Finding Coefficient of Kinetic Friction in a Pulley/Tension Problem

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Homework Help Overview

The problem involves two masses, one on a table and the other hanging off a frictionless pulley, with friction present between the mass on the table and the surface. The system is released from rest, and the hanging mass falls a certain distance in a specified time. The goal is to find the coefficient of kinetic friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations used to relate the forces acting on the masses and the acceleration of the system. There are inquiries about the arithmetic involved in calculating the coefficient of friction and the necessity of drawing free body diagrams (FBDs) to clarify the forces at play.

Discussion Status

Some participants have provided guidance on the importance of free body diagrams and have pointed out potential errors in calculations. There is an acknowledgment of a mistake in the arithmetic leading to a negative value for the coefficient of friction, which was later corrected. However, there is no explicit consensus on the correctness of the free body diagrams used by the original poster.

Contextual Notes

There is a mention of the original poster's confusion regarding free body diagrams, indicating a possible gap in understanding that may affect their analysis. The discussion also reflects differing opinions on the original poster's approach to the problem.

cdenne
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Homework Statement


I have a problem about two masses: m1 sitting on a table and m2 hanging off the end of th table across a frictionless pulley. There is friction between m1 and the table, however. The system is released from rest and m2 falls 1 meter in 1.2 seconds. m1=10.0 kg, m2=5.00 kg. find the coefficient of kinetic friction.

Homework Equations


I used x=.5at^2 to find a. I used m2g - T = m2a for m2 with down being positive and up being negative. And I used T - (mu)m1g = m1a for m1 with positive being the direction toward the end of the table.

The Attempt at a Solution


I used x=1 meter, t = 1.2 seconds in the first equation and solved for a to get a = 1.39 m/s^2. Then I solved the next two equations for T, set them equal to each other, then solved for mu to get mu = (m2g - m2a - m1a) / (m1g). This gave me a negative value for mu (-0.163) and I know that mu should be positive. Thanks for your help.
 
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Please show your arithmetic for this: mu = (m2g - m2a - m1a) / (m1g)

Chet
 
Lets see some FBD's first.
 
ME_student said:
Lets see some FBD's first.
He obviously used some correct FBDs, or he couldn't have written down the two force balance equations correctly.

Chet
 
Chestermiller said:
Please show your arithmetic for this: mu = (m2g - m2a - m1a) / (m1g)

Chet
(5*9.8 - 5*1.39 - 10*1.39) / (10*9.8) = - 0.163

Alright. I just found my mistake. I don't know what I did on the calculator last night, but I kept getting -0.163. Now I get the correct value of 0.287. Thanks for your help. By the way, what are FBDs?
 
cdenne said:
By the way, what are FBDs?
Free Body Diagrams. You obviously used these to get your force balances.

Chet
 
Chestermiller said:
Free Body Diagrams. You obviously used these to get your force balances.

Chet
Thank you very much for your help.
 
Chestermiller said:
Free Body Diagrams. You obviously used these to get your force balances.

Chet

OP didn't know what a FBD was so I was going to start helping him/her by having him/her draw some FBD. BTW Mr. Chet if he used correct FBD's he wouldn't be here asking us questions. Obviously OP must have analyzed something wrong on his/her free body diagram.
 
ME_student said:
OP didn't know what a FBD was so I was going to start helping him/her by having him/her draw some FBD. BTW Mr. Chet if he used correct FBD's he wouldn't be here asking us questions. Obviously OP must have analyzed something wrong on his/her free body diagram.
Dear ME_student,

I stand by what I said, to wit: If he hadn't drawn proper free body diagrams, he would not have been able to write down the force balance equations correctly (which he was able to do). His difficulty was not in drawing free body diagrams and developing the force balance equations, but in solving the force balance equations. So, obviously OP must not have analyzed something wrong on his/her free body diagram. Based on 50 years of engineering experience, this is the determination that I made. When you have 50 years of experience, come back and we can discuss this further. In the meantime, this thread is Closed.

If you want to discuss the issue further, you can contact me via private conversation. But be advised that, as a Mentor, I'm beginning to get an itchy Warning finger.

Chet
 
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