How can i find the coefficient of kinetic friction?

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Homework Help Overview

The discussion revolves around finding the coefficient of kinetic friction for a crate sliding down an inclined plane at a 16-degree angle. Participants provide various known values, including the distance traveled (2.85 meters) and the final velocity (2.71 m/s), while also calculating the static friction coefficient.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between static and kinetic friction, with some attempting to derive the kinetic friction coefficient using kinematic equations and free body diagrams. Questions arise regarding the role of mass in the equations and how to handle multiple variables.

Discussion Status

There is ongoing exploration of the problem, with participants providing insights and suggestions for drawing free body diagrams and checking algebraic steps. Some participants express confusion about the implications of mass in their calculations, while others encourage further attempts to derive the kinetic friction coefficient.

Contextual Notes

Participants note the absence of mass in their calculations, which complicates their attempts to find the acceleration and subsequently the kinetic friction coefficient. There is also mention of specific algebraic errors that need to be addressed.

ciara0682
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Homework Statement



I know incline is 16 degrees
I know distance traveled is 2.85 meters
I know Vf is 2.71 meters/second I calculated the meu static and got .287

Homework Equations

The Attempt at a Solution


Tried every thing so lost
 
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ciara0682 said:

Homework Statement



I know incline is 16 degrees
I know distance traveled is 2.85 meters
I know Vf is 2.71 meters/secondI calculated the meu static and got .287

Homework Equations

The Attempt at a Solution


Tried every thing so lost

Welcome to the PF.

Please fill out the Template a bit more. What is the exact problem statement? What are the relevant equations? Please show us your work where you say you got the static μ to be 0.287
 
A flatbed truck slowly tilts its bed upwards to dispose of a crate. When the tilt angle is 16 degrees or above the create begins to slide. When the create reaches the bottom of the flatbed, it has slide 2.85 meters, its speed is 2.71 m/s.
I found meu static by summing the forces in both the x and y direction which all and all gave me the formula of meu static = tan of the angle. Tan(16)=.287 now I'm stuck on how to calculate meu kinetic because you need acceleration. But with no mass...I have two variables. I'm just very confused.
 
ciara0682 said:
But with no mass...I have two variables.
Give it a try anyway. Sometimes (when the net force is proportional to the mass) you don't need to know the mass. For example, if I drop my computer right now, even though I have no idea what it's mass is, I can tell you it accelerates at 9.8 m/s/s
 
But in the formula I reach...the masses just can't cancel? I'm sorry, I'm not quite understanding. I get your analogy, but still confused on how it pertains to a frictional problem.
 
ciara0682 said:
A flatbed truck slowly tilts its bed upwards to dispose of a crate. When the tilt angle is 16 degrees or above the create begins to slide. When the create reaches the bottom of the flatbed, it has slide 2.85 meters, its speed is 2.71 m/s.
I found meu static by summing the forces in both the x and y direction which all and all gave me the formula of meu static = tan of the angle. Tan(16)=.287 now I'm stuck on how to calculate meu kinetic because you need acceleration. But with no mass...I have two variables. I'm just very confused.

ciara0682 said:
But in the formula I reach...the masses just can't cancel? I'm sorry, I'm not quite understanding. I get your analogy, but still confused on how it pertains to a frictional problem.

Draw a free body diagram (FBD) for the crate sliding down the ramp. Show the component forces due to gravity and due to the sliding friction. Then use F=ma and sum the forces... Please show that work...
 
I went back to the basics and started with the kinetic formula. Xf=x+volt+1/2at^2... I used Vf=Vo+at solved for t...plugged that into kinematic equation.
Now I have Xf=1/2a(Vf/a) ^2 solved for a. Then plugged that into...meu k(mgcos(theta))-(mgsin(theta))=ma but I have two variables. This is where I'm lost. And who knows of I am even doing that part right.
 
ciara0682 said:
But in the formula I reach...the masses just can't cancel?
Have you actually tried it, or are you just guessing?

ciara0682 said:
I get your analogy, but still confused on how it pertains to a frictional problem.
The analogy was not really supposed to give you any insight; the point of my post was to just try finding μk and see what happens!

If μk depends on mass like you assume it does, then you should at least determine exactly how it depends on mass.
 
I provided you with the work above that I have done so far.
 
  • #10
ciara0682 said:
meu k(mgcos(theta))-(mgsin(theta))=ma but I have two variables.
What happens if you divide both sides by m?
 
  • #11
μk=a/g(cos(theta))-(sin(theta)) ?
 
  • #12
ciara0682 said:
μk=a/g(cos(theta))-(sin(theta)) ?
Double check your algebra. If you get the same answer show your steps.
 
  • #13
μk=a/(gsintheta)-(gcostheta)...I solved for uk and I still got the incorrect answer. grrrrrrrrrr. sorry for growling.
 
  • #14
ciara0682 said:
μk=a/(gsintheta)-(gcostheta)
Your algebra is still wrong. Triple check it :smile:

ciara0682 said:
grrrrrrrrrr. sorry for growling.
As long as you don't bite we're okay.
 
  • #15
a+gsin(theta)/gcos(theta) ...
 
  • #16
From your equation, I get \mu_k=\frac{a}{g\cos\theta}+\tan\theta

Also when you plug in the value for a, remember which directions you chose to be positive and negative (you said that downhill was negative, so the acceleration will be negative)
 
  • #17
trig is not my strongest area. whoops. I realized my mistake in algebra... gracias mi amigo!
 
  • #18
de nada
 
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