1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Slide loving pig force

  1. Oct 3, 2004 #1
    slide loving pig force....

    ok, heres a problem that is supposed to be easy, and i thought i knew how to do it, but obviously not, cause i cant 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. heres what the equation i used looked like:
    2(mgsin24)=mgsin24-MUkmgcos24.... so that is the same as
    then sin24=-MUkcos24
    so tan24=-MUk
    buuuut that didnt work, so what do i do wrong?
  2. jcsd
  3. Oct 3, 2004 #2
    Two times the time doesn't translate to two times the force IIRC
  4. Oct 3, 2004 #3
    whats a way to relate the force to time? i cant find a formula to do that?
  5. Oct 3, 2004 #4
    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?
  6. Oct 3, 2004 #5
    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.
    Last edited by a moderator: Oct 3, 2004
  7. Oct 3, 2004 #6
    uhhhh in lamen's terms? i cant figure out what you're saying
  8. Oct 3, 2004 #7
    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?
  9. Oct 3, 2004 #8
    so your saying make it 1/2 mgsin(24) yadda yadda yadda?
  10. Oct 3, 2004 #9
    not quite.. Take a close look at the relationship of force to time given its units above.
  11. Oct 3, 2004 #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?
  12. Oct 3, 2004 #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.
  13. Oct 3, 2004 #12
    heck yeah man, that worked.... thanks a lot! now how bout my other thread, any ideas there?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook