Maximum acceleration of a front-wheel drive car

[Kaevan807]

Homework Statement

In the badly loaded front-wheel-drive minibus shown in Figure Q26 the centre of gravity
has been moved away from the driving wheels, determine the max acceleration possible in
m/s² if l=3.0 m, h=1 m, and the static co efficient of friction μ_s = 0.3.

(Image is attached below)

A - 2.12
B - 0.89
C - 1.34
D - 1.63
E - 1.15

Homework Equations

Ok well two possible equations, the one I used first of all was a = μ_s * g * fraction of g on front wheel (using moments)

Although I have another equation which says a = μ * g * (distance from back wheel/(distance between wheels - μ * height))

The Attempt at a Solution

Using the first equation I get a = 0.3 * g * 1/3 = 0.98 ~ 0.89 Which is B
Using the second equation I get a = 0.3 * g * (1/3-0.3) = 1.09 ~ 1.15 Which is E

Are either of these equations correct? And if so which one?

Edit Just realized that if I change the sign of the second equation to a + (1/3+0.3) I get exactly 0.89? Coincidence?

Any help would be great

 

[PhanthomJay]

In your first attempt, the normal forces on the wheels do not split 2/3, 1/3 when the car is accelerating. In your second attempt, are you using some equation you found somewhere, or doing some calculation which you don't show?

 

[haruspex]

if I change the sign of the second equation to a + (1/3+0.3) I get exactly 0.89? Coincidence?

No coincidence. Can you derive the correct form of the equation? Have you perhaps taken the equation you quoted from a different context?