CATAGORY: Mechanics
QUESTION: What is the maximum acceleration a car can undergo if the
coefficient of static friction between the tires and level ground is 0.80?

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The forward push on the car is provided by the reaction of the car tires
pushing on the road. To push the car forward the tires must not slip so
the forward force must equal the static friction force.

(1)         Friction force = Forward force

The Friction Force is Uk, the coefficient of static friction, times the
normal force  which in this case is just the weight of the car mg, where
m is the mass of the car and g =9.8m/s^2 is the acceleration due to
gravity. The Forward Force is the mass of the car times the acceleration
of the car. Thus

(2)          Uk mg = ma

dividing by m gives

(3)          Uk g = a


So

(4)          a = 0.80 x 9.8m/s^2 = 7.84 m/s^2

                                 = 25.6 ft/s^2

If a car could continue to accelerate like this it would be going about
175 mph after 10 seconds!

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                  *                  *
                  *  a = 7.84 m/s    *
                  *                  *
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CATAGORY: Mechanics
QUESTION: A box is given a push so that it slides across the floor. How
far will it go, given that the coefficient of kinetic friction is 0.30 and
the push imparts an initial speed of 3.0 m/s?

The acceleration of the box, a(box), (or perhaps I should say
deceleration) is given by the coefficient of sliding friction Uk times the
weight of the box mg (g = acceleration due to gravity =9.8 m/s^2) divided
by the mass of the box m. In effect we are just saying that a(box) = F/m
where m F is the friction force that slows the box down.

                         Uk mg
(1)          a(box) = - ------- = -Uk g
                           m

The minus sign means the box is slowing down.

The force on the box is then F = ma(box) so

(2)          F = -Uk g m

The work done on the box as it slides across the floor is FL where L is
the distance the box slides.

(3)          W = -Uk g m L

The minus sign means that energy is taken away from the box's kinetic
energy. The box slows to a halt when the initial kinetic energy equals
the energy lost from the box due to friction thus

(4)       1
          - m v^2 = Uk g m L
          2

          or after cancelling  m

(5)       1
          - v^2 = Uk g L
          2

Solving for L gives

                 v^2
(6)       L =  -------
                2 Uk g

                   (3.0m/s)^2
            =  -------------------
               2 x 0.30 x 9.8m/s^2

            = 1.53 m

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              *          *
              * L = 1.5m *
              *          *
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