specific values related to the system's masses, distance, time, and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of free body diagrams and equations relating the sum of forces to acceleration for each block. There is an exploration of the relationship between the forces acting on the system, particularly focusing on the equation for the force of friction in terms of the normal force.

Discussion Status

Participants are actively engaging with the problem, offering guidance on how to approach the calculations. There is acknowledgment of the need to rewrite equations and clarify terms, such as the applied force. Some participants express uncertainty about the necessity of creating a graph for the final calculation.

Contextual Notes

There is mention of imposed homework rules and specific procedures suggested by the professor, which may influence how the participants approach the problem and their calculations.

Her0
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Homework Statement


In a lab assignment we created a Fletcher's trolley system and were asked to calculate the normal force, the force of friction, and then create a graph of both to find the coefficient of kinetic friction.

So far, the variables I have are:

Hanging mass: 0.100 kg
Sliding mass: 0.1553 kg
Distance traveled by the sliding mass: 0.398 m
Time taken for the sliding mass to reach the ground: 0.89 s
Acceleration of the system: 1.01 m/s^2
Normal force: 1.52 N
Force of friction : Unknown

Homework Equations


The acceleration of the system was found using the equation a=(2(d-vit))/t^2 which i simplified to a=2d/t^2 since vi was 0 m/s.

The normal force of the sliding object was found using Fn = mg

The force of friction would be calculated using Ffk = Fn*μk but I don't have μk.

The Attempt at a Solution


As stated above, I tried using Ffk = Fn*μk but i didn't have all the variables.
 
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Did you draw free body diagrams for each of the 2 blocks and then write equations to relate sum of the forces to acceleration for each block?
 
TomHart said:
Did you draw free body diagrams for each of the 2 blocks and then write equations to relate sum of the forces to acceleration for each block?
Okay so I tried doing that and essentially what I got was that Ffk = Mg - ma where Mg is the force of the hanging object and ma is the force of both masses of the system multiplied by the system's acceleration because Fnet = Fapp -Ffk.
 
I'm not used to working with systems. I would normally write separate equations for each block and then solve the two equations with the two unknowns. But I think your equation below makes sense.
Her0 said:
Ffk = Mg - ma
So I think you have everything you need to solve that equation, provided you rewrite Ffk in terms of the normal force. You know the masses and acceleration.

I did not know what Fapp was.

Edit: P.S. Welcome to Physics Forums.
 
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TomHart said:
I'm not used to working with systems. I would normally write separate equations for each block and then solve the two equations with the two unknowns. But I think your equation below makes sense.

So I think you have everything you need to solve that equation, provided you rewrite Ffk in terms of the normal force. You know the masses and acceleration.

I did not know what Fapp was.

Edit: P.S. Welcome to Physics Forums.
Fapp was a formula for applied force which I used to derive my new formula for Ffk. Other than that, I now have everything I need to calculate the force of friction. Thank you!
 
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Let me know what you get for a final answer. I always like to check my work.
 
So I subbed in the values resulting in the equation becoming Ffk = (0.100 kg(9.81m/s^2))-((0.100kg+0.1553kg)(1.01m/s)) which resulted in the answer being 0.723 N rounded to three significant digits.
 
Okay, that looks right. For some reason, I was thinking that the goal of the problem was to find the coefficient of friction. And as I now re-read the problem statement, finding the coefficient of friction is indeed the final question. I just don't know why you would need to create a graph to do that.
 
It's just a procedure my professor told us to use even though we could find the coefficient using μk = Ffk/Fn after Ffk was determined. Either way, thanks again for the help.
 
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