When a person bends forward, does normal reation force change as a component of his weight is applied in torque? My thoughts :i read that normal reaction is equal to the weight of the person. As  in the case above, the effective vertical force in the system is zero. But, I'm confused, cuz a component of the weight vector is used in torque and other towards the point of contact.
I read that bending forward also increases the horizontal force hence causing friction on a rough surface, but if this horizontal force comes from weight , how can the vertical normal reaction force still be equal to weight.
Kindly point out the error in my thought process and correct me. :)
 A: Consider to configurations below:

Necessary condition of equilibrium is $x_N=x_G$
When you bend and still are in equilibrium (configuration 1 and 2), then certainly $x_N=x_G$. When you bend, your center of mass ($G$) displaces. But, until $x_G\le x_U$ you can be in equilibrium because application point of resultant normal reaction $N$ displaces with $G$'s displacement.
If you bend more (configuration 3), so that $x_G\gt x_U$; then you will rotate and cannot be in equilibrium (you can check this by calculating resultant torque about an arbitrary point).
Until you are in equilibrium, magnitude and direction of the resultant normal reaction $N$ don't change ($N=mg$). What that changes, is the application point of $N$.
A: You're correct: bending over causes a person to temporarily "weigh" less, or exert less force on the ground at least.
Considering the mechanics of the human body, the muscles in the torso pull down on the upper body and up on the lower body. Since the lower body is accelerated upward by this action, it applies less force to the ground, and consequently the balancing normal force from the ground is also less.
The reverse of this is when somebody is already bent somewhat and straightens up. They temporarily apply more force to the ground, and if they apply enough extra force quickly enough they can even accelerate themselves off the ground entirely. We call this "jumping."
A: Yes, when you drop the soap and bend forward to pick it up, your center of mass changes and so does the location of application of the net normal force applied by the floor on you.
A: This is a complex problem because the moment of inertia of the body changes so consider the motion of the centre of mass.
When upright taking down (y-direction) as positive
$mg - N_i = 0$  where $N_i$ is the normal reaction at the start.
Now have the centre of mass accelerating downwards and taking to the right (x direction) as positive.
$mg - N = m a_y$ and $\mu_s N = ma_x$
So the normal reaction does decrease and the frictional force which depends on the normal reaction provides the horizontal acceleration.
A: After you come to a stop in the bent forward position, the ground has to be exerting a torque on your feet to balance the torque associated with your center of mass no longer being directly over your feet.  The force that the ground exerts on your feet is distributed over the contact area with the ground.  If you are standing straight up, the force is more or less uniformly distributed over your feet.  But, when you bend forward, the regions of higher force per unit area develop near your toes and the regions of lower force per unit area develop near your heels.  The average over the entire foot is still your weight, but the new force distribution produces a torque.
