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velikii [3]
3 years ago
9

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. I

f 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.
Physics
1 answer:
natulia [17]3 years ago
5 0

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)    \frac{g}{W_van + W_load}

      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 = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

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A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it
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Answer:

t = 1.82

Explanation:

Given

u = 7.70m/s -- initial velocity

s = 30.2m --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

s = ut + \frac{1}{2}gt^2

Where

g = 9,8m/s^2

So, we have:

30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2

30.2 = 7.70t + 4.9 * t^2

Subtract 30.2 from both sides

30.2 -30.2  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9t^2 - 30.2

7.70t + 4.9t^2 - 30.2  = 0

4.9t^2 + 7.70t - 30.2  = 0

Solve using quadratic formula:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Where

a = 4.9;\ b = 7.70;\ c = -30.2

t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}

t = \frac{-7.70\±\sqrt{651.21}}{9.8}

t = \frac{-7.70\±25.52}{9.8}

Split the expression

t = \frac{-7.70+25.52}{9.8} or t = \frac{-7.70-25.52}{9.8}

t = \frac{17.82}{9.8} or t = -\frac{33.22}{9.8}

Time can't be negative;  So, we have:

t = \frac{17.82}{9.8}

t = 1.82

Hence, the time to hit the ground is 1.82 seconds

7 0
3 years ago
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