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This question is a follow up to the following question asked by esdoublelef

Free body diagram on a rack of wooden blocks

To repeat, consider a rack of blocks (each block is a cube), that are held up in a horizontal line without collapsing, by being stuck between 2 vertical immovable walls which apply two equal and opposite horizontal forces at the left and right ends. The whole thing is over a lake of lava.

enter image description here

Let us say, that there are 5 blocks in total and block A is the central block. There are 2 blocks to the left of it and 2 to the right of it.

Now, If we are forced to stand on ANY ONE of the blocks, which block would we choose to stand on, the central block A or the one near the ends ?

For the sake of this question, the assumption is that the blocks will not deform or the rack will not sag, so the only forces of concern are the horizontal reaction forces and the vertical frictional forces.

Also, to clarify, I am asking what would happen IN REAL LIFE, not what would happen only under these assumptions? The assumptions I make about rigidity, non bending, non-deformation have been made only assuming that they would not change drastically what would be the result. If in real life, center block would be weakest, but these assumptions lead to a "drastically opposite" conclusion that center block would be strongest, then clearly I want people to reject this assumption and say "X is what happens in real life, but here is why your assumptions are leading to opposite conclusions". Work with only those assumptions that do not drastically change the conclusion from the real life conclusion.

To address some comments seeking clarification, i wanted to elaborate a bit more on the intention behind this question and what i am looking for. This question was motivated by me playing with some toys with my nephew and noticing this phenomenon where the central block is the weakest. This was not surprising to me, it was what i would have expected intuitively. I thought, it would be obvious by drawing a free-body diagram why that is the case, but i could not do it. I just wanted to understand why the central block was the weakest. Obviously, it might not be a universal thing and there might be cases where the central block is the strongest . In which case , i would be happy to hear about that as well.

But, to sum up i just want an explanation of what is going on here, this is the reason i am not giving concrete numbers about weight, coefficient of friction etc. I am just using variables for everything , and if it turns out, there will be different conclusions for different ranges of these parameters, i would be glad to hear about that as well.



The following is what I have figure out till now

If we were to draw the free body diagrams of the block A and the 2 blocks to the right of it, it would be as shown by Farcher's answer as follows and we can similarly draw the 2 blocks to its left as well

enter image description here

The problem I have with this is, it seems to suggest that the friction force between the blocks at the end is greater than the friction force between the blocks near the center. So, the end blocks are closer to the limit of friction than the central block. So, if I were to ask the question as to which block can support more additional weight on top of it before the structure collapses, this would seem to suggest that the central block A would be able to support a higher additional weight than the end blocks before it all collapses, since the friction force on its surfaces are lower and hence have more leeway before they reach the threshold friction limit

But from intuition, if the whole thing was on top of lava, and we were told we could choose to stand on any one of the blocks, would the central block not be the one block that we would want to avoid? Which is wrong here? Does the free body diagram drawn here have a mistake or is it simply a flawed intuition and will the central block indeed be able to support more weight than the end blocks?

In short, if the whole structure was on top of a lake of lava, would we be actually better off choosing to stand on the central block (even though my intuition says we would be safer standing on the blocks near the ends)?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 19, 2021 at 14:05
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    $\begingroup$ Please do not let posts look like revision histories. Also, please note that the first person pronoun 'I' is capitalized in English and there usually should be no whitespace before punctuation marks. Conforming to standard orthography makes posts much easier to read. $\endgroup$
    – ACuriousMind
    Apr 19, 2021 at 14:06
  • $\begingroup$ Okay, i will try to incorporate edits more smoothly into the question $\endgroup$ Apr 19, 2021 at 14:23
  • $\begingroup$ If you are asking this question again, please would you specify more details: In particular whether the walls can move, whether the normal force $N$ changes if they do move, the number of blocks, (is it 5, 1m cubes), Youngs modulus of each, the weight of each block, the weight of the person - also the co-efficient of friction between blocks and between the blocks and the walls. It's going to need to be really specific to reach a definite conclusion. (Oh, and the value of $N$!) $\endgroup$ Apr 26, 2021 at 7:57
  • $\begingroup$ I was just using variables, to reach a general conclusion. I am not actually building an actual structure where i have definite values, and need to find the conclusion for those definite values. So, just take variables N, youngs modulus, W, etc, as required . And if the math says, it will be conclusion X for these ranges of N and W or something along those lines, then that should be fine. I am assuming walls do not move and no. of blocks = 5 . For everything else, i am deliberately not specifying any values $\endgroup$ Apr 26, 2021 at 8:26

6 Answers 6

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

Let us replace this problem

enter image description here

with this one

enter image description here

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 $(W_{board} + W_{you})/2$.

enter image description here

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.

enter image description here


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 $W_{block} + W_{you}$.

enter image description here

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?


