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I started learning rotational mechanics and while looking at a solution for one of the problems in the book, I came across this statement:

In the absence of any external force in horizontal direction, the normal reaction N passes through the centre of mass of the block; when force F is applied, normal reaction shifts in the direction of applied force.

Why does this happen?

Vincent Thacker
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GBaelish
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    Your quote from the text may be taken too much out of context. As stated, I don't think that the statement is necessarily true. Could you include more verbiage from your source, such that the context and any qualifying statements are more readily apparent? – David White Feb 25 '17 at 16:34
  • It is true that in general the reaction force shifts, but the direction it shifts depends on the vertical position of the applied force relative to the height of the center of mass of the block. Consider a stunt motorcycle rider pulling a wheelie, for example. You can experiment with this in a more controlled manner by applying the force at the top of a tall object like a pack of breakfast cereal, and observing which way it falls. If you push at the bottom, it will fall towards you, not away from you! – alephzero Feb 26 '17 at 06:30

2 Answers2

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Yes, it does shift.

The normal force $N$ and friction force $F$ are reaction forces. They adjust according to applied forces.

These forces are in fact many forces spread across the surface of contact. It is only for convenience that we resolve them into normal and tangential components and combine them into single forces which act through the centre of the surface of contact. enter image description here

Without any applied force on the block the total gravity force (weight) is balanced by the normal force : $W=N$. The object does not turn, so the resultant normal force and the resultant gravity force must be aligned, the normal force passing through the centre of gravity.

If a horizontal force $P$ is applied to the upper corner of the block this is resisted (up to the limit of $P=\mu N$) by the friction force on the base : $P=F$. These 2 equal and opposite forces are not aligned so there is a torque on the block. If the block does not topple, there must be an opposing torque. This is created by the resultant normal force $N$ shifting towards the pivoting corner (away from the applied force), out of line with the gravity force $W$. This happens by the individual normal forces decreasing closer to the applied force and increasing further from the applied force, in such a way that $N=W$ still holds.

If the the block and plane are perfectly rigid and perfectly flat then the corner nearest to $P$ immediately lifts off the plane and the normal and friction forces $N, F$ move to the furthest corner. Real blocks and planes deform to some extent before contact is broken; the normal and friction forces increase gradually toward the pivoting corner.

enter image description here

On a microscopic level the bodies deform. Forces and deformations are two aspects of the same interaction : force $\propto$ deformation. Where the deformation is largest, the reaction force is greatest. (It is no co-incidence that this resembles an object floating in a fluid : here the solid object "floats" on an elastic surface.)

