# Equilibrium of a tower of 2-d blocks

In the image below,

• All the blocks are frictionless & identical with side of unit length, height $h$, weight $w$ & center of gravity at their geometric centers.
• The 2 lowest blocks are on solid ground.
• The distance from the corner of each block to the midpoint of the bottom side of the box above it is given (namely $a1,a2,b1,b2$).
• forces $F_{ac}$ and $F_{bc}$ are the resultant reaction forces exerted by blocks $A$ & $B$ on $C$. I am interested in the behavior of these blocks immediately after setting them in this configuration and releasing them, or more specifically: For what relation between $a1,a2,b1$ & $b2$:

• do $A,B$ and $C$ move?- case 1
• do only $A$ and $C$ move?- case 2
• do only $B$ and $C$ move?- case 3
• is the configuration stable (does not change at all once set under this condition and then left.)?- case 4

For those interested, here's my approach & what (I think) I know already:

In an attempt to find the limiting conditions (the borderline conditions between equilibrium and non equilibrium), I assumed that initially $C$ will tend to be in equilibrium.(I have no rigorous justification for this assumption, just a hunch that "this isn't where the trouble is").Under this condition, $F_{ac}$ & $F_{bc}$ can be calculated, and the moments due to their "equal and opposites" ($F_{ca}$ & $F_{cb}$) about $P1$ and $P2$ can be obtained as: $$M_{ca}(x,y)=w(.5+a1-a2-x)/(x/y+1)$$ $$M_{cb}(y,x)=w(.5+b1-b2-y)/(y/x+1)$$ where $x$ & $y$ are the perpendicular distances of the respective forces from the center of $C$.

With some ad-hoc and shaky logic here's what I arrived at:

• when $$w\cdot a_2 <M_{ca}(.5,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,.5)$$ case-1 occurs with $A$ & $B$ touching $C$ only through it's vertices.

• when $$w\cdot a_2 =M_{ca}(x1,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,x1)$$ where $a1\leqslant x1< .5$ case-1 occurs with $C$ rotating with $A$ maintaining a surface of contact with $A$ but only a point contact with $B$, but if $$\boldsymbol{w\cdot a_2 =M_{ca}(x1,.5)\quad \& \quad w\cdot b_2 \geqslant M_{cb}(.5,x1)}$$ the configuration is stable (case-4).

• when $$w\cdot a_2 >M_{ca}(a1,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,a1)$$ case-3 occurs, but if $$\boldsymbol{w\cdot a_2 >M_{ca}(a1,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,a1)}$$ the configuration is stable (case-4).

• The above conditions with $a$ & $b$ exchanged along with their corresponding variables.

But I have no way to check this or provide a satisfactory argument for these conditions, especially the parts written in bold (I arrived at it by putting various combinations of arguments for $M_{ca}$ & $M_{cb}$ & thinking about what would happen in each case). Is this set of conditions right?. What would be a good approach with a logical progression of steps to solve it?

• The forces always appear on the edges of the contact. – John Alexiou Oct 17 '16 at 16:41 Blocks A, B and C are identical, so cases 2 & 3 are the same. In case 1 blocks A and B turn about P1 and P2 at the same time, while C is supported equally at P5 and P3. In case 2 (illustrated in the diagram above) block C turns about P4 while also touching A at P5. I suppose case 4 means that no blocks move, so this is the same as saying that neither of cases 1 or 2 applies.

In case 1, if the 2 blocks at the base are not further apart than $1+2h$, where $h$ is the height of each block - ie if $(a1+b1)-(a2+b2) < 2h$ - then block C will not fall through to the ground. Likewise in case 2 blocks A and C may become jammed when the end face of C is flat against the top face of A. These complications add further restrictions if "unstable" means that block C reaches the ground. For simplicity, I assume that "unstable" means that the blocks move from the initial position to some different configuaration.

Your approach is correct but you need to complete it. Eliminate $w$ from your inequalities by substituting for $M$, assign $x$ and $y$ their extreme values as in the 1st paragraph, and rearrange to get inequalities relating only $a1$, $a2$, $b1$ and $b2$.

