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"A block of side length $a$ and mass $m$ resting on a ground with a coefficient of friction $\mu $. A force $F$ is applied on the upper edge of the block (think of the cross-section and one of the topmost vertices). How much force should be applied to topple the block?"

In the solution key, they showed that the normal reaction acts on one of the vertices rather than the center of mass. After reading a stack exchange article, I came to know that normal force does indeed shift. But I can not understand how to know where it will shift ( I need to know where exactly this is so I can take moments of force i.e: torque) ref: Does the normal reaction shift when a force is applied?

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  • $\begingroup$ In the solution key is the block only in contact with the surface it is on at the vertex of the block? $\endgroup$ Apr 12 '20 at 23:17
  • $\begingroup$ Are you essentially asking how to get from the diagrams on the left to the diagrams on the right in the first figure of the accepted answer in your linked question? $\endgroup$ Apr 12 '20 at 23:25
  • $\begingroup$ Yes how to know where the forces are acting exactly? and also I'm pretty sure the whole block is on contact, I'll check and verify $\endgroup$
    – 666User666
    Apr 13 '20 at 9:25
  • $\begingroup$ Yeah it is infact stated the the normal force shifts to a corner due to force application $\endgroup$
    – 666User666
    Apr 13 '20 at 9:39
  • $\begingroup$ If the whole bottom surface is in contact with the ground then the normal force cannot be just located at the corner. The normal force will only be at the corner if the corner is the only part of the base that is in contact with the ground. $\endgroup$ Apr 13 '20 at 14:29
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For an ideal block standing on an ideal surface, each contacting point carries some load (counteracts a portion of the burdening weight). At each such point a normal force comes into existence to counteract the pressure at that point.

If horizontal and with a symmetric block with its centre-of-mass (CoM) right in the middle, then all the contact points carry equally much and experience the same small normal force. Regarding the torques they cause about the CoM, all those normal forces on the left side of the CoM cancel all those out on the right side. If the CoM is not in the centre, then some normal forces will have to be bigger than others (not evenly distributed).

You rarely need to care about this normal-force distribution over the contacting surface. Instead, how do we in an easier way note the normal forces' influence? We'll just sum them all up (so it still has the same influence in Newton's laws) and let them pass through the CoM (so it still has the same influence on the rotational version of Newton's laws). It has to be drawn at the contacting surface of course, since this is a contact force. And that's why we often only draw one normal force passing through the centre-of-mass. (Of this same reason, we also only draw one weight from the CoM.)

If you now push at the box from, say, the right side, then you are creating a torque which pushes particles to the right up and particles to the left down. In other words, you are influencing the load and disturbing the normal-force balance. Surely, the contact points farthest to the left will experience a larger load and thus a larger normal force. This corresponds to the total normal force being shifted towards the left.

At the very moment where the box is about to tilt over, then all points are about to lift off from their contact with the surface - except for one point: the pivot point at the left corner. At this moment, the load has been lifted off from all those points, so their normal forces are zero, and only the corner pivot point carries any load - and it now carries the whole load. Thus the total sum of all normal forces now equals the normal force at this corner pivot point.

And this happens only at the tipping moment - before then, the total normal force is anywhere between the point under the box' CoM and the corner, depending on when in the pushing phase you are looking.

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  • $\begingroup$ Oh so you're saying right when it starts tipping, all the normal force gets shifted onto the corner? $\endgroup$
    – 666User666
    Apr 13 '20 at 15:37
  • $\begingroup$ What will the normal force per length look like before tilting? $\endgroup$ Apr 13 '20 at 15:38
  • $\begingroup$ @DDD4C4U Yes, indeed. When tipping, only that corner is in contact with the surface - a normal force is a contact force, so it can be nowhere else. How large that single normal force in the corner then is, is a matter for Newton's 1st law, and it might turn out to be equal to the weight of the box. $\endgroup$
    – Steeven
    Apr 13 '20 at 18:49
  • $\begingroup$ @AaronStevens I can't say for sure. But I doubt it is linear. The known conditions are during before the push and after tilting has begun: 1) Before pushing, with no force pushing sideways, the normal-force-per-length is constant, so the same at each point (assuming symmetric and centeret CoM as described). 2) After pushing the box to tilt, the normal force is concentrated at the corner point with a value of normal-force-per-length being zero at all other points than this corner. Had the normal-force distribution been linear, ... $\endgroup$
    – Steeven
    Apr 13 '20 at 18:57
  • $\begingroup$ ... then the neighbour point to the corner should experience a large normal force as well, just slightly smaller, and so on. This is clearly not the case with the final situation, where all force is at the corner. I thus do not think a linear model is fitting - at least not across the tilting. $\endgroup$
    – Steeven
    Apr 13 '20 at 18:57
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The magnitude of the normal force is sufficient to prevent the block from accelerating downward. As long as the material does not yield, it will equal the weight of the object.

The location of the normal will shift to prevent rotation. If you apply a torque $\tau$, the normal will shift sufficiently to oppose the rotation. The limit of this process is when the normal has shifted to the edge of the object. If the external torque were to increase at that point, the object would have an angular acceleration (topple).

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