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Forces on a car skidding to a halt.

  • Thread starter RobertE
  • Start date
  • #1
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Homework Statement



The following is an instructive problem which I've been unable to solve.

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.


Homework Equations



See below.



The Attempt at a Solution



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.)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
960
0
heres my take,
(11,000)g=2Nf+2Nr where these are the normal forces of front and rear tires.
Because of unequal lever arms, to maintain 0 rotation, from lengths of lever arms,

Nr=1.8/2.4* Nf substitute...
 
  • #3
3
0
That's what I thought originally, but it's evidently incorrect

Thanks, but one must take into consideration that the car is decelerating, not standing still, so this becomes more complicated. Your solution (as was mine) is for a stationary car.
The decelerating car would have a tendendency to flip over (ie in the extreme case, back wheels over front) and this tendency serves to reduce Nr and increase Nf, though their sum remains 11000N. So I still dont know how to solve this, but the fact that the center of gravity is .75m above the road must come into play. I suppose the precise location of the center of gravity would determine whether or not the car indeed does flip over.
 

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