Object moving down an inclined plane and an object moving along a banked road Why is the normal force of an object sliding down an inclined plane equal to mgcos(θ), whereas for an object moving along a banked road, the normal force of the object is mg/cos(θ)? Also, in some textbooks I see that the frictional force on the object along a banked road is down the inclined plane and contributes to the centripetal force, but in other textbooks I see that the frictional force is in the opposite direction? Great thanks for any clarity! :)
 A: The vertical component of the normal force plus tire friction must support the weight of the car on the banked track if it is to remain at a given height on the banked track.  This vertical component is $mg=F_N cos(\theta)$.  This means that the normal force becomes $F_N=mg/cos(\theta)$.
The horizontal component of the normal force supplies the centripetal force that takes the car in a circle around the track.  This force is $F_x=F_Nsin(\theta)$.  If this force component equals the centripetal force required to keep the car on a circular path, the car will not try to move up the track or down the track, and the frictional force on the tires will be zero in the upward-downward direction on the track.  If this force component is "low" due to a car velocity that is low, the friction force on the tires will be pointing up the track as the car tries to slide down the track.  If this force component is "high" due to a car velocity that is high, the friction force on the tires will be pointing down the track as the car tries to slide up the track.  Thus, the magnitude and direction of the frictional force on the car tires will depend on whether or not the horizontal force component of the normal force is greater than, less than, or equal to $mv^2/r$, which is the centripetal force on the car as it travels in a circular path.
For the block sliding down an inclined plane, the net force on the block is zero (the forces balalnce) only until the static friction force pointing up the incline reaches its maximum value as given by the static friction coefficient.  Once this maximum value is exceeded, the forces on the block are no longer balanced, and the block slides down the incline.
A: The motion is in different directions for the two cases; see the figure below.

For motion down an inclined plane, the motion is sliding down the plane which is motion in the direction parallel to the plane surface.  For motion on a banked road, you are interested in motion in a circle due to the centripetal force, with no sliding motion.  So, you resolve the components of the normal force differently for the two cases.
You asked whether the force of friction for the circular motion is up or down the plane. Friction always opposes the actual or impending motion, and in this case can be either up or down the road, or zero, depending on the speed of the car.  The answer by @David White explains this.  In my drawing I took friction down the plane, therefore implicitly assuming a high velocity.  For a low velocity just replace $f$ with $-f$ in the equations.
This can be seen by considering the motion of the car from a observer in the car.  For the observer, there is a centrigugal force $F_{cent}$ of magnitude ${mv^2 \over r}$ acting in the outward horizontal direction such that the acceleration relative to this observer is zero; that is for this observer $-F_{cent} + Nsin\theta + fcos\theta = 0$, taking the positive direction as inwards and friction as down the road. As $v$ is sufficiently large for a given $m, r, N,\theta$, the centrifugal force can be large enough such that $f$ is zero or negative (up the road).
