Free-body diagram of a rack of blocks revisited a.k.a. if it were on top of lava, which block would you choose to stand on?

The blocks are a distraction. You can take an intuitive shortcut to solve the problem.

Let us replace this problem

with this one

Either way, the total weight is $N cdot W_{block} = W_{board}$. The upward force of friction on each end must be $W_{board}/2$.

Now you are added to the board. If you are in the center, the weight on each end is $(W_{board} + W_{you})/2$.

If you are near the left end, the forces of friction must be approximately $F_{left} = W_{board}/2 + W_{you}$ and $F_{right} = W_{board}/2$. $F_{left}$ is bigger than before. So this configuration is more likely to fail. Standing on the center is safest.

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So far so good. But if you want credit for a homework problem about free body diagrams, you will need to use free body diagrams to justify your intuition.

Previous answers have shown that the upward force of friction must be just strong enough to hold up the weight of the blocks. Outer blocks must hold up inner blocks. So the outermost blocks have the highest friction forces.

Here is a diagram that shows the forces on each block when you are in the center. Note that the upward forces on the center block are big enough to counteract $W_{block} + W_{you}$.

I don't want to solve a homework problem completely. Can you see how it would change if you were centered over the left block?

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Edit - Addressing torques and intuition that the center is the weakest.

The intuition that the center is the weakest is perfectly reasonable. If you laid a plank across a gap, the center would be the weakest. But that is a different situation that this.

A plank would break if the force became to big. We can understand this in terms of torques. Consider the plank to be two half planks strongly attached to each other. Each half does not rotate if the total torque on it is $0$. We suppose the ends are supported well enough that they do not move. See Toppling of a cylinder on a block for more about this.

This shows the forces on the left half plank.

The blue force is the weight of the half plank. The black force is the reaction of the support holding half the weight of the plank. If the other half plank was not present, this torque would rotation the left half clockwise.

But it is present. The top of the left half cannot rotate without compressing both itself and top of the the right half. The two halves press against each other strongly. The top red force is the force the right half exerts on the left.

The bottom is under tension. The two halves pull on each other. The bottom red force shows the right have pulling on the left.

The plank breaks if the red forces are stronger than the internal forces that hold molecules together.

This diagram is an idealization. The red forces would really be distributed through the interior of the plank, much like the weight is really sum of the weights of each atom.

A perfectly rigid plank would be entirely under compression. A real plank would sag. The amount of compression and tension would depend on the amount of sag. For example, a plank made of taffy or like a slinky would be entirely under tension and would sag into a deep U shape.

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Suppose you stand on the left half plank. You add a torque that the plank must resist.

If you stand near the support, most of your weight is held by the left half. But the torque is small because the distance from the support is small. You maximize the torque by standing in the middle of the plank.

There should be no preferential block: if the system can hold extra weight, it does not matter where it is placed; and if it cannot hold it, the system will collapse regardless.

This is because given Newton's 3rd law, all stones share the same horizontal force and hence the same static friction. The assumption for that is that all the materials are the same (walls and stones) and rigid.

The only way that frictions of different blocks would differ is if they have different vertical positions, because the touching area with neighbors would be different. In this case, the weakest block is that with the smallest contact area with the neighbors.

I think the intuition fails because it is trying to derive information from U the shape of suspended bridges. However, this is not possible in the blocks' case.

As mentioned in the question, the friction force on the end block, $F_3$, is most likely to be near limiting friction. In reality the beam would bend, but if we presume that it fails only due to the limiting friction force being exceeded, it'll happen like this:

Farcher found that $F_3$ is $frac{5W}{2}$. We can get the same result by doing moments around point P.

Without the person (of weight $kW$)

$0.5W+1.5W+2.5W+3.5W+4.5W - 5F_3=0$

$F_3 = frac{5W}{2}$

Person standing on block A

$$0.5W+1.5W+2.5(kW+W)+3.5W+4.5W - 5F_3=0$$

$$5F_3 = 12.5W+ 2.5kW$$

Person standing on block B

$$0.5W+1.5W+2.5W+3.5(W+kW)+4.5W - 5F_3=0$$

$$5F_3 = 12.5W+ 3.5kW$$

Person standing on block C

$$0.5W+1.5W+2.5W+3.5W+4.5(W+kW) - 5F_3=0$$

$$5F_3 = 12.5W+ 4.5kW$$

$F_3$ is lowest for block A, so it's safest to stand on block A and least safe to stand on block C.