I think your concern is that there are not enough equations to enable you to find limiting values of each of the 4 variables independently. This is inevitable because there are not enough restrictions in the problem. Each of the 4 variables can be adjusted independently, whereas there are only 2 inequalities for each mode (case) of toppling, arising from the balancing of moments at P1 and P2.

For case 1 the ratios $r_a=\frac{a1}{a2}$ and $r_b=\frac{b1}{b2}$ must each be less than the same constant integer.

For case 2 the ratios $r_a$ and $r_b$ must each be less than a critical value which depends on $b1$ alone, a different critical value for each.

The final question is how to represent the stability conditions. This can be done either on a 3D plot of $(r_a,r_b,b1)$, or on 2D plots of $(r_a,r_b)$ for selected values of $b1$, which can vary from $0$ to $\frac12$, or 2D plots of $(r_a,b1)$ and $(r_b,b1)$. The boundaries are provided by the above inequalities. The stable region lies between the boundaries and the origin.

• I agree with and am aware of most of what you have said (will the 3-d case really be a polyhedron or something more complicated with curved surfaces?). But I think what I meant to ask was how exactly do we go about finding and representing these 4-d inequalities? – alex Oct 16 '16 at 15:44
• also there's another problem of how $x$ and $y$ approach the limiting conditions. for e.g what happens when $w\cdot a2 <M_{ca}(.5,b1)$ and $M_{cb}(.5,.5) <w\cdot b2 < M_{cb}(b1,.5)$ I feel case 1 may occur but how do I formulate a mathematical argument for this and other such situations? – alex Oct 16 '16 at 15:44
• i.e. I feel that $y$ approaches $b1$ instead of $.5$ but have no concrete argument for it. – alex Oct 16 '16 at 15:47
• I think you should 1st eliminate $M$ and $w$ so that you can see your inequalities in terms of only the 4 variables $a1$ to $b2$. Then it will be clearer what the problems are. – sammy gerbil Oct 16 '16 at 16:48
• I haven't posted it online for the sake of brevity (my question is already too long) but as I said while working it out for myself I eliminated $w$ and $M$ and know that it is really only the four quantites$a1,a2,b1$ & $b2$ that are at play. – alex Oct 16 '16 at 16:56

This is very interesting problem, but I think finding the configurations where the "back" contact point lift is not sufficient for instability. For example the configuration below is stable: This is because the middle block on the right is "pinched" by the other blocks and friction can hold it up. Is there a reason you are looking at frictionless only? I think this is too much of a simplification without friction.

You also have to consider that small angle rotations change the problem considerably. Again, in the situation above if the top block is not pushing down on the middle block on the right then it is going to fall. But, if the top block has a small angle and establishes contact you can end up in a stable scenario.

• By stability I mean does not change at all from the configuration shown in the question if I somehow set manage to set it up that way and then leave it all of a sudden. (Just for the sake of argument though, it doesn't need friction for it to be stable in your picture). – alex Oct 18 '16 at 18:56
• Without friction the middle right block will slide left and eventually fall out of place. Hence, unstable! – John Alexiou Oct 19 '16 at 12:42
• if it's c.g. lies beyond the edge then its definitely unstable. But if not then the moment due to its own weight about the pivoting point may cancel the moment due to the force exerted by $C$, the horizontal components of the reaction forces from the box below it and box $C$ may cancel out and their vertical components may counter its weight so that its stable, no? – alex Oct 19 '16 at 13:55
• @alex You are describing the situation on the middle left block. My point is for the middle right block: no-friction = unstable, with-friction = stable. So understanding stability without friction is unrealistic and misleading. – John Alexiou Oct 19 '16 at 14:00
• I don't see why the same logic doesn't apply to the middle right block. I also want to make sure that it's clear that outside this discussion my definition of stable is "doesnt move even slightly the instant after it is left in the configuration shown in the question." during this infinitessimally short time i doubt any slipping will play any role. – alex Oct 19 '16 at 15:12