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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?

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    $\begingroup$ By adding "more weight on top", you are also increasing the volume, so density probably isn't the right way to think about it. $\endgroup$
    – Kyle Kanos
    Commented Oct 24, 2013 at 20:05
  • $\begingroup$ I found a great answer to 3 in the following article: dallaswinwin.com/Newtons_Laws/Newton%20Third%20Law.htm. The reaction is the same. $\endgroup$
    – pran
    Commented Nov 1, 2013 at 4:39
  • $\begingroup$ Regarding 1, N(t) is higher initially if you are pushing the mass on ground vs. pushing it on a bed of cotton/foam/sponge ... because the body comes to rest quicker on a dense object. That is why I talk about density. $\endgroup$
    – pran
    Commented Nov 1, 2013 at 4:42

3 Answers 3

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No.

If you take all your masses and and squeeze them into a tiny volume, although the density increases, the normal force is the same.

Loosely speaking consider the following:

  • 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, $\delta = \frac{1-\nu^2}{E} \iint \frac{P(\vec{r})}{|\vec{r}|}\,{\rm d}A$
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1:No it is wrong to say so.in fact N depends on volume and density and thus ultimately on mass.when you add objects on top of one another volume changes with density.

2:yes,truly so.But you talk as if the ground is elastic and takes a considerable amount of time.But since the ground is rigid the time will be negligibly small accelaration will change to 0 instantaneously.

3:The "kick" you say is actually the reaction force. so just vanish the term Fb and you will see the result.

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  • $\begingroup$ 1. What you say is not accurate because I could take the same mass of cotton and wood. The Normal for cotton will take some time to become the same as that of wood (until the cotton gets to the same density as wood. $\endgroup$
    – pran
    Commented Oct 25, 2013 at 18:17
  • $\begingroup$ 2. I am just presenting a case of a body depressing another body. For my discussions in this case, change of acceleration to 0 in a smaller interval or larger interval is not that important. The point is that N will eventually equal the force applied on the body but not initially. This seems to happen faster for dense objects. $\endgroup$
    – pran
    Commented Oct 25, 2013 at 18:21
  • $\begingroup$ 3. If I attempt to push a body, I will find myself kicking backwards to move forwards. This is not the reaction on me. This is an active action on my part on an external object - the ground. $\endgroup$
    – pran
    Commented Oct 25, 2013 at 18:25
  • $\begingroup$ @pran 2.Correct.Now i think question 2 is clear. $\endgroup$
    – soumyadeep
    Commented Oct 26, 2013 at 9:01
  • $\begingroup$ @pran pls elaborate what you wanna say in Qs.1 $\endgroup$
    – soumyadeep
    Commented Oct 26, 2013 at 9:01
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You start off by saying that normal would take some time before becoming equal to weight of placed body, this is wrong because you are assuming that the bodies will sink in the ground due to their weight, but that is the practical case, in that case the tensile strength of ground gives in and it shows strain till it develops enough stress to counter balance the total weight, the question you are asking is idealisitic and in this case the bodies will not sink in the ground and will remain above the surface of ground at all times.

1st : it is not accurate to say so, since normal force actually depends on massof the body,and mass is a function of both density and volume of the bodh, remeber $ m = \rho V $ so as the volume need not be constant it is inaccurate to say that it depends on density

2nd : since in accordance with the realm of question, the bodies will not sink into the ground, it can be said as soon as they touch the ground the normal equals their toa weight without any time delay.

3rd : I see that you have used $F_b$ applied by the person on ground in the equations for the block, this is not valid untill you are using for the entire system at once, which wouls also include the reaction from ground to $ F_b$ applied by the person. Individually if you see the reaction from ground would equate $F_b$ the force $F$ applied by person on block and block's reaction $R$, then the rest can be equated as $F - μmg = ma$

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