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Applications of Newton laws

  1. Mar 12, 2004 #1
    An automobile is going up a grade of 15 degree at a speed of 30m/s. the coefficient of static friction between the tires and the road is 0.7. a) What minimum distance does it take to stop the car? What minimum distance would it take if the car were going down the grade?

    I know how to get the distance v^2 = v^2 + 2ax Deltax
    how to get the minimum??
    can anyone help me with this pls?
    Last edited: Mar 12, 2004
  2. jcsd
  3. Mar 12, 2004 #2


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    Staff Emeritus
    Science Advisor

    I presume you have the brakes on and so the tires are sliding down the road- that's where the friction comes in.

    When saying the "minumum" distance, they are really just asking for the distance in which the velocity becomes 0.

    You will need to calculate the components of force down the slope(gravity) and back up the slope (friction).

    You need to know the total gravitational force: that's just the car's weight but you didn't give it here. Are you given that?

    Draw a picture of the slope: a right triangle with one angle 15 degrees. Draw the "force" diagram: that will be a right triangle with hypotenuse downward one leg along the slope and one leg perpendicular to it. You should be able to determine that the angle away from the slope of that triangle is also 15 degrees. Take the "length" of the hypotenuse to be the weight of the car. The component of force along the slope is then "weight times sin(15)". That's the component of force pulling the car down the slope.

    Since you know the "coefficient of friction", the friction force due to the tires is that times the "normal component" of weight- the other leg in your right triangle above. The friction force back up the slope is .7 times weight time cos(15).

    Now compare those. Hopefully, the friction force will be greater than the gravitational force (other wise the car won't stop!) and the "resultant" force is the difference between the two. Use that to find the acceleration (F= ma) and then the distance necessary to stop.
  4. Mar 12, 2004 #3

    Doc Al

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    Staff: Mentor

    A few notes to Halls's answer:

    They ask for minimum stopping distance. The way to stop in the shortest possible distance is to apply the brakes hard, but not hard enough to lock them. Once the brakes lock, the tires start slipping, and you are now relying on kinetic friction to slow you down. Kinetic friction is a weaker force that static friction. That's why you want to keep your wheels rolling without sliding.

    Also, while it may appear that you need the weight of the car, you'll find that it cancels.

    Follow Halls's method and you'll be fine. Another way to approach the problem is to use "energy methods", if you've covered that. The total mechanical energy (KE + PE) decreases since mechanical energy is converted to thermal energy by friction. So: ΔKE + ΔPE = - (work done against friction).
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