# Silly Question [single large force versus several small forces]

Hi All,

I'm a first year physics student. This seems like a ridiculous Newtonian question, but I'm trying to mathematically prove this.

If an object takes 5N to break, why is it that multiple six 1N events do not summate to cause that object to break?

I'm thinking it's just conservation of energy, since the energy of each event, of the six, does not summate, but rather dissipates?

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Simon Bridge
Homework Helper
Welcome to PF;
I'm going to keep it simple.

Think of the object as being like a spring - it's not a bad analogy as it sounds: many objects are well modelled as a bunch of small masses joined by springs. If they don't seem very springy, they just have very stiff springs.

In order to break the object, you have to apply enough energy to break the spring... i.e. you deform the spring beyond it's elastic limit (if you remember Hook's Law).

But if you don't break it, then the spring quickly returns to normal and ready for the next go.

If it take 5N to deform the spring past it's elastic limit - then 1N won't get anywhere near.
Release the 1N force and the spring just bounces around for a bit and settles down.
Applying another 1N force just repeats this.

There is an in-between conditon though, where a little bit of damage gets done but not enough to break it.
The obvious example is that a tree may spring back fro a hammer blow but an axe will cut a chunk out of it ... repeated axe blows will chop it down.
So it's not all that simple - but you get the idea.

Bob33
Welcome to PF;
I'm going to keep it simple.

Think of the object as being like a spring - it's not a bad analogy as it sounds: many objects are well modelled as a bunch of small masses joined by springs. If they don't seem very springy, they just have very stiff springs.

In order to break the object, you have to apply enough energy to break the spring... i.e. you deform the spring beyond it's elastic limit (if you remember Hook's Law).

But if you don't break it, then the spring quickly returns to normal and ready for the next go.

If it take 5N to deform the spring past it's elastic limit - then 1N won't get anywhere near.
Release the 1N force and the spring just bounces around for a bit and settles down.
Applying another 1N force just repeats this.

There is an in-between conditon though, where a little bit of damage gets done but not enough to break it.
The obvious example is that a tree may spring back fro a hammer blow but an axe will cut a chunk out of it ... repeated axe blows will chop it down.
So it's not all that simple - but you get the idea.

Thank you.

Does the "in-between conditon...where a little bit of damage gets done but not enough to break it" occur with the 1N adding up in a summation though? If repeated 1 N events caused the spring to bounce back and forth, would that cause deformation in the spring elasticity? Or, were you referring to a totally different scenario, which is why you mentioned the axe example (i.e. The spring is twisted with 1N, like the axe, rather than stretched with 1N, like the hammer?)

Nugatory
Mentor
Does the "in-between conditon...where a little bit of damage gets done but not enough to break it" occur with the 1N adding up in a summation though? If repeated 1 N events caused the spring to bounce back and forth, would that cause deformation in the spring elasticity?
An ideal spring would not fail, but real objects will under some circumstances. Google for "metal fatigue".

Bob33
sophiecentaur
Gold Member
An ideal spring would not fail, but real objects will under some circumstances. Google for "metal fatigue".
And Work Hardening?

Bob33
An ideal spring would not fail, but real objects will under some circumstances. Google for "metal fatigue".
Thank you. It is a lot more complicated than just saying it summates.

Thanks again for pointing me in the right direction.

And Work Hardening?
Thanks for pointing me in the right direction.

Doug Huffman
Gold Member
LOL N. N. Taleb makes much of this observation in his 'Antifragility'

And Work Hardening?
Doesn't work hardening only occur if you exceed the elastic limit?

sophiecentaur
Gold Member
Doesn't work hardening only occur if you exceed the elastic limit?
You could be right there. But I think the 'elastic limit' may be hard to define with some materials.

Hi All,

I'm a first year physics student. This seems like a ridiculous Newtonian question, but I'm trying to mathematically prove this.

If an object takes 5N to break, why is it that multiple six 1N events do not summate to cause that object to break?

I'm thinking it's just conservation of energy, since the energy of each event, of the six, does not summate, but rather dissipates?
keep all the things aside and just compare the fact to this:
if a slap you hard, you just feel the pain but you are intact...
i slap you...many more times with the same force but you are just intact...
and then after a while, all of a sudden, i punch u with my maximum ability... and the effect is that your face changes the shape and no body is thereafter able to recognize you...
like wise small force arent able to break the object...
:)

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