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    Maria A Gonzale's Avatar
    Maria A Gonzale Posts: 13, Reputation: 1
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    #1

    Sep 29, 2008, 08:33 PM
    Confusing Physics
    How about the formula for this problem when the greatest acceleration a runner can muster if friction between her shoes and pavement is 65% her weight? When acceleration of gravity is 10 m/s^2:confused:
    follow please's Avatar
    follow please Posts: 64, Reputation: -1
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    #2

    Sep 29, 2008, 08:36 PM

    Uhm I'm pretty sure that the acceleration of gravity is 9.8 m/s^2, and you need to know the runners weight, particularly in kg.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #3

    Sep 29, 2008, 11:53 PM

    Again F=ma

    Let the weight be 'W'. Don't forget that weight is mass x acceleration due to gravity. So, weight is equal to 'F'. Friction = 65% of F. Therefore the resultant force forward will be 'F - 65%F' or '35% of F'.

    So I think that her maximum acceleration will be 35% the acceleration she should have if there were no friction.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Sep 30, 2008, 05:48 AM

    I'm afraid there's lots of bad "help" in all the previous posts.

    Start with F=ma. The runner's acceleration is therefore a = F/m. So you need to know the force moving the runner forward. This force comes about from the friction between the runner's shoes and the ground. This force is given as 65% of the runner's weight, or 0.65*mg.

    Hence: a = 0.65 mg/m = 0.65g.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Oct 1, 2008, 09:05 AM

    Ok, sorry if I made mistakes. But it's through mistajes that we do progress, right?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Oct 1, 2008, 11:13 AM

    You were using Newton's 3rd law...

    But two things I'm not understanding... that means that in an environment (a race where all the conditions are the same for each competitor) with friction as 65% of a body's weight, then, the maximum acceleration will always be 0.65g, that is 6.5m/s^2, even if all the forces applied by each body is different?

    Secondly, if the forward force and the friction are equal, how can there be any acceleration?

    Please, help me understand!!
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #7

    Oct 1, 2008, 04:09 PM

    The issue is that this problem asks the student to apply a model of friction force = weight * coefficient of friction, which is really only valid under conditions of sliding friction. It is not a very good model for what happens in real life when you run. When you run you push forward off the ground with a force that depends on the strength of your legs and the angle of attack of your foot to the floor. As long as your shoe doesn't slip, the coefficient of friction doesn't enter into it.

    As for your second point - here friction is helping the runner accelerate, not slow down. It's analogous to a car that is trying to get started in snow - you will get a some forward force even though the wheels are spinning like crazy. The amount of friction between the tire and the ground allows the car to accelerate, just as the friction between the runner's shoes and the ground in this problem allows her to accelerate. The silly thing is that the runner's feet in this problem would be spinning, like a character in a cartoon.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Oct 5, 2008, 09:28 PM

    Ok...

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