Why don't we consider centrifugal force on a mass placed on earth? Let us say a block of mass is placed on the surface of earth. Then while drawing the forces on that body, we say:


*

*Force $F = mg$ acting towards the center of Earth.

*Normal reaction $N$ offered by the surface of Earth.


If no other forces are acting, we say $F=N$. But what about the centrifugal force $m\omega^2R$ . Why don't we ever bring that into picture? What am I missing?
 A: I think that all the right physics is contained in Martin Beckett's answer and the comments on it, but I'd like to restate it in a way that may bring out what I think the key point is.
In practice, when we do experiments in a lab near Earth's surface, we use a value of $g$ that's been determined empirically at that location. For instance, we might determine it by dropping something in vacuum and measuring its acceleration with respect to our lab. That value of $g$ already includes the centrifugal contribution, so we don't need to (indeed we must not!) include it separately.
We often tell introductory physics students that $g$ is the "acceleration due to gravity," but strictly speaking we're telling a small lie when we do this: $g$ is really the acceleration due to gravity and inertial forces. 
Of course, that lie is only a lie in the context of Newtonian mechanics: when we get to general relativity the distinction between gravity and inertial forces goes away anyway! The acceleration of a falling object in general relativity is most naturally thought of as being all inertial force: the falling object is moving along a geodesic, and the reason we see it as accelerating is that our lab is not in an inertial reference frame.
A: Because it's effect is smaller than the variation in $g$ due to earth's bulge (caused by the same centrifugal force) or the local geology - when you use $9.8m/s^2$ that's just an approximation.
The effect of the bulge and centrifugal force mean that $g$ at the equator is about 0.5% lower than $g$ at the poles
edit: velocity at equator   $40,000 km / 24 h = 1666.7 km/h = 0.463 km/s$
'centrifugal g' =  $(0.463 km/s)^2 )/ 6375 km = 0.03 m/s^2$  or 0.3% of 'g'  
