- #1

bbbl67

- 212

- 21

## Homework Statement

Okay you have a rear-wheel-drive car, we can model it as a uniform rectangular block, 12 ft long, 4 ft tall, and weighs 4000 lbs. The front and rear axle centerlines are positioned within 2 ft from the far edges of the block. What kind of forward acceleration (in g's) must it achieve to lift its front wheels off the ground at the start? If it maintained the same level of acceleration throughout the run, what kind of 0-60 mph times would it achieve?

## Homework Equations

-Center of mass equations

-torque equations

-Pythagorean theorem

## The Attempt at a Solution

We need to split the car into 2 sub-blocks. The front block is everything from the rear axle centerline to the front, while the rear block is everything from the rear axle centerline to the rear.

Front block:

center of mass:

x = (12 - 2) ft / 2 = 5 ft

y = 4 ft / 2 = 2 ft

d^2 = x^2 + y^2

d = (x^2 + y^2)^1/2

= (5^2 + 2^2)^0.5

= 5.385 ft

m = 4000 lb * 10 / 12

= 3333 lb

Rear block:

center of mass:

x = 2 ft / 2 = 1 ft

y = 4 ft / 2 = 2 ft

d = 2.236 ft

m = (4000 - 3333) lb

= 667 lb

So now we have to add up the balance of rotational forces:

t1 = force of gravity on front block

= 3333 lbf * 5 ft

= 16665 ft * lbf

f = engine linear force

= ?

t2 = engine torque on front block

= f * 5.385 ft

t3 = engine torque on rear block

= f * 2.236 ft

t4 = force of gravity on rear block

= 667 lbf * 1 ft

= 667 ft * lbf

The sum of torques on the overall body:

t1 = t2 + t3 + t4

16665 = 5.385f + 2.236f + 667

15998 = 7.621f

f = 2099 lbf

a = f / m

= 2099 lbf / 4000 lbm

= 9337 N / 1814 kg

= 5.146 N/kg

= 5.146 m/s^2

= 11.511 mph/s

(= 0.52475 g)

I converted to metric in the middle because it was easier to work with, before converting back to US units.

0-60 mph:

T = 60 mph / 11.511 mph/s

= 5.2 s

So the car will pop a wheelie, if it accelerates slightly faster than 0.52475 g or 5.2 s in the 0-60 mph trap. Is there any thing wrong with this methodology? Did I miss anything?