# [Physics] When a person bends forward, does normal reation force change as a component of his weight is applied in torque

newtonian-mechanics

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. ðŸ™‚

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 $N$.