The sports car has a weight of 4500-lb and a center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are us=0.5 and uk=0.3, respectively. Neglect the mass of the wheels.

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opudodennis opudodennis · Answer 1 · 2019-12-12T11:02:15+01:00

Answer:

Time=2.72 seconds

Front wheel reactions= 1393 lb

Rear wheel reactions= 857 lb

Explanation:

The free body diagram is assumed to be the one attached here

The mass, m of the car is

[tex]M=\frac {W}{g}[/tex] where W is weight and g is acceleration due to gravity

Taking g as [tex]32.2 ft/s^{2}[/tex] then

[tex]M=\frac {4500}{32.2}=139.75 lbm[/tex]

Considering equilibrium in x-axis

[tex]Ma_G-f=0[/tex]

[tex]Ma_G-(\mu_g\times 2N_B)=0[/tex]

[tex]139.75\times a_G-(0.3\times 2\times N_B)=0[/tex]

[tex]0.6N_B=139.75a_g[/tex]

[tex]N_B=232.92a_g[/tex]

At point A using the law of equilibrium, the sum of moments is 0 hence

[tex]-2N_B(6)+4500(2)=-Ma_G(2.5) [/tex]

[tex]-12N_B+9000=-139.75a_G\times 2.5[/tex]

[tex]-12(232.92a_G)+900=-349.375a_G[/tex]

[tex]a_g\approx 3.68 ft/s^{2}[/tex]

The normal reaction at B is therefore

[tex]N_B=232.92a_G=232.92\times 3.68\approx 857 lb[/tex]

Consider equilibrium in y-axis

[tex]4500-2N_A-2N_B=0[/tex]

[tex]N_A+N_B=2250[/tex]

[tex]N_A+857=2250[/tex]

[tex]N_A=1393 lb[/tex]

To find time that the car takes to a speed of 10 ft/s

Using kinematic equation

V=u+at

10=0+3.68t

[tex]t=\frac {10}{3.68}\approx 2.72 s[/tex]