Question

A 2600 lb vehicle has a drag coefficient of 0.35 and a frontal area of 22 ft2. What is the minimum tractive effort required for this vehicle to maintain a 60 mi/hour speed on a 5% upgrade through an air density of 0.002045-slugs/ft3

Answer 1

Answer:

minimum tractive effort  = 232.54 lb

Step-by-step explanation:

Given data;

vehicle weight  is 2600 lb

drag coefficient is 0.35

frontal area 22 ft^2

vehicle speed is 60 mi/hr = 88 ft/sec

gradient - 5%

air density = 0.002045 slugs/ft^3

drag force is given as

$F_D = \frac{1}{2}C_D \rho AV^2$

        $=\frac{1}{2}\times 0.35\times 0.002045\times 22\times 88^2$

$F_D = 60.97 lb$

force due to vehicel weight

$F_W = 0.01 (1+\frac{v}{147})W$

        $= 0.01(1+\frac{88}{147}) 2600 = 41.565 lb$

force due to gradient

$F_G = W\times g$

        $= 2600 \times 0.05 = 130 lb$

minimum tractive effort

$F =F_D+ F_W+ F_G$

  =60.97 + 41.656 + 130 = 232.54