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In this figure, a car is shown, driving away from the viewer and turning to the left on a level surface.  The following forces are shown on the car: w pointing straight down, N pointing straight up, and f which equals F sub c which equals mu sub s times N, pointing to the left. The forces w and N act on the body of the car, while f acts where the wheel contacts the ground. The free body diagram is shown to the side of the illustration of the car and shows the forces w, N, and f as arrows with their tails all meeting at a point.
This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

Strategy

  1. We know that F c = m v 2 r . Thus,
    F c = m v 2 r = ( 900.0 kg ) ( 25.00 m/s ) 2 ( 500.0 m ) = 1125 N .
  2. [link] shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is μ s N , where μ s is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so N = m g . Thus the centripetal force in this situation is
    F c f = μ s N = μ s m g .

    Now we have a relationship between centripetal force and the coefficient of friction. Using the equation
    F c = m v 2 r ,

    we obtain
    m v 2 r = μ s m g .

    We solve this for μ s , noting that mass cancels, and obtain
    μ s = v 2 r g .

    Substituting the knowns,
    μ s = ( 25.00 m/s ) 2 ( 500.0 m ) ( 9.80 m/s 2 ) = 0.13 .

    (Because coefficients of friction are approximate, the answer is given to only two digits.)

Significance

The coefficient of friction found in [link] (b) is much smaller than is typically found between tires and roads. The car still negotiates the curve if the coefficient is greater than 0.13, because static friction is a responsive force, able to assume a value less than but no more than μ s N . A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that, in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less, as discussed next.

Check Your Understanding A car moving at 96.8 km/h travels around a circular curve of radius 182.9 m on a flat country road. What must be the minimum coefficient of static friction to keep the car from slipping?

0.40

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Banked curves

Let us now consider banked curve     s , where the slope of the road helps you negotiate the curve ( [link] ). The greater the angle θ , the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle θ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for θ for an ideally banked curve and consider an example related to it.

In this figure, a car is shown, driving away from the viewer and turning to the left on a slope downward and to the left. The slope is at an angle theta with the horizontal surface below the slope. The free body diagram is superimposed on the car. The free body diagram shows weight, w, pointing vertically down, and force N, at an angle theta to the left of vertical. In addition to the force vectors, drawn as bold red arrows, the vertical and horizontal components of the N vector are shown as thin black arrows, one pointing vertically up and the other horizontally to the left. Two relations are given: N times cosine theta equals w,  and N times sine theta equals the centripetal force and also equals the net force.
The car on this banked curve is moving away and turning to the left.

For ideal banking    , the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

Practice Key Terms 6

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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