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A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. If the van is traveling at 40 ft/s, determine the distance it skids before stopping. The brakes cause all the wheels to lock or skid. The coefficient of kinetic friction between the wheels and the pavement is . Assume that the two rear wheels are one normal, NB, and the two front wheels are one normal, NA.

Sagot :

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    [tex]\frac{g}{W_van + W_load}[/tex]

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = [tex]\frac{v_o^2}{2a}[/tex]

        x = [tex]\frac{v_o^2}{2 \mu g}[/tex]

        x = [tex]\frac{40^2}{2 \ 32 \ \mu}[/tex]

        x = 25 / μ     [ ft]

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