# Homework Help: Slide loving pig force

1. Oct 3, 2004

### HobieDude16

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
2sin24=sin24-MUkcos24
then sin24=-MUkcos24
so tan24=-MUk
buuuut that didnt work, so what do i do wrong?

2. Oct 3, 2004

### vsage

Two times the time doesn't translate to two times the force IIRC

3. Oct 3, 2004

### HobieDude16

whats a way to relate the force to time? i cant find a formula to do that?

4. Oct 3, 2004

### HobieDude16

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?

5. Oct 3, 2004

### vsage

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
6. Oct 3, 2004

### HobieDude16

uhhhh in lamen's terms? i cant figure out what you're saying

7. Oct 3, 2004

### vsage

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?

8. Oct 3, 2004

9. Oct 3, 2004

### vsage

not quite.. Take a close look at the relationship of force to time given its units above.

10. Oct 3, 2004

### HobieDude16

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. Oct 3, 2004

### vsage

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. Oct 3, 2004

### HobieDude16

heck yeah man, that worked.... thanks a lot! now how bout my other thread, any ideas there?