Friction & Forces: Calculating Time for Moving 3.45m on Tile Surface

[mkwok]

Homework Statement

To meet a U.S. Postal Service requirement, footwear must have a coefficient of static friction of 0.500 or more on a specified tile surface. A typical athletic shoe has a coefficient of 0.755. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.45 m on a tile surface if she is wearing the following footwear?

(a) footwear meeting the Postal Service minimum

(b) a typical athletic shoe

Homework Equations

Fs\leq (less than or equal to) \musn (static frictional constant times the vertical force of the object)
Fk=\mukn (kinetic frictional force times the vertical force of the object

\sum F = ma

The Attempt at a Solution

I don't even know how to start this problem... suppose I use n=1 as the weight of the runner/person... I would get F=0.5N for the force needed to over come to static frictional force of the U.S Postal Service shoes.. but how do I use that to calculate the time?

 

[mjsd]

this is a rather confusing questions, or some assumptions which should be included is not listed .. one of the thing that concerns me is that how should we model the walking person is this context? or was the persion sliding (in which case you need kinetic fricition too)?

 

[Dick]

I think you can assume the person is not sliding. You aren't going to 'overcome' the static friction, you are going to use it. What's the maximum force the person can attain and what does that tell you about the maximum acceleration?

 

[mjsd]

what I don't get about this is wouldn't the person "make a few steps" before reaching 3.45m?? and hence wouldn't there be additional burst of acceleration? I am kind of confused

 

[Dick]

It's an idealized question, it asks for a minimum time. The minimum time is achieved when the runner is constantly in contact with the surface and exerting maximum force without sliding.

 

[mjsd]

yeah.. very idealized... and only if you know what your answer is would you know how to tackle it unless they state those assumptions clearly or you've done this type of idealised question before... still the context actually makes the question harder than it should be.
thanks

 

[Dick]

True. But as it doesn't even specify how many legs the average postal worker has, you can probably forget about the detailed mechanics of running. At least it gives a gender. And we didn't even use that! :)

 

[Baldealge79]

The Setup

This looks like an interesting problem. Some things to remember.

1) it's the frictional force that is doing all of the moving in this case. So the sum of the forces should look something like this:
\sumFx = ma - fk
..
ma = f(force of friction)
ma = \muN
ma = \mu(9.8)(m)
the masses cancel
solve for a
a= 7.399m/s^2

Then use two kinematic equations to find out how long it takes to go from rest a distance of 3.45m at an acceleration rate of 7.399m/s^2

Hint:
v(final)^2 = v(initial)^2 + 2a(xf - xi) ---> xf is 3.45m, xi is 0
v(final) = v(initial) + at

Hope it helps. I get 0.966s w/ appropriate sig digs. I could be missing something though . You would do the same thing for the minimum coeffecient of friction value of 0.500...

 

[Dick]

This looks like an interesting problem. Some things to remember.

1) it's the frictional force that is doing all of the moving in this case. So the sum of the forces should look something like this:
\sumFx = ma - fk
..
ma = f(force of friction)
ma = \muN
ma = \mu(9.8)(m)
the masses cancel
solve for a
a= 7.399m/s^2

Then use two kinematic equations to find out how long it takes to go from rest a distance of 3.45m at an acceleration rate of 7.399m/s^2

Hint:
v(final)^2 = v(initial)^2 + 2a(xf - xi) ---> xf is 3.45m, xi is 0
v(final) = v(initial) + at

Hope it helps. I get 0.966s w/ appropriate sig digs. I could be missing something though . You would do the same thing for the minimum coeffecient of friction value of 0.500...

The object of these forums is to help people learn to solve problems, not to solve their problems for them. Read the forum rules. Posting complete solutions is not helpful.

 

[mkwok]

Thank you Baldealge79, that helped me a lot
thanx for the detailed explanation too =]

 

[Baldealge79]

No problem

MKWOK,

No problem. It's good to see how the concept is supposed to work sometimes, so that you can see how to apply it. Apply that concept to the remainder of the problem, and you will be able to get your final answer.

Good luck.

 

[Baldealge79]

Answers

The object of these forums is to help people learn to solve problems, not to solve their problems for them. Read the forum rules. Posting complete solutions is not helpful.

Dick,

In the future if you have something negative to say, do so in a private message. Common civility affords me that right, and I shouldn't have to state that so bluntly to another adult. I gave an example of only half of the problem at most. I can understand how the site doesn't want to promote cheating, but your subjective opinion of whether or not I gave the "complete" answer is debatable.

 

[Dick]

There was no offense meant to you, I know you were just trying to help. I don't consider the question of whether that constitutes a complete solution to be "debatable".

 

[mjsd]

Dick,

In the future if you have something negative to say, do so in a private message. Common civility affords me that right, and I shouldn't have to state that so bluntly to another adult. I gave an example of only half of the problem at most. I can understand how the site doesn't want to promote cheating, but your subjective opinion of whether or not I gave the "complete" answer is debatable.

there is a fine line (however thin) between posting a complete solution to a problem, and the displaying a type of solution as an example. While learning by examples is ok sometimes, it is most oftenly abused. Because ppl like to take "short cut". And we certainly do not encourage taking short cut in technical subject like physics and maths, because in the longer run, it is true understandings that matter, especially in research when things can be so open ended.

 

[MechaMZ]

This looks like an interesting problem. Some things to remember.

1) it's the frictional force that is doing all of the moving in this case. So the sum of the forces should look something like this:
\sumFx = ma - fk
..
ma = f(force of friction)
ma = \muN
ma = \mu(9.8)(m)
the masses cancel
solve for a
a= 7.399m/s^2

Isn't the total force equal to zero if we let the ma equals to force of friction? there is a acceleration, so it should have a resultant force. I'm confuse, please help me.
ma = f(force of friction)
ma = \muN

thank you =)

 

[MechaMZ]

any idea?