Normal reaction on wheels when car is taking a turn I recently read in a book that for a car taking a turn on a horizontal surface, the normal reaction of the road on the outer wheels is always greater than the normal reaction on the inner wheels. 
However, I feel it should be the other way round. For example, in real life situations the outer wheels are lifted off the ground when taking a really sharp turn at high speed. This clearly shows that inner reaction is greater. 
Any help would be appreciated.
 A: I think you'll find it's the other way around to what you think, as described in the book. A car going around a bend must be pushed around the bend by the friction force on its tyres directed at right angles to its motion and towards the centre of the curve (there is a nett centripetal force - I'm looking at this problem from a momentarily commoving inertial frame). So this centripetal force is acting towards the centre of the bend and acting lower (vertically) than the car's centre of mass. The nett torque (axis along the car's direction) of this force about the centre of mass therefore tends to tip the car outwards. 

So above is a drawing of my car. Don't laugh - I wanted to be a designer for Lamborghini once and can't for the life of me think why I didn't get the job :)
Anyhow, the car is driving away from us and rounding the bend to our right. All the forces on the body are shown in orange. There are: (i) the reaction forces from the road $n_{o,1},\,n_{o,2}$ on the forward and hinder outer wheels and $n_{i,1},\,n_{i,2}$ on the inner wheels, (ii) the sideways frictions $c_{o,1},\,c_{o,2}$ and $c_{i,1},\,c_{i,2}$ on the outer and inner wheels that supply the centripetal force to push the car around the bend and (iii) the  car's weight. I show the angular momentums in green: small ones $L_{o,1},\,L_{o,2}$ and $L_{i,1},\,L_{i,2}$ from spin of the wheels and the spin  $L_C$ of the car body as it rounds the bend. 
As the car begins to round the curve, there must be an imbalance between the sideways friction on the forward and hinder wheels to accelerate the body's angular momentum to a uniform value to make the car's heading rotate to follow the bend. However, we assume that steady state has been reached, so that $L_C$ is constant and given by $I_x v / R$ where I_x is the mass moment of inertia about the $x$ axis, $v$ the car's forward speed and $R$ the bend radius. Therefore no torque in the $x$ direction is needed to keep the body rotating. The angular momentums of the wheels are rotating with the car, so one needs a small torque in the $z$ direction (along the car's motion) to make these rotate with the car. Hence, at the steady state, there is no difference between the forces on the forward or hinder wheels and we can treat this as a two dimensional problem. There is, of course, a difference between the forces on the inner and outer wheels: let's now let $c_i, c_o$ and $n_i, n_o$ stand for the total forces from the inner and outer wheel pairs.
Describing the centre of mass's motion by Newton's second law yields:
$$c_i + c_o = m \frac{v^2}{R}\qquad(1)$$
$$n_o + n_i = m g\qquad(2)$$
where $m$ is the car's mass. Here we have written equations for the horizontal ($y$) and vertical ($x$) forces, respectively. Now for the description of the car's rotational dynamics by Euler's second law: let $2 a$ be the car's width and $h$ the height of its centre of mass above the ground (here we assume the centre of mass is in the middle of the car). The torque equation for torques in the $z$ direction (along the car's motion) is:
$$(n_o -n_i) w - (c_o + c_i) h = L \frac{v}{R}\qquad(3)$$
where $L$ is the total, constant magnitude angular momentum of the wheels. The $L$ vector rotates at angular speed $v/R$ uniformly with the car, so its rate of change is $L v/R$. Now we have three equations in three unknowns $n_o, n_i, c_t$ where $c_t$ is the total sideways force $c_o + c_i$ - how it distributes itself between the inner and outer wheels cannot be determined by this first order analysis. We find:
$$n_o=\frac{g m R w+h m v^2+L v}{2 R w}\qquad(4)$$
$$n_i=\frac{g m R w-h m v^2-L v}{2 R w}\qquad(5)$$
and, since the chosen directions in the picture are such that all the quantities are positive, we see that $n_o$ is always greater than $n_i$.
To end the analysis, we would need to check that the sideways force is not bigger than that which can be supplied by the tyres, given co-efficients of friction (which are roughly equal to 1 for rubber tyres) and the solutions in (4) and (5).
A: You're getting confused about normal reaction. The normal reaction is a "reactive" force, so it depends on the actual active force. 
When you're taking a sharp turn, centrifugal force pushes your car outwards. Since the center of mass of the car is above the level of the wheels, it tends to create a moment which pushes the outer wheels down and the inner wheels up. Since the outer wheels are being pushed downwards, the ground exerts a greater normal force on the outer wheels. 
