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a couple of physics questions i truly don't understand...?

1. A 1200kg car rounds a curve of radius 70m banked at 12 degrees at 25 m/s. is there a (static) friction force? if so, how much and in what direction?

I think it's asking if a friction force is needed to contribute to the centripetal force or if a component of the normal force is enough.

2. Similar question:

A car of unknown mass traveling at 19.4 m/s is rounding a curve with radius 80m that's "perfectly banked" for the car. (I think "perfectly banked" means no friction force is needed to keep the car at a constant height. A component of the normal force alone does that.)

If the car speeds up to a constant 25 m/s what coefficient of static friction will be needed to keep the car in its circular path and not skid?

1 Answer

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  • 1 decade ago
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    1. I believe you are correct. Part of the centripetal force is parallel to the bank (and driving the car out). Part of it is perpendicular to the bank (and contributing to the normal force). Gravity is also working in two directions (pulling the car down the bank, contributing to normal force). Friction (if present) is pulling the car down the bank. Question: does uphill component of centripetal force match downhill component of gravity? (Probably not.) If not, what's the difference?

    2. Well, according to what I stated above, "perfectly banked" means

    component of centripetal accel uphill = component of gravity downhill, or

    v²/r • cosΘ = g•sinΘ

    (19.4m/s)² / (80m•9.8m/s²) = 0.48 = sinΘ / cosΘ

    I've never been good w/trig functions, and I don't know a rigorous way to solve this. A little trial and error gives Θ = 25.6º

    At 25m/s, centripetal accel:

    v²/r = (25m/s)²/80m = 7.8 m/s²

    normal accel = 7.8m/s²•sin25.6º = 3.38 m/s²

    uphill accel = 7.8m/s²•cos25.6º = 7.05 m/s²

    gravity:

    normal accel = 9.8m/s²•cos25.6º = 8.84 m/s²

    downhill accel = 9.8m/s²•sin25.6º = 4.23 m/s²

    net up-/down-hill accel = 2.82 m/s² UP

    net normal accel = 12.22 m/s²

    therefore the required µ = 2.82/12.22 = 0.23

    the way I see it.

    (I didn't bother multiplying by mass to get forces. Why bother?)

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