Action and reaction - not the same effect If I  press on something, it presses me back with the same force. if I  press a pillow with my fist, l do not feel pain but if I  press a rock with my fist with the same force, I  do feel the pain. why?
Regardless of the fact that my force gets distributed over the pillow surface, I should still get a reaction on my fist that should equal the force I applied.
To be able to compare the two well, please assume that my force on the rock is also over a similar area to the pillow and not concentrated on narrow points that do not cover the whole fist area.
 A: I think that what you feel as pain is mainly the shearing forces in your hand. 
With a pillow, you get a pressure spread evenly on your hand so even though the force is high, it is not sheared.
With a rock, the force is localized to a very small contact zone around which the forces (stresses) are strongly sheared. That's bad for your hand (don't experiment too much with this !) and that's what your nerves try to draw your attention to.
Note that even if the overall area of the contact zone is the same, in the case of the rock the contact is in fact limited to a small number of points until either the rock or your hand changes shape.
A: The mass and hardness of the pillow are much lower than those of your hand, while those of the rock are much higher. Thus the force of the impact mostly deforms the pillow when you punch it, and your hand when you punch the rock.
A: I wouldn't put it in terms of force on a pillow but rather look it from this point of view. To throw a fist to something, means that you first gave your fist a certain quantity of motion, say $p$. Then you can model the act of throwing a fist as a collision beetween your fist and the object you are fighting with. 
For example, suppose that the rock has a fixed position, a skerry maybe, and also assume a skerry of marshmallows. The rock is almost a rigid body, while the marshmallows mountain is not. That means that when you hit the rock, your quantity of motion changes from $p$ to $0$ very rapidly, while when you hit the marshmallows mountain it is changing gradually from $p$ to $0$, since the marshmallows will deform and offer a resistance that increases more gradually than in the first case. This, ofcourse, implies that the force exerted from the rock on your fist is much higher than the force exerted by the marshmallows mountain.
From this discussion, it's clear that you don't feel the same force because, actually, you're not applying the same force.

Since there seems to be some confusion, I'll try to put it in an other way: let's think about a mass $m$ free falling on a spring of constant $k_1$ and then on a spring of constant $k_2>>k_1$. Let's analyze the forces on the mass $m$ and on the springs $k_1$ and $k_2$.
At instant of contact: the deformation of the springs (supposing they were initially at rest) are zero, so there's no force exerted by the spring on $m$ in both cases. The only force on $m$ is, if we neglige friction, $m\mathbf g$. As you know, this force is applied by the earth on $m$, so the reaction will be applied by $m$ on the earth (and not on the spring as you seem to mean in your comments). OK, so at the instant of contact the presence of a spring is irrelevant. 
One istant later: it's easier if we think of it in a quantized manner, as a computer does when solving numerically a differential equation. So suppose that for a little interval $\text d t$ the situation is the same as above. If $\text d t$ is small enough, then we can say that the vertical displacement of $m$ is:$$\text d x\approx v \text d t,$$
where $v$ is the speed at the instant of contact. What are the forces now?
The mass is still feeling her weight and exerting her reaction to the Earth. Plus, both springs are compressed by an amount $\text d x$, so the force felt in both cases is:$$F_1=k_1 \text d x,\qquad F_2=k_2 \text d x$$
and, since $k_1<<k_2$, also $F_1<<F_2$. The forces exerted by the mass on the springs are $F_1$ and $F_2$, and have nothing to do with the weight of $m$: they would be the same if $m$ was $M$ or $\cal M$! What would change, would be the center of oscillations, but this is another thing.
One last thing: you may be arguing that the mass is exerting its weight on the spring because of the analogy with ground exerting our weight on us when we stand still. I want to make you note that in that case, you don't know a priori the force exerted by you on the ground and it's not automatical that it is $mg$. What you do is saying: "I'm not accelerating, so the total force on me is zero". The total force on you is $m\mathbf g+\mathbf R$, so the reaction $R=-m\mathbf g$. As you see, you can't find the reaction of the ground as a reaction to the force exerted by you using third law: it's the other way around.
A: 
"please assume that my force on the rock is also over a similar area to the pillow and not concentrated on narrow points that do not cover the whole fist area"

Well, ok, but it's pretty hard to achieve that by pushing against a rock. To do so I guess you would need a rock with a perfect smooth inverted image of your fist, as if you pushed your fist into wet plaster of Paris and let it dry. If you push your fist into such a fist-shaped depression, I think you will not feel much pain.
Of course, a real rock isn't shaped like that. It's either a flat, non-inverted-fist-shaped surface, or else it's rough and jagged. Either way, the force gets concentrated into narrow points that do not cover the whole fist area, and that's why the same amount of force causes pain.
