# Homework Help: Friction Force of a dancer

1. Oct 4, 2007

### klm

A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer's foot are mu_s and mu_k, respectively. n represents the normal force exerted on the dancer by the bridge, and Fg represents the gravitational force exerted on the dancer, as shown in the drawing For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges.

Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, Ff?

Express your answer in terms of some or all of the variables n, mu_s, and/or mu_k.

i thought the answer should be -mu_s X N , but that is wrong.

Last edited: Oct 4, 2007
2. Oct 4, 2007

### PhanthomJay

Have you tried applying Newton's laws in the horizontal direction?

3. Oct 4, 2007

### klm

yeah i have friction force pointing horizontally on the -x axis. so then i did Fnetx=-Fs=max=0 and Fnet y= n-w=0, so n=mg. but since the answer is suppose to be in terms of Us, Uk , or n. i thought -UsN

4. Oct 4, 2007

### PhanthomJay

Yes, I also don't understand how you can express 0 as a function of any of those variables.

5. Oct 4, 2007

### klm

i dont mean that the variables are zero, i just mean that there is no net force. so -Fs=0 and n-w=0

6. Oct 4, 2007

### PhanthomJay

So are you sayinfg F_f = 0?

7. Oct 4, 2007

### klm

actually yes i think that would be right b/c there is no Ff

8. Oct 4, 2007

### PhanthomJay

Well maybe, but why not no magnitude at all?

9. Oct 4, 2007

### klm

yeah i think you are right, it shouldnt have a magnitude

10. Oct 4, 2007

### PhanthomJay

Well, maybe yes, and maybe no, I may have spoken too quickly. In general, what is the static friction force formula?

11. Oct 4, 2007

### klm

Fs=UsN
i think i understand, thank you

12. Oct 4, 2007

### PhanthomJay

No, that is close, but that is not the formula.

13. Oct 4, 2007

### klm

that is the formula that we learned

14. Oct 4, 2007

### klm

that is the formula that we learned fsmax=usn

15. Oct 4, 2007

### PhanthomJay

Yes fsmax=usN. But fs in general is ???

16. Oct 4, 2007

### klm

i dont know

17. Oct 4, 2007

### PhanthomJay

Well, you think in this problem that fs might be zero. And you already know that in this problem fs=usN is the wrong answer. So can you venture to guess what the formula for Fs is in general? (HINT: think in between).

18. Oct 4, 2007

### klm

do you mean like fs=ma ? so fs=0 ? i dont really understand what you are getting at sory

Last edited: Oct 4, 2007
19. Oct 4, 2007

### PhanthomJay

Oh OK, I guess you can look it up and you will find that the static friction force is less than or equal to usN. In other words, it can vary from 0 to usN. So back to the problem at hand, is the friction force 0, usN, or somewhere in between (or none of the above).

20. Oct 4, 2007

### klm

oh um okay. yes i know that it can vary, but thank you for clarifying what you mean. i understand how to do the problem now, so thank you very much for all your help.

21. Oct 4, 2007

### PhanthomJay

err, I was hoping you'd help me! Looking at the figure, that chap is standing erect and still, so certainly fs=0 looks like a good answer; however, there could be some leg muscle action that would counteract the friction force and still cause no motion, so I guess that the correct answer is that fs is somewhere in between 0and usN. And if this is a webassign problem, chuck that minus sign.

22. Oct 5, 2007

### klm

ooh sorry! no i checked and the friction force is suppose to equal 0

23. Feb 20, 2009

### mansari

Ff=0. This shows a very important point. When you are not told that an object is slipping or on the verge of slipping, then the friction force is determined using Newton's laws of motion in conjunction with the observed motion and the other forces on the object.