$F=ma$, so if acceleration is zero the force must also be zero? I was just pushing a huge rock which I couldn't move at all. Given that the acceleration was zero as a result of my efforts, since $F=ma$, doesn't this mean that the force must also have been zero?
But, in reality, I was applying a force $F$, there is a huge rock with mass $m$, but the acceleration $a$ I observed was zero.  So does this mean physics denies the existence of my applied force? How can I reconcile this contradiction?
 A: It is the net force, not force.
A: The confusion comes from how you have written the equation. If you write it like this $$F_{net} = ma$$ it will be easier to see your error. You are exerting a force on the boulder, but net force is zero. This means that other forces such as friction are canceling your force out.

In this case. Friction equals applied force. 
A: Force = mass * acceleration is the basic simplified version of the equation.
There are more complected formulas available for this that take into account more complicated scenarios; like taking in the account of different forces coming from different directions like in your scenario. 
sum(forces) = mass * acceleration
force = the derivative of its linear momentum
and so on ...
http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_second_law
A: The main thing is that the total force you are applying on the body is not enough to move the rock.This means that the total force on the box is zero because the force of static friction is grater than that of your applied force,which cancels out the effect your force.As your forces increases the static friction also increases until a point comes when you are able to overcome that force and the body also moves.Now the friction is kinetic friction which has a magnitude lower than that of static friction.If your draw the graph of force vs static and kinetic friction you will see that the static friction line increases steeply until a certain force after that when it starts to move it decreases a little and continues to remain constant(not fully constant,only to an approximation).
A: This type of question keeps reccuring. I suppose either teachers don't teach this correctly, or students do not pay attention in class. $F=m a$ is really $$\sum F = m a_{cm}$$ These two distinctions (the sum, and the acceleration of the center of mass) make all the difference in the world (At least read on Newton's Laws of motion).
You push on the rock and the ground is pusing on the rock with a net result zero.
At all starts from the following principles:


*

*Momentum is mass times velocity of the center of mass $P=m v_{cm}$

*The effect of a single force over time is to change the momentum $F \mathrm{\Delta} t = \mathrm{\Delta} P$

*Multiple forces can be combined algebraicly for a net (effective) force which describes the net change in momentum $\sum F \mathrm{\Delta} t = \mathrm{\Delta} P$

*For a constant mass the change in momentum becomes a change in speed of the center of mass only $\mathrm{\Delta} P = m \mathrm{\Delta} v_{cm}$

*The change in speed over time is called acceleration $\sum F = m \frac{\mathrm{\Delta} v_{cm}}{\mathrm{\Delta} t} = m a_{cm} $


Understand all these steps and you will be able to understand any situation involving linear motion.
