Question

A 3200 lb vehicle is being designed with a drag coefficient of 0.35 and a frontal area of 50 ft2. If the vehicle is traveling at 70 miles per hour on a downhill grade of 2% at an air density of 0.002378 slugs/ft3, what tractive effort is required to overcome the vehicle resistance forces (in lb)?

Answer 1

Answer:209.73 lb

Explanation:

Given

Weight of vehicle=3200 lb

frontal area$=50 ft^2$

Velocity$=70 mph\approx 102.67 ft/s$

$\rho =0.002378 slugs/ft^3$

Tractive effort is the sum of aerodynamic, Rolling and grade resistance

$F_D=\frac{\rho }{2}C_dA_fv^2$

$F_D=\frac{0.002378}{2}\times 0.35\times 50\times (102.67)^2$

$F_D=219.33 lb$

Rolling Resistance

$F_R=F_{rl}\omega$

$F_{rl}$=Rolling resistance coefficient

$F_{rl}=0.01\left [ 1+\frac{v}{160}\right ]$

where v is in km/hr

$F_{rl}=0.01\left [ 1+\frac{112.65}{160}\right ]$

$F_{rl}=0.017$

$F_R=0.017\times 3200=54.4 lb$

Grade resistance

$F_G=\omega G$

$F_G=3200\times \frac{2}{100}=-64 lb$ (beacuse of downward slope)

Total resistance$=F_D+F_R+F_G=219.33+54.4-64=209.73 lb$