sammy gerbil
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  • sammy gerbil, it would be helpful if you also included the case of the horizontal force at the bottom of the block. – David White Feb 26 '17 at 02:31
  • @DavidWhite : Thank you for the suggestion. In the static case $P\le\mu N$ this is not interesting (I think). It only becomes interesting for the dynamic case of $P\gt \mu N$ when the block accelerates. I am not sure the OP wants to get into that. – sammy gerbil Feb 26 '17 at 10:46
  • The OP may not want to get into that, but it is apparent to me that if P is large enough, the block would tend to rotate clockwise in the given picture, which means that the normal force would move to the right edge of the block. This means that the answer depends on where the force P is located relative to the center of mass. – David White Feb 26 '17 at 16:12
  • What do you mean by "large enough"? In that case the applied force exerts no torque about the lower corner, and the normal force will not shift unless the block is accelerating. ... The purpose of my answer is to explain why the normal force shifts position, not to examine all cases comprehensively. – sammy gerbil Feb 26 '17 at 16:45
  • @sammygerbil But what happens at the atomic scale that causes this shifting. – Shashaank Mar 29 '17 at 18:41
  • Deformation and reaction forces occur together. Without any shearing force P the load, deformation and reaction forces are spread uniformly over the interface. The shearing force causes a torque which lifts the applied end a little, increasing the load and deformation towards the pivot end and reducing them towards the other end. – sammy gerbil Mar 29 '17 at 18:54
  • @sammygerbil So would it be totally right if I put it like this ( because I had thought this way)-On applying a force ( like in the above picture) the successive layers of màsses get a bit moved towards the "pivot" point ( type of a deformation , I suppose) ,which causes a bit more mass in increasing order towards the pivot which causes a greater Reaction force (N) at the pivot. The only thing I feel wrong here is that it would also signify the change in position of center of gravity (because of deformation). BTW I have made it a habit to go through a complete answer of yours(whichever I spot) – Shashaank Mar 29 '17 at 21:10
  • I would not go so far as totally right but it seems ok to me. I would say that the mass of a column above any point on the elastic surface changes, increasing near the pivot point and decreasing elsewhere, hence deformation also changes. The position of the CM within a solid body does not change. – sammy gerbil Mar 29 '17 at 21:45
  • @sammygerbil Thanks! But I can't see why center of gravity will not change. While more mass " shifts" towards the pivot ( than originally) while it's coordinate remains same , the CG shall change ( due to a change in internal mass distribution). – Shashaank Mar 29 '17 at 21:54
  • The deformations - inc. change in position of CG - are usually very small, on a microscopic level. However, the shifts in the normal and frictional forces are significant and measurable. In my last diagram the black block is rigid so its CG does not move position; only the horizontal surface deforms. – sammy gerbil Mar 29 '17 at 22:22
  • @sammygerbil So the point of application of these forces shifts to the farthest corner when it starts to topple? – Kunal Pawar Mar 30 '17 at 00:57
  • Yes, the line of action of the resultant shifts towards the pivot. Eventually the pivot is the only point in contact with the ground - the reaction forces are then at the pivot. – sammy gerbil Mar 30 '17 at 01:58
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    @sammygerbil I am getting confused. You said that the mass of successive columns increase on moving towards the pivot. That's the Reaction force increases at the pivot. How is the Reaction force going to shift such a lot WHEN THE THING CAUSING IT SHIFTS AT MICROSCOPIC LEVEL .... – Shashaank Mar 30 '17 at 06:37
  • Sorry, you are right, my comment about the mass of a column is incorrect. – sammy gerbil Mar 30 '17 at 15:11
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    @sammygerbil Then all that I said about the shifting of mass was wrong. There has to be altogether another explanation for this shifting of the Reaction force ( not that to balance torques such a thing shall happen , but why shall such a thing happen ). What happens at that scale that causes this shifting ? I guess the answer shall be in " why the Reaction force acts through the center when there is no force". It's been perplexing because I have been trying to look for a reason that explains this shifting at that scale .. – Shashaank Mar 30 '17 at 18:27
  • I kept thinking about this till now but I have not quite understood it intuitively , why the normal reaction shifts ? Further my reasoning in the above comments and even your last comment say that only the only reason I can see is wrong too. Would you like to provide an intuitive reason as to why the normal reaction shifts when the normal reaction it self comes from the fact that a mass is resting over something which will provide a greater Reaction force if a greater mass is resting on it in constant g – Shashaank May 11 '17 at 11:34
  • @Shashaank My doubts are exactly the same as yours! I mean, after reading a comment your reply is exactly the same as my thoughts! Did u get an explanation to the shifting of the Normal reaction. – think__tech Oct 05 '18 at 08:07
  • Does that mean tension force shifts as well!!??? – Maddy Jan 16 '24 at 08:49
  • @Maddy What tension force? – sammy gerbil Jan 18 '24 at 12:08
  • @sammygerbil Nah My bad... I was thinking A Normal Force is a reaction force, but so is tension. So perhaps tension self-adjusts as well. But a string would only make contact with an object at one point, so unlike normal force, the tension in a string can't change the maximum strength. – Maddy Jan 21 '24 at 20:04
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I think that since there is force being applied horizontally, the internal forces also change and there is some horizontal internal forces acting in order to exert a force on whatever is exerting the horizontal force. The masses inside don't shift or anything, but these internal forces and the normal force etc adjust themselves to make the body inertial and oppose the torque generation as much ad possible. If the external horizontal force becomes large enough, the body ends up toppling over. Basically, to get a detailed analysis, we will need to break down the body into small particles and analyze the forces, however, importantly, all these forces generated on the particles individually are independent of each other and all particles exert forces on each other as well. Due to regular physical restrictions, it so happens that the gross effect is as we see it.

user155871
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