# How much force does it take for a egg to break?

1. Nov 11, 2012

### WMalik81

1. The problem statement, all variables and given/known data
the egg of mass 59g breaks from a height of approximately 0.35m and breaks
How much force does it take for the egg to break

We know the egg broke at exactly 0.35m and any height less then that it was not as effectively broken?

2. Relevant equations
eg = mgh
fg=mg

3. The attempt at a solution
eg = (0.059)(9.8)(0.35)

fg=(0.059)(9.8)

What do i do?

2. Nov 11, 2012

### Simon Bridge

Welcome to PF;
The question cannot be answered exactly from the information supplied.
By Newton: F=ma ... you know m but you don't know a... for which you need speed-time data during the impact. In an impact the force is not a constant - so the acceleration is not a constant.

You'll have to come up with some sort of approximation.

A common way to analyze impacts is via "impulse" (you can look it up) which is the change in momentum... $impulse = \int F.dt$ ... the graph of F(t) is usually an inverted quadratic (if the object does not break that is) ... for the minimum drop to break, $F_{peak}$ for the quadratic will be the force needed to break the egg. Higher than that and the egg breaks before the $F_{peak}$ is reached. Less than that and the egg bounces.

But if none of this sounds familiar to you, then you need to look through your class notes and see what sort of thing you need to try.

3. Nov 11, 2012

### PAllen

I see Simon made the main point. Somehow you must have information about the 'stopping time' of the egg. Then, the force will be proportional to the momentum divided by the stopping time. Pretend both the eggshell and the surface it hits are perfectly rigid, and the force will be infinite from any height, and the egg will break from any height.

4. Nov 11, 2012

### WMalik81

Thank you both for your replies,

I am a bit confused in one point, seeing as I am not throwing the egg with any initial speed the only force acting on the egg is gravity?

could i not say that mg=f(d)
this corresponds to the total work done?
and then i could solve for my force?

5. Nov 11, 2012

### Simon Bridge

Well, if d=distance fallen and f=force of impact, that would give you f=mg/d ... which means that the impact force decreases with height. Does this make sense?
If you mean f(d) is some function of distance having something to do with total work... then it is too vague: be specific.

Try working the ideas through to some conclusion before asking about them. For eg.
The total work done by gravity pulling the egg a distance h without resistance is $W=F_gd_{fall}=mgh$
The force of falling is mg (gravity) we know that: but what is the force of impact?

Usually an object hitting the ground does not entirely stop falling - a rigid object, for eg, will make a small hole in the ground. In which case we can take an approximation by making $d$=depth of the hole then $F_{impact}=mgh/d$ is just conservation of energy. The deeper the hole the slower the deceleration on impact, the lower the force of impact.
Knowing the depth of the hole is basically the same as knowing the time-frame for the impact.
The approximations here are that all the impact energy went into making the hole and the force of the impact was constant over the time of the impact.

Your egg generally does not make a hole in hard surfaces.
What it does is flex (deform) like a ball. If it flexes too much it breaks. If it does not break, then the unflex creates the bounce.
So you still need to know the time-frame of the impact or some information about the flexibility of an egg.
That is sort-of doable incidentally... you could model the egg as a spring and measure the spring constant the usual way.

However, if your homework problem is about designing a safe egg-drop then your best bet is to over-engineer the thing.
The problem with asking us is that we are aware of many more issues that you are likely to be expected to take into account.

Last edited: Nov 11, 2012