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.

enter image description here

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.


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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 20, 2021 at 15:38
  • $\begingroup$ It is not a beam because the blocks cannot transmit tensile forces. Also there is a limit to the shear force due to the friction limitation. $\endgroup$
    – JAlex
    Apr 30, 2021 at 20:07
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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.

There are two types of instabilities that can occur in the system. The first one is the is the insufficiency of the friction force to prevent the bridge from sliding down (which is what people mostly considered here). The second type of instability is the failure of the normal forces to supply the necessary torque. This instability was not considered by other people here and that's what this answer is mostly about.

Concerning the friction force failure, as many people noticed, the highest demand onto the friction force is at the edges, and the closer the person is to the wall, the larger portion of the person's weight that wall is going to carry, and the easier the friction at that wall is going to fail. To prevent this kind of instability it is better to stand in the middle, to let both walls share the same portion of the weight.

As of torque instability, consider a very similar problem to the OP's. Suppose the bridge consists of only two blocks of unequal length $l$ and $L-l$ installed exactly like on the picture: enter image description here

This problem is very similar to the original problem with blocks, but some of the cubes are effectively "glued together". Even if the friction coefficient is arbitrarily high, such system can still break by forming a crack between the blocks, and both blocks "flipping down", like on the picture. The stability of the original system of many cubes can be analyzed by looking for the weakest point, i.e. by looking for the easiest place for the crack to form in the two-block picture.

In the system of two blocks, there is a maximum weight of the guy the bridge can carry. We can find the maximum weight 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): $$ \begin{cases} -W\frac{l}{L}\frac{l}{2}+Nd+F_m l-x \omega=0;\\ -W\frac{L-l}{L}\frac{L-l}{2}+Nd-F_m (L-l)=0; \end{cases} $$ 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 $$ \omega=\frac{l}{x}\left[Nd\left(\frac{1}{l}+\frac{1}{L-l}\right)-\frac{W}{2}\right].\quad (1) $$

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). Also, as one can see, if $x\rightarrow 0$, the maximum weight is going to infinity, which makes sense, since the person does not create any torques around the pivot point at $x=0$.

We can also notice that if we are standing on a crack, i.e. $x=l$, the critical weight is $$ \omega=Nd\left(\frac{1}{l}+\frac{1}{L-l}\right)-\frac{W}{2},\quad (2) $$ 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.

To summarize everything so far: we have a competition of two instabilities: one is the friction instability, for which it is safer to stand in the middle of the bridge. Another instability is the torque instability, for which it is safer to stand near the edge. Determining which of the two is more important in general case is quite involved, because the answer will depend on the ratio of $N/W$, on friction coefficient $\mu$ and on the total number of cubes.

In the case when the bridge is very long, $L\gg d$, it is clear that it is safer to stand near the edge, since the normal force required to keep the bridge stable from the torque instability is $W L/8d$, while the normal force required to prevent the bridge from sliding is only $W/2\mu\ll W L/8d$. In other words, the friction force is less on an issue for long bridges. On the other hand, being exactly at the edge, $x=0$, is never the best spot, since the critical weight due to torque instability is infinite, and friction is the primary instability cause, which means one can increase the critical weight by moving a little bit away from the edge.

