Calculating Friction Force on a Stationary Dancer on a Drawbridge

[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.

 

[PhanthomJay]

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.

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

 

[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

 

[PhanthomJay]

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

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

 

[klm]

i don't mean that the variables are zero, i just mean that there is no net force. so -Fs=0 and n-w=0

 

[PhanthomJay]

i don't mean that the variables are zero, i just mean that there is no net force. so -Fs=0 and n-w=0

So are you sayinfg F_f = 0?

 

[klm]

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

 

[PhanthomJay]

well no b/c i mean it has to have some magnitude...i don't really know?

Well maybe, but why not no magnitude at all?

 

[klm]

yeah i think you are right, it shouldn't have a magnitude

 

[PhanthomJay]

yeah i think you are right, it shouldn't have a magnitude

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

 

[klm]

Fs=UsN
i think i understand, thank you

 

[PhanthomJay]

Fs=UsN
i think i understand, thank you

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

 

[klm]

that is the formula that we learned

 

[klm]

that is the formula that we learned fsmax=usn

 

[PhanthomJay]

that is the formula that we learned fsmax=usn

Yes fsmax=usN. But fs in general is ?

 

[klm]

i don't know

 

[PhanthomJay]

i don't know

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).

 

[klm]

do you mean like fs=ma ? so fs=0 ? i don't really understand what you are getting at sory

 

[PhanthomJay]

do you mean like fs=ma ? or fs=0 ? i don't really understand what you are getting at sory

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).

 

[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.

 

[PhanthomJay]

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.

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.

 

[klm]

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

 

[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.