# 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 in the case above as the effective vertical force in the system is zero.but im confused cuz a component of the weight 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 equally to weight.

Kindly point out the error in my thought process and correct me. :)

• for a complex body we always first find the center of mass of whole body then apply the forces on the center of mass. – Anni Feb 5 '16 at 10:01

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 $N$.

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."

• Well, my text book says that in the position mentioned above , the normal reaction is equal to weight of the person ! – Joe Feb 5 '16 at 6:40
• @Joe how does your textbook define "weight?" If it defines it as the force an object exerts on the ground, the logic becomes a bit circular... At any rate, the person forces the ground and the ground forces the person and bending over affects the former - and thus the latter as well. Most people understand these as 'weight' and 'normal force,' respectively. – Asher Feb 5 '16 at 6:48
• well 'weight ' here means (mass * gravity ) of the body – Joe Feb 5 '16 at 6:51
• farside.ph.utexas.edu/teaching/301/lectures/node133.html refer to this situation . its somewhat similar ,the normal reaction force is equal to the sum (Mg + mg ) – Joe Feb 5 '16 at 6:52

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.

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.