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  • $\begingroup$ In the Wmax equation, does W represent the weight of the left block only ? And also i assume you got that equation by balancing the torques around the bottom left corner. If so, then shouldnt there also be a term having Fm i.e. the friction between the 2 blocks ? You said finding the equation for Wmax is a straight forward exercise. But i cannot figure out how you did it . $\endgroup$ Apr 21, 2021 at 5:36
  • $\begingroup$ $W$ represents the total weight of both blocks. You need two equations for zero torque, one for each block, written around bottom corners that push against the wall. They will contain $F_m$ and $\omega$ as unknowns, and my first equation for $\omega_{max}$ is the solution. $\endgroup$
    – Pavlo. B.
    Apr 21, 2021 at 6:20
  • $\begingroup$ Also, why is the horizontal force at bottom left Nbar , if all other horizontal forces are N. Would not that lead to disequilibrium if we consider left and right forces ? And how do you reach this conclusion from your equation " if the chain of cubes is long enough I will prefer to stand at the edge, but if short, then in the middle " ? I am uncertain how your answer changes when there are 5 blocks or 100 blocks. $\endgroup$ Apr 21, 2021 at 6:27
  • $\begingroup$ The equation you have worked out seems to suggest that " If there are 2 blocks, then a man who is standing right at the meeting point of the 2 blocks would be safer if one of the blocks is longer, than if both blocks are identical. " I am not sure how you extrapolate that conclusion to the question i have asked . I am sure, the basic method of using torques etc. will be the same, but how can we say, we will reach same conclusion, when the guy stands at a different spot, rather than at the meeting point ? $\endgroup$ Apr 21, 2021 at 6:34
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    $\begingroup$ @silverrahul "immovable walls" (infinite spring stiffness) will means the bridge can only fail by blocks sliding. There is no room for failure by rotation. And, yes, "spring stiffness" refers to Hook's law (change in force per change in position). I think what you want is zero spring stiffness so that the normal force is fixed and the separation of the walls adjusts to keep the force constant. $\endgroup$
    – Roger Wood
    Apr 26, 2021 at 17:08
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Pavlo, in the accepted answer, pointed out that there are two ways for the bridge to collapse. One is that the maximum friction is exceeded at an interface, the other is that a crack would apppear. Part 1) is about friction. In part 2) the crack appearing case is considered.

Part 1) Collapse due to exceeding maximum friction.

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:

enter image description here

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 $kW$)

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

$F_3 = 2.5W$

Person standing on block A

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

$$F_3 = 2.5W + 0.5kW\tag1$$

Person standing on block B

$$0.5W+1.5W+2.5W+3.5(W+kW)+4.5W - 5F_3=0\tag2$$

$$F_3 = 2.5W + 0.7kW$$

Person standing on block C

$$0.5W+1.5W+2.5W+3.5W+4.5(W+kW) - 5F_3=0\tag3$$

$$F_3 = 2.5W + 0.9kW$$

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

It just remains to check that $F_1$ and $F_2$ are lower than this. When on A the two $F_1$ forces support the person and one block

$$F_1 = 0.5W+0.5kW$$

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

$$F_2 = 1.5W+0.5kW$$

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

Roger Wood points out in an answer to this question, that the answer depends on the assumption of whether the collapse requires one interface to exceed the maximum allowed friction, or two interfaces to exceed it.

The above assumed that if one interface exceeds the maximum friction, then the collapse would occur. In reality this could happen as there will be some compressibility or wearing that would cause the collapse. Roger points out that if the maximum friction must be exceeded at only one interface, then the middle block, A, is safest - and if it's two interfaces, then it doesn't matter where you stand.

So in theory, for incompressible blocks, the conclusion seems to be that A is safest or that it doesn't matter where you stand.

Part 2) Collapse due to cracking.

It's now assumed that the walls can be forced back enough to give room for the bridge to collapse.

As a crack starts and grows, the total horizontal length of the bridge increases and work is done by gravity to force the walls further apart.

enter image description here

At first, as block $A$ drops a very small distance $h$, the total horizontal length of the bridge becomes $1+4\cos\theta + 2 \sin\theta$ and $\sin\theta = \frac{h}{2}$. For small angles $\cos\theta = 1$, so the increase of the length of the bridge is $1+4+h-5 = h$. The walls are forced apart and work is done against the compressing force $N$, there might also be a gain in kinetic energy , $K.E.$.

