Calculating Coefficient of Kinetic Friction for a Slide Loving Pig

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Homework Help Overview

The problem involves calculating the coefficient of kinetic friction for a pig sliding down a 24° incline, where the pig takes twice the time to slide down compared to a frictionless scenario. The original poster expresses confusion over their approach and the relationship between force and time in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to equate forces in their calculations but struggles with the implications of time on force. Other participants question the validity of translating time into force and explore the relationship between these variables.

Discussion Status

Participants are actively discussing the problem, with some offering insights into the relationship between force and time. There is a recognition of potential misunderstandings in the original poster's approach, and a suggestion that a different factor might yield a correct result.

Contextual Notes

There is mention of the problem being rated on a difficulty scale, indicating that it may be perceived as more challenging than expected. Participants also note the need to clarify the relationship between force and time, which remains a point of confusion.

HobieDude16
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slide loving pig force...

ok, here's a problem that is supposed to be easy, and i thought i knew how to do it, but obviously not, cause i can't get the right answer...


In figure 6-21, a slide loving pig slides down a certain 24° slide in twice the time it would take to slide down a frictionless 24° slide. What is the coefficient of kinetic friction between the pig and the slide?

what i did:
i set 2 times the force without friction (mgsin24) equal to the force without friction minus friction (mgsin24-MUkmgcos24) and tried to solve for MUk... but that didnt work, and i have no idea what happened. here's what the equation i used looked like:
2(mgsin24)=mgsin24-MUkmgcos24... so that is the same as
2sin24=sin24-MUkcos24
then sin24=-MUkcos24
so tan24=-MUk
buuuut that didnt work, so what do i do wrong?
 
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Two times the time doesn't translate to two times the force IIRC
 
whats a way to relate the force to time? i can't find a formula to do that?
 
this problem is supposed to be a 1 of 3 dot problem (How hard it is, 3 is hardest) so why is it so hard for me?
 
This doesn't really follow up with what I just said though but the resultant force of the slide with friction is less than the no-friction slide (and wouldn't go UP by any factor > 1) To figure the factor by which the resultant force is different from the no-friction force consider the units of force and how time plays into it.
 
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uhhhh in lamen's terms? i can't figure out what you're saying
 
Here's what your equation above said to me when you said 2 * mgsin(24) = mgsin(24)-mg(cos(24))*u or whatever:

Twice the force of mgsin(24) is created by applying friction to the same force which of course doesn't make sense.

F = kg*m/s^2 right? What effect does making s 2s have?
 
so your saying make it 1/2 mgsin(24) yadda yadda yadda?
 
not quite.. Take a close look at the relationship of force to time given its units above.
 
  • #10
ohhhhh 1/4? that seems to make sense, so if i made it 1/4 (instead of 2), then did the rest the same way i did it, it shoudl work?
 
  • #11
I'm fairly sure it would. You won't know for certain until you try it out I guess. I'm not exactly a physics expert just happened to see something possibly wrong.
 
  • #12
heck yeah man, that worked... thanks a lot! now how bout my other thread, any ideas there?
 

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