The reason is very simple. The coefficient of frictions needs to be higher than 0.5 for a square body to topple (assuming that the force acts at the topmost point). This does not happen in reality because if µ = 0.5, the surface must really be very rough. Most surfaces are smoother than this. Hence a square body prefers to move by translation and not by toppling.
But now let us consider a hexagon. a hexagon.Again assume that the coefficient of friction is µ and that the force is being exerted at the top.Assume the width of the sides to be a.
A gain we can take torque about A.The side of the hexagon is a.Therefore the height will be a tan 60. You can easily see that this is the case, by looking at figure.
The torque about A is,
Now tan 60 ~ 1.732.
Therefore the body will topple for F > 0.21 mg You can now see the huge difference. for µ < 0.21, the body will translate and for µ > 0.21, the body will topple.
The situation is again reproduced below
Again the same conditions that were discussed before,regarding toppling hold true. When a body just topples, the coefficient of friction is less than µmg. The point of constant is a rest (i.e. point A). Different points have different accelerations.
These are typical characteristics of toppling and you should now be clear about these.
Note that we can make an interesting observation. The value of µ required to make a square topple is 0.5, where as the value of µ required to make a hexagon topple is 0.21. this obviously bears out the fact that it is easier to topple a hexagon than a square. Similarly you will find that the minimum value of µ required to make a body topple will decrease if the body is an octagon.It will reduce still further for a decagon. This means that a decagon will move by toppling more easily than by pure translation. The force required to move it by toppling is less.
