Normal Force, Friction Force, Accel problem

[Etsang2005]

Hi everyone. :) My first post here! Anyway, I have a Physics12AP question that I would like some help with. I've been thinking about it for a long time but I have not been able to solve it.

Here's the problem:
A car on a horizontal road makes an emergency stop such that all four wheels lock and skid. The coefficient of friction is m=0.40. The wheels are 4.2 m apart. The centre of mass is 1.8 m behind the 0.75 m above the front axle (we are looking at two wheels at the side of a car). The car weighs 11,000 N. Calculate acceleration of the car, and normal force and friction force of each wheel.
Hint: The 11,000 N of weight is NOT evenly distributed.

There's the question, any help would be much appreciated! :) Thanks in advance guys.

EDIT: The work I have done so far is not helping at all, I've tried using pythagorus, FN=ma, FF=MFN. I don't really know how to get started with it...

 

[Päällikkö]

Welcome to PF, one of the very best sites on the internet .

The car must not rotate, and there must be no acceleration in vertical direction.
Can you put these two thoughts into equations?

 

[RobertE]

I'm having difficulty with the identical problem, which I describe in more detail under the thread "Forces on a car skidding to a halt," which reads as follows:

A car on a horizontal road makes an emergency stop such that all four wheels lock and skid. The coefficient of friction between tires and road is mu=0.40. The distance between the front and rear wheel axles is 4.2 m. The centre of mass is 1.8 m behind the front axle and 0.75 m above the road. The car weighs 11,000 N. Calculate acceleration of the car, and normal force on each wheel.
Hint: Although the car is not in translational equilibrum, it is in rotational equilibrium.MY WORK SO FAR.

Once the wheels lock, the only (?) horizontal force on the car is that of friction.
F = (mu)(M)(g). But F=Ma. Therefore (mu)(M)(g) = Ma. So a = (mu)(g) = (.40)(9.8) = 3.9 m/s^2. (This is correct)

To calculate the normal force (Fr) on the pair of rear wheels, one would think that by using rotational equilibrium one could sum all the external torques about any point and set that sum equal to 0. A convenient point is where the front wheels meet the road.
Then: (11000N)(1.8m) - (Fr)(4.2m) = 0, which yields Fr = 4720 N, ie 2360 N/wheel

But this is incorrect! The proper rear normal force is 2000N/wheel.

My question is: what's wrong? If the car were standing still, 2360N would be correct.
But the fact that it is skidding to a halt has increased the normal force on the front wheels and decreased the normal force on the rear ones (as there is a tendency for a car in such a situation to flip over forward). So how does one set up the equations for this? They must incorporate the fact that the centre of mass is 0.75 m above the road (which I didnt use in the above calculation.)