Calculating Coefficient of Kinetic Friction for a Slide Loving Pig

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A problem involving a slide-loving pig sliding down a 24° incline with friction was discussed, focusing on calculating the coefficient of kinetic friction. The initial approach incorrectly equated forces, leading to confusion about the relationship between time and force. It was clarified that the time taken to slide down the incline should be related to the acceleration, suggesting a factor of 1/4 instead of 2 for the force calculation. After adjusting the equation accordingly, the correct approach yielded a solution. The discussion highlighted the importance of understanding the relationship between force, time, and acceleration in physics problems.
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 teh 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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