A rear-wheel-drive car 2600 lb, has an 80-inch wheelbase, a center of gravity 20
ID: 1828464 • Letter: A
Question
A rear-wheel-drive car 2600 lb, has an 80-inch wheelbase, a center of gravity 20 inches above the road surface and 30 inches behind the front axle, a drivetrain efficiency of 85%, 14-inch-radius wheels, and an overall gear reduction of8 to 1, The car
A rear-wheel-drive car 2600 lb, has an 80-inch wheelbase, a center of gravity 20 inches above the road surface and 30 inches behind the front axle, a drivetrain efficiency of 85%, 14-inch-radius wheels, and an overall gear reduction of8 to 1, The car's torque/engine speed curve is given by Me = 6ne - 0.045ne2 If the car is on a paved , level roadway surface with a coefficient of adhesion of 0.75, determine its maximum acceleration from rest.Explanation / Answer
Lets start with the easy items first: Lets assume the car has unlimited power with the given coeficient of adhesion and weight lets see what the max acceleration is:
F = m_car*a = m_car*g* .75
a = g * .75
a = 24 ft/s^2
Lets see if the car can handle this with out flipping over:
C_g is 30" above road surface and 50" infront of the rear axle.
we need to make sure the moment due to gvaivty is greater than the moment at maximum acceleartion or else the car will flip over.
m_car * g * 50" >= m_car * a * 30
50*g > 24 * 30
this is obviously ture so we dont need to worry about the car flipping
Lets see if the engine has enough power for this.
Torque at the wheels needed to accelearte at a= 0.75*9.81= 7.35m/s^2 is:
T = F*r = 24*(2600/32)*(7/12) = 1137.5 ft-lbs
75% efficency --> T/.85 = 1338.235 ft-lbs
8:1 gear ration, engine must generate 1338.235/8 = 167.28 ft-lbs
167.28 = 6*n - .045n^2
n = 39
Well an engine speed of 39 seems extremely low but I'm not sure what the units are for that equaiton. I assumed n was rpm and Me was ft-lbs. Basically your maximum physical acceleation is 24 ft/s^2 since if you go higher than that your wheels will slip. The car will not wheelie and flip due to that acceleration so you are OK in that respect. You then just need to calculate your enine torque needed for that acceleration. Keep in mind that the car's weight is given not its mass. You need to divide the weight by 32 ft/s^2 to get its mass.
I would guess something with that equation is incorrect or because my math is right for the requried engine torque. If your engine can produce 167.28 ft-lbs of torque then the limiting factor is the wheels grip on the road.
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