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Why do objects with big size break easily? For example: if I drop a chalk of length $L$ from height $h$ then there is a greater probability that it might break, when compared it to a chalk of length $\frac{L}{2}$ dropped from that same height $h$. And if I repeat the same experiment with same chalk after it gets broken many times, I have also observed after a certain length it doesn't break at all. It's just a physical phenomenon I am curious about which I have observed in daily life.

I have observed a similar phenomenon with glass too. For example, a glass cup gets shattered into many pieces but when some of its shattered pieces fall from that same height they don't break at all. Also in this whole experiment I am considering all objects free falling.

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  • $\begingroup$ The greater fragility of large objects versus smaller ones was the subject of the first of Galileo's "Two New Sciences". en.wikipedia.org/wiki/Two_New_Sciences $\endgroup$ – John Doty Jun 12 at 17:22
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    $\begingroup$ In short, due to $E_p = mgh$, heavier objects has more gravitational potential energy which gets converted into greater kinetic energy upon impact with ground. $\endgroup$ – Agnius Vasiliauskas Jun 12 at 21:35
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    $\begingroup$ All the answers here seem to overlook the fact that the 'weight to the surface area' ratio is mainly responsible for this phenomenon. This ratio is responsible for determining the pressure the object experiences on hitting the ground. $\endgroup$ – Display name Jun 13 at 1:54
  • $\begingroup$ I added a discussion on what soft spots are and why objects have them. $\endgroup$ – Botond Aug 17 at 14:34
  • $\begingroup$ Air resistance is irrelevant in explaining why larger objects break easier since this is a phenomenon independent of the type of the load (falling/kicking/hitting/pushing/bending) $\endgroup$ – Botond Aug 17 at 14:41
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The bigger (and longer) the object, more will be the torque experienced by it. Let's say the length of the chalk we have is $\frac{L}{2}$ (Chalk 1) and $L$ (Chalk 2).

When the chalk falls on the floor, it's most likely to hit on one of its edges. Given that it is dropped from the same height, the force on the heavier mass (Chalk 2) will be more than the one faced by Chalk 1 at one of its edges and on top of that, if we bring torque into the picture, Chalk 2 faces more torque than Chalk 1 on an average because torque is directly proportional to the product of its length from its axis of rotation and force.

Also, the damage from a collision is approximately proportional to momentum aka inertia which is proportional to mass and velocity and proportional to its kinetic energy, which is proportional to its mass and the square of its velocity.

EDIT:

  1. About objects falling flat on the ground, the potential energy of the object is used up in breaking the intermolecular bonds in the solid. As larger objects have greater mass, their potential energy tends to be greater so they tend to break the bonds holding the solid together.

  2. If we include air resistance, then it's intuitive that objects with more mass fall harder than a light object. Given that larger objects, in general, are heavier than small objects, we could say the momentum imparted to the larger object is much greater than momentum imparted to the smaller one. So, that could probably explain as to why larger objects break more frequently than smaller objects.

  3. Here's what I think could be the plausible answer (do share your views on it): Smaller objects, in general, have more surface area than volume (magnitude wise). So, the bonds holding the atoms on the surface is well spread which sort of protects the insides pretty well compared to larger objects. When the object falls, due to the larger surface area, the energy transferred to the object is more spread (due to greater surface area to volume ratio). A certain amount of energy is spread out on a larger surface area, therefore the energy density isn't enough to break the intermolecular forces. The object, as a whole, would be relatively safe as to damage the object, we first have to break the surface and given that the surface protects the insides petty well, things are fine for smaller objects. So, I suspect this to be the reason as to why, on average, larger objects tend to break easily.

In the end, it's all about how much mass the object has (which depends on its shape and mass density), the ratio of surface area to volume and how long/big the object is. These all contribute to the severity of the damage faced by the objects.

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    $\begingroup$ Air resistance is also important. Chalk dust takes ages to fall. $\endgroup$ – PM 2Ring Jun 12 at 6:57
  • $\begingroup$ I knew the answer to this question, but could never have explained it that well. Thanks for sharing. $\endgroup$ – TCooper Jun 12 at 16:41
  • $\begingroup$ @PM2Ring I thought to include that too but the question clearly stated that all objects are free falling. $\endgroup$ – Pratham Hullamballi Jun 12 at 17:07
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    $\begingroup$ I think this is not a valid explanation. It is true that both momentum and torque are larger for a larger object, but they are distributed over a larger volume (or mass). The density of momentum and torque is the same. What is important to reach the breaking point is the density of these quantities, not the total values. $\endgroup$ – sintetico Aug 16 at 17:01
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Big objects break because they are heavier than small objects, so they hit the ground harder.