There is a loss of gravitational potential energy of $(k+1)Wh + 2\times2W\frac{h}{2}$

$$(k+1)Wh+2Wh = Nh + K.E.$$

There can be spare energy for kinetic energy, i.e. a downward movement if

$$N< 3W + kW\tag4$$

(when a person stands on block $A$).

If we compare this with the 'friction collapse' formula (1), that is equivalent to $$\mu N < 2.5W + 0.5kW\tag5$$

Imagine the $N$ being reduced until the bridge collapsed. Would it be a 'friction collapse' or a 'crack collapse'? If $\mu$ was $1$ we can see from (4) and (5) that the crack collapse condition would be met first (as $N$ reduced), but if $\mu$ was $0.5$ it would be the friction collapse that occurred first.

To find out under what conditions we would expect 'friction collapse' or 'crack collapse' we can put (4) and (5) equal and make $k$ the subject.

$k= \frac{2.5 - 3\mu}{\mu-0.5}$

A graph of $k$ against $\mu$ is here,

https://www.desmos.com/calculator/tbriugxbep

to the left of the red line is friction collapse and to the right is crack collapse.

Doing similar equations to 4) and 5) for standing on $B$ and $C$ and finding $k$ in terms of $\alpha$ where $\alpha= \frac{W}{N}$

for A: crack collapse if $k>\alpha - 3$ , friction collapse if $k>2\mu\alpha-5$

for B: crack collapse if $k>\frac{4}{3}\alpha - 3$ , friction collapse if $k>\frac{10}{7}\mu\alpha-\frac{25}{7}$

for C: crack collapse if $k>\frac{5}{2}\alpha - 5$ , friction collapse if $k>\frac{10}{9}\mu\alpha-\frac{25}{9}$

A graph of these 6 lines is here

https://www.desmos.com/calculator/qnhueum2zv

The coefficient of friction is set to $0.4$ and can be changed.

Imagine we increased $k$ for a given $\alpha$. By looking up from the required $\alpha$ on the $x$ axis, the line reached first means that if $k$ were increased to that value, the bridge of that colour would collapse in a way depending on which of the two lines was met. Reds are for standing on block $A$, blues are for block $B$ and greens are for block $C$.

The colour met last (going up from the $x$ axis), shows which position would allow the greatest value of $k$ before collapse.

For low $\mu$ it's best to stand on $A$, when $\mu$ reaches $0.7$ the red and blue are parallel, but $A$ is best. For $\mu= 0.8$, $A$ is best if $\alpha<4$, but $B$ is best if $\alpha>4$. From $0.9 - 1.2$, $B$ is best and for $\mu > 1.2$, $C$ is best.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Apr 25, 2021 at 22:22
  • $\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$
    – Chris
    Apr 25, 2021 at 22:22
  • $\begingroup$ The assumptions made are meant to be realistic: In part 1) the bridge collapses after maximum friction is exceeded at one interface - due to some compressibility that could happen (it's certainly safest to assume that). In part 2) it's the walls that can compress, not the blocks - it has to be one or the other or both for a crack to appear. Since the walls can push, some movement is possible, so it's assumed that they can be moved back slightly if the force on them exceeds $N$ $\endgroup$ Apr 28, 2021 at 8:24
  • $\begingroup$ @silverrahul , not sure if you've seen the update, hope the assumptions seem reasonable and that it deals with the crack possibility. $\endgroup$ Apr 28, 2021 at 15:50
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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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 19, 2021 at 16:11
  • $\begingroup$ "all stones share the same horizontal force and hence the same static friction." - this only means that all stones will collapse if this same maximum friction force is exceeded. But the force required to hold up each stone varies and needs calculating, depending on where the person is standing, and it's position in the row $\endgroup$ Apr 20, 2021 at 21:38
  • $\begingroup$ Yes, i agree that each surface has the same threshold of maximum friction after which it will give way. " But the force required to hold up each stone varies and needs calculating, depending on where the person is standing, and it's position in the row " The point of my question was to ask how to calculate this . $\endgroup$ Apr 22, 2021 at 5:30
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[Edit: I'll leave this answer for interest, but it addresses a different situation where the blocks are supported by a force applied through two vertical walls separated by a fixed distance.]