It just remains to check that $F_1$ and $F_2$ are lower than this. If the person is on block A, it causes $F_3$ to be $2.5W+0.5kW$. Whilst on A the two $F_1$ forces support the person and one block $2F_1 = W+kW$

$$F_1 = 0.5W+0.5kW$$

and the two $F_2$ forces support the person and three blocks

$$F_2 = 1.5W+0.5kW$$

both are lower than the lowest possible $F_3$, so A is safest.

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Due to the question edit about horizontal forces: These vary in height along the blocks and provide the necessary torque to stop each block spinning. If the horizontal contact forces (N) between blocks A and B were at the top, the horizontal contact forces (still N) between B and C is W/N further down and those between C and the wall are another 2W/N further down again.

Due to symmetry the conclusion from moments about P, when the person stands on A is unaffected. When the person stands on C, there may be a moment from the height difference of horizontal forces, it's going to get complicated, but initial investigation showed that this effect is small and may even require an increase in F3 for equilibrium to be maintained and so it still seems safest for a person to stand on A.

Wow, I saw this problem being discussed for several days and I was thinking "why is this high-school block problem is so popular?", and never bothered to read it carefully. But when I finally did it, I think the problem is excellent. My conclusion is that if the chain of cubes is long enough I will prefer to stand at the edge, but if short, then in the middle.

All force considerations that other people did here are valid, but one point was missing. Namely, can forces onto the blocks supply the necessary torque to counteract the torque from friction forces?

Consider a very similar problem. Suppose the bridge consists of only two blocks of unequal length $l$ and $L-l$ installed exactly like on the picture:

This problem is very similar to the original problem with blocks, but some of the cubes are effectively "glued together". Such system allows us to analyze the cubes against a formation of a crack at a particular spot. When we go back to the problem with cubes, the crack would form at a weakest point.

Let us assume the friction coefficient is arbitrarily high, so friction is not a problem. The system of the two blocks can still break by forming a crease between the two blocks.

We can find the maximum weight of the guy by by writing down the balance of torques and forces for the two blocks. The following system is the two zero-torque equations around the bottom corners of the blocks that push against the wall (left and right correspondingly): $$ begin{cases} -Wfrac{l}{L}frac{l}{2}+Nd+F_m l-x omega=0; -Wfrac{L-l}{L}frac{L-l}{2}+Nd-F_m (L-l)=0; end{cases} $$ Here $W$ is the total weight of the bridge, $N$ is the force we squeeze both blocks with, $d$ is the height of the bridge and $x$ is the position from the right where the person stands. Solving for $omega$ yields $$ omega=frac{l}{x}left[Ndleft(frac{1}{l}+frac{1}{L-l}right)-frac{W}{2}right].quad (1) $$

For a fixed crack position $l$, the critical weight $omega$ is larger the smaller the $x$ is, i.e. the further from the crack the person stands. In the case where there are many places for the crack to appear (as in the case of cubes), it is safer to stay away from all cracks (which goes along with the common sense).

We can also notice that if we are standing on a crack, i.e. $x=l$, the critical weight is $$ omega=Ndleft(frac{1}{l}+frac{1}{L-l}right)-frac{l W}{2}, $$ which is the smallest for $l=L/2$, and increases towards the edges. In other words, cracks near the edges are safer.

Interesting. We see from Eq. (1), sometimes the right part of the equation may get negative, which means we have to support the bridge in order for it to not fall apart. If the number of cubes is even, the weakest crack is in the middle, and for $W>8N d/L$ the bridge is going to fall apart by itself.

We can also find when friction will be the cause of failure, rather than the torque (assuming there is no human standing on the bridge). Maximum weight of the bridge which is not sliding down is $2Nmu$ ($mu$ is friction coefficient), which, when comparing this to the maximum "bridge splitting weight", yields that if $mu>4 d/L$ the bridge will split and if $mu< 4 d/L$ the bridge will slide down. For a friction coefficient $mu=0.5$ (typical for wood), this yields that For $L<8d$ the bridge is more likely to slide, and for $L>8d$ - to split.