2.8 A rear-wheel–drive car weighs 2600 lb and has an 84-inch wheelbase, a center of gravity 20 inches
above the roadway surface and 30 inches behind the front axle, a drivetrain efficiency of 85%, 14-inch–
radius wheels, and an overall gear reduction of 7 to 1. The car’s torque/engine speed curve is given by
𝑀𝑒 = 6𝑛𝑒 − 0.045𝑛𝑒 2
If the car is on a paved, level roadway surface with a coefficient of adhesion of 0.75, determine its
maximum acceleration from rest.
Solution
1- Determine the maximum tractive effort for the front wheel drive 𝐹max using the relation:
𝜇𝑊 (𝑙𝑟 + 𝑓𝑟𝑙 ℎ)
]
𝐿
𝐹max =
𝜇ℎ
[1 + ( )]
𝐿
[
Substitute 0.75 for 𝜇 , 2600 Ib. for 𝑊, 0.01 for 𝑓𝑟𝑙 , 20 in. for ℎ ,54 in. for 𝑙𝑟 , and 84 in. for 𝐿 :
0.75 × 2600(54 + 0.01 × 20)
]
𝐿
𝐹max =
= 1067 Ib
0.75 × 20
[1 + (
)]
84
[
2-
Determine the maximum tractive effort for the rear wheel drive 𝐹max using the relation:
𝜇𝑊(𝑙𝑓 − 𝑓𝑟𝑙 ℎ)
[
]
𝐿
𝐹max =
𝜇ℎ
1−( 𝐿 )
Substitute 0.75 for 𝜇 , 2600 Ib. for 𝑊, 0.01 for 𝑓𝑟𝑙 , 20 in. for ℎ ,30 in. for 𝑙𝑓 , and 84 in. for 𝐿 :
0.75 × 2600(30 − 0.01 × 20)
] 691.87
84
𝐹max =
=
= 842.71 Ib
0.75 × 20
0.821
1−(
)
84
[
The lesser value is taken for the maximum tractive effort, therefore, the maximum tractive effort is
842.71 Ib.
3- Determine the torque in the engine 𝑀𝑒 using the relation.
𝑀𝑒 𝜀𝑜 𝜂𝑑
𝐹𝑒 =
𝑟
Substitute 842.71 Ib for 𝑀𝑒 , 7 for 𝜀𝑜 , 85% for 𝜂𝑑 , and 14 in. for 𝑟.
𝑀𝑒 × 7 × 0.85
842.71 =
14
(12)
842.71 × 1.167 = 𝑀𝑒 × 5.95
983.17
𝑀𝑒 = ( 5.95 ) = 165.23 ft-Ib
4- Determine the engine speed 𝑛𝑒 using the relation:
𝑀𝑒 = 6𝑛𝑒 − 0.045𝑛𝑒 2
Substitute 165.23 ft-Ib for 𝑀𝑒 :
165.23 = 6𝑛𝑒 − 0.045𝑛𝑒 2
Solve using quadratic equation:
The engine speed is 94.46 rev/s
5- Determine the rolling resistance 𝑅𝑟𝑙 using the relation:
𝑅𝑟𝑙 = 𝑓𝑟𝑙 𝑊
Substitute 0.01 for 𝑓𝑟𝑙 and 2600 Ib for 𝑊:
6- Determine the acceleration mass factor using the relation:
𝛾𝑚 = 1.04 + 0.0025 𝜀0 2
Substitute 7 for 𝜀0
𝛾𝑚 = 1.04 + 0.0025 (7)2 = 1.162
7- Calculate the maximum acceleration (𝑎) using the formula:
𝑎=
𝐹 −∑𝑅
𝛾𝑚 𝑚
Here ∑ 𝑅 is the summation of resistance and 𝑚 is mass of vehicle.
Substitute 842.71 Ib for 𝐹, 26 Ib for ∑ 𝑅 , 1.162 for 𝛾𝑚 , and 2600 Ib for 𝑚 :
842.71 − 26
816.71
𝑎=
=
= 8.70 ft/s 2
1slug
93.82
(1.162) × (2600 Ib ×
)
32.2Ib
Thus, the maximum acceleration from the rest is 8.70 ft/s2