You might think that a big object is also stronger than a small object. That's true, but it's not enough to compensate for the heaviness.

To see why, imagine two objects of the same shape, one twice as long as the other. Since the big object has twice the height, twice the width, and twice the depth, it will weigh eight times as much as the small object. But it's strength is roughly in proportion to how thick it is - twice as wide and twice as deep. It is only four times as strong.

J.S. Haldane memorably wrote about animals falling down mine shafts: "A rat is killed, a man is broken, a horse splashes."

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    $\begingroup$ Not an experiment that would get past the IRB today $\endgroup$ – pericynthion Jun 13 at 0:46
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    $\begingroup$ I take it that's meant as an increasing progression of severity—being broken is worse than merely being killed, and splashing is worse than merely being broken. $\endgroup$ – Tanner Swett Jun 13 at 5:58
  • $\begingroup$ Addendum: a flightless insect depending on size could be unharmed, while an even smaller insect is too good a flier to fall that far. $\endgroup$ – J.G. Jun 13 at 11:26
  • $\begingroup$ @TannerSwett Yes and the progression starts with the mouse, which Haldane says "gets a slight shock and walks away". $\endgroup$ – bdsl Aug 16 at 17:38
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One of the reasons stems from extreme value statistics. Objects break at their least resistant (call it softest) spot. The probability of having a softer spot is larger in a larger object.

You could think of a chain with $N$ links. Each link has a maximum force it can bear, $F$. Since links are not all the same, $F$ comes from a probability distribution, $P(F)$. Then the resistance to tear of the entire chain is the minimum $F$ out of $N$ values. So now you have $F_1, F_2, ..., F_N$ values but the overall force the chain can bear is the minimum out of those. The larger the number of the links $N$, the larger the probability you'll find a weaker link. The weakest link hypothesis and the resulting extremal statistics is widely used in mechanical engineering to estimate the yield strength of various materials and structures.

If you code a little, you can play around yourself: throw $N$ random numbers according to any distribution and take the minimum of these. You can average over several independent runs, and get the average minimum value out of $N$ random numbers. Then see how this average minimum value changes with $N$. Below is a small Python code that just does that:

import numpy as np
import pylab as pl

min_N = []
for N in range(10,1000):
    min_current = 0
    for realizations in range(100):
        min_current+=np.min(np.random.rand(N))/100.0
    min_N.append(min_current)

pl.loglog(range(10,1000), min_N)
pl.xlabel('N', fontsize=22)
pl.ylabel('min(N)', fontsize=22)    

and the result:

enter image description here

So now you can see that the minimum of $N$ uniformly distributed random numbers (i.e. the strength of the chain) decreases with $N$. This is a log-log plot so looks like it decreases as a power law.

Edit: why do objects have soft spots? There are multiple reasons:

  • Objects are typically inhomogeneous at scales larger than a few tens of atoms/molecules. Crystalline objects have defects such as dislocations or disclinations which cause non-homogeneous stress fields in the materal; wherever the stress is the largest, the object is softer and tends to break there. Amorphous materials are heterogeneous by definition.
  • Even if materials were completely homogeneous, the external load on them is heterogeneous: a hit from the floor is not an evenly distributed load on the boundaries so within the material stresses will be inhomogeneous.
  • Finally, even if the load was evenly distributed on the boundary, objects' boundary shape is irregular which again causes non-uniform stress fields in the material.

To summarize, the fracture nucleation is an interplay of two effects: soft spots in the material and non-uniform stresses throughout the material. This is the reason why materials break at different spots depending on the external load they experience: a spot might be soft (susceptible to break, for instance, due to atomic ordering defects), but it ultimately depends on the load (and the associated non-uniform stress field) whether it will break at that spot or somewhere else.

In a simple minded model, you could think of the material as lattice sites, each of them having a $\sigma_Y(\vec{r})$ yield stress they can bear (note that this yield stress depends on the position and is related to the local atomic structure). Then the external load (coming from a hit from the floor or other strain) causes a stress $\sigma(\vec{r})$ (again, non-uniform due to the reasons stated above) in the material. The material will break at the spot where $\sigma_Y - \sigma$ is the smallest (out of all spots).