The answer depends on the form of the friction force vs. applied load at each interface: Friction force vs. applied shear force

The assumptions are that the blocks are perfect cubes and can only move vertically. At each interface between blocks there is a given normal compressive force that is the same for all interfaces and consequently a given maximum friction shear force that can be sustained by any interface. The actual friction shear forces (and the moments present) are different for each interface. The bridge fails only if the shear force exceeds the maximum value on two interfaces in which case all the blocks between the two interfaces (plus anyone standing on them) fall to their doom in the fiery lava!

The answer depends on the form of the friction force vs. applied load at each interface. Two extremes are shown above. In the left picture, the friction force disappears entirely if the applied force exceeds the maximum. In the right picture, the friction force simply stops increasing when the maximum is reached. Real systems act somewhere between these two extremes.

In case A, the blocks fall as soon as the shear force on one interface exceeds the maximum because the entire load suddenly gets transferred to second interface which then immediately fails. So standing on one end of the bridge is a bad idea. The force will be a maximum on that end interface which will fail and transfer the full load to the far-end interface which will also necessarily fail.

In case B, the blocks will fall when the shear forces on two interfaces both exceed the maximum. When the first interface reaches its maximum, the excess load will be transferred to the other interfaces. (The moments across the interfaces will readjust accordingly as the forces shift.) The system will not fail until a second interface also reaches the maximum load. The total load (blocks plus person) cannot exceed twice the maximum friction force. It doesn't matter where the person stands.

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  • $\begingroup$ So, if i got this right. then you are saying, in case A, standing in center is best. In case B, standing wherever does not matter. If these are the 2 extremes, what happens in real systems ? what is between these 2 extremes of "center is best" and "it does not matter" ? $\endgroup$ Apr 25, 2021 at 19:00
  • $\begingroup$ @silverrahul I have no idea. Unless the friction has a very simple form, you would have to calculate it numerically. It may also depend on the sequence in which the load is applied in time and position. e.g. is it best to step straight to the middle of the bridge, or creep in gradually from one end, or get yourself lowered very gradually from a helicopter? - it's certainly a fun question! $\endgroup$
    – Roger Wood
    Apr 25, 2021 at 20:38
  • $\begingroup$ Yeah, a fun question that no one seems to be able to give a straight answer to. Most answers seem to be unable to explain what would happen in real life $\endgroup$ Apr 25, 2021 at 20:46
  • $\begingroup$ @silverrahul Hah, yes, real life is tricky - too many variables. It's usually the case that you have to conjure a simple enough version that you can actually solve it, An added advantage of a simple model is that it can often be parameterized and can give some insights into the behavior over a range of parameters. $\endgroup$
    – Roger Wood
    Apr 25, 2021 at 21:01
  • $\begingroup$ " It's usually the case that you have to conjure a simple enough version that you can actually solve it, " Yes, in most examples, a relatively simple version with few assumptions and idealisations is able to explain the real result. But in this case, no one is able to explain the real life result with any kind of simple version ( or with a complicated version for that matter ) $\endgroup$ Apr 25, 2021 at 21:04
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After centuries of building viaducts and bridges and churches and igloos, we are back to square one.

Real life lava lake. The arch that supports the structure runs from the lowest point at either end to the highest at wherever you are standing. No matter what eventually makes the bridge collapse, either the 'long' end is going to go first due to compression, because it has the smallest angle between the arch and the horizontal, or the 'short' end is gonna give way first, because it is the first one that overcomes the friction due to having the smallest angle between the arch and the friction surface. Not knowing which is going to be first to give way, compression or friction, keeping both angles equally long would theoretically give you the best chance to survive, which leaves you standing in the middle.

My advice? Don't stand on unreliable structures over lava lakes.

This would mean the safest place to stand is on the middle block.

Edit: Apparently I've been shuffling things around, but it doesn't change the main idea. Standing on one of the end blocks is a fifty/fifty gamble, whereas standing in the middle gives the best average. If you win the gamble, your chances of survival are better than in the middle. Surviving in the middle is only better if you gamble wrong. What makes it still better is the chance of going down or surviving, no matter where you stand.

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  • $\begingroup$ So, your claim is, it is better to stand over the central block rather than the end blocks ? $\endgroup$ Apr 30, 2021 at 19:23

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