# [Physics] Is the Normal force dependent on the density of the object the body is resting on

forcesnewtonian-mechanics

If a body is resting on the ground, the Normal force equals its weight. If I keep putting another body on top of it, eventually the body will start moving into the ground until the new Normal equals the total weight of the body. I can add more weight on top and the body will move further into the ground. So it appears that the denser the object gets, the more the Normal force.

So my question is 3 folds:

1. Is it accurate to say that the normal force depends on the density of the object that the reference object is sitting on?

2. It appears that when the body starts moving into the ground, the equation at any given time t will be Mg – N(t) = Ma where M is the total mass which essentially says that the force that the ground feels on it at time t is N which will keep changing with time and become Mg at some point where acceleration becomes 0?

3. If I draw a free-body diagram of an object moving horizontally forward(just to distinguish from backward) because of force applied on it, I will have two forces – The external force applied on it and the force of friction resisting its motion and the equation will be
F – $\mu$mg = ma or
F = m($\mu$g + a)
When I draw the free body diagram of the person pushing the object, I will have three forces – the reaction from the body that is being pushed, the force of friction in the forward direction and the force applied by the person in the opposite direction to his/her movement (a kick backwards). The equation for the person will be
$\mu$Mg -R -F$_{b}$ = Ma or
R = M($\mu$g -a) – F$_{b}$
It would seem that perhaps the reaction to Force F will not be the same as F. Shouldn't action and reaction be the same?

• Density + Volume = Mass, $m = \rho V$
• Mass + Gravity = Weight, $W= m g$
• Weight + No Motion = Equal and Opposite Normal Force, $N=W$
• Normal Force + Area of Contact = Contact Pressure, $N=\iint P(\vec{r})\,{\rm d}A$
• Contact Pressure + Elasticity = Contact Deflection, $\delta = \frac{1-\nu^2}{E} \iint \frac{P(\vec{r})}{|\vec{r}|}\,{\rm d}A$