Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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]

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.