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    $\begingroup$ Thank you for your answer. Just one more thing whenever i google search about soft spots , i get nothing . If you have a reference for your answer that would be great. $\endgroup$ – Binod Aug 18 at 19:11
  • $\begingroup$ That is probably not the best terminology, I just called it that for simplicity. In crystals you are looking for lattice defects (dislocations, for instance), whereas in amorphous materials such as glasses, they are called shear transformation zones (STZs). Strictly speaking, both dislocations and STZs are means of plasticity rather than cracking, but cracking usually happens where plasticity is prominent. $\endgroup$ – Botond Aug 18 at 19:42
  • $\begingroup$ P. S. Generally speaking any spot can be a soft spot if the load is such, but due to structural ingomogeneities some spots may bear larger stress then others. $\endgroup$ – Botond Aug 18 at 22:20
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APPROACH 1:- Development of cracks and breakage points pave the way to breaking. An important process in breaking is the propagation of those cracks and breakage sites through the lattice (or the propagation of dislocations through the lattice) Now, larger objects have more probability of development more irregular arrangements or more defects in a direction as compared to small objects as they would have a longer or more expanded arrangement. So, it creates more probability for defects. As large objects have more defects, they will be more suitable for the propagation of cracks or dislocations. Hence, they are more likely to break. Similarly, you can ask why some materials break more than other. Here is an additional info. for you:- enter image description here

APPROACH 2:- Now let's talk about it on the basis of breaking stress. Imagine the stress vs strain graph. Suppose we are comparing two objects (of same material but different dimensions). The $Y$(Young's modulus) will be the same for both. And the curve after the yield point upto the breaking point will almost be similar for both. So if we could compare the yield points for both objects, we can somewhat get an idea of the breaking stress.Now, $${Y}=\frac{stress}{strain}$$ $${Y}=\frac{F/A}{∆L/L}$$ Its obvious that the larger will exert more force on the ground and the ground will also exert more force on larger object than smaller object. But there is also a factor {A}(area of the object). Larger objects also have more area(which also accounts for more air resistance). So the overall $\frac{F}{A}$ term(stress) is almost same for both. Now talking of strain; $${strain}=\frac{∆L}{L}$$ This ∆L is a very small change for normal rigid objects. Now when we divide this small term by $L$, we get a much smaller term. Now for large objects (longer) L is large as compared to the small objects. So, the net strain produced in larger objects is less than that produced in the small objects. But $Y$ is the same for both the objects (since they both have same material). So to keep $Y$ constant, the maximum stress(within which it remains in elastic limit) should be smaller(for the large object) since it's strain is small. So, in the graph of larger objects, the yield point will be obtained at a lower stress as compared to that of small objects. It shows that the break point of larger objects is at a lower stress condition. Thus, larger objects will break easily as compared to the smaller objects. $${stress} \propto {strain}$$ So less the maximum possible strain(till elastic limit) less is the maximum tolerable stress.

It can also be understood from the formula:- $${\mu} =\frac {{\sigma}^2}{2Y}$$ Where, $\mu$ is the modulus of resilience, ${\sigma}$ is the yield strain. This yield strain can be compared for the two objects using the similar approach I have done above.

Other concepts like torque have already been posted. So, I am not going more into it.

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  • $\begingroup$ Insertion of images with text is not encouraged by the website. Type in the information you want to include or make link. This way answers and concepts within remain searchable. $\endgroup$ – ohneVal Aug 20 at 13:23
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Different effects scale differently.

The energy of a chalk of length $l$ dropped at $h$ is $mgh$, which scales with its mass, which scales with volume, so $E∝l^3$.

The strength of an object though will typically scale with its thickness ($∝l$) or its cross-sectional area ($∝l^2$), depending on how it breaks.

So a larger piece of chalk will be stronger, but will have to withstand even more energy dissipation.

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The main reason why a chalk of length $L/2$ is more robust than a chalk of length $L$ is that when you break a chalk into two halves, the diameter of the chalk remains constant. The ratio between length $L$ and diameter $d$ changes. The fragility of the chalk increases if the ratio $L/d$ increases.

If you consider 1) a chalk of length $L$, diameter $d$, falling from a height $h$, and 2) another chalk of length $L/2$, diameter $d/2$, falling from a height $h/2$, the outcome will be the same, for symmetry reasons (scale invariance).

Same reasoning applies to glass. In this case, the relevant ratio is length (or width) over thickness.

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  • $\begingroup$ I wanted to upvote this but I realised the argument in second paragraph doesn't work: chalk -- the material -- is not scale invariant. $\endgroup$ – alexarvanitakis Aug 16 at 19:35
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    $\begingroup$ @alexarvanitakis Why chalk is not scale invariant? Of course we are talking about a regime where all relevant lengths are larger then the interatomic distances and structure details of the object. $\endgroup$ – sintetico Aug 17 at 8:00

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