Calculating force of an object...

[beansbaxter]

Some friends and I were talking during the Lakers game last night, and we are having no luck with this, curious to see if and how this can be done.

Let's say there are two SOLID objects, both made of the same material.... (let's say iron).

The first object (A) is a cuboid that is 3 cm long with sides of 1 cm each.

The second object (B) is a cuboid with sides 3 cm x 3cm and is 1 cm thick.

The latter object (B) has a round hole in the center of it's 3x3 side of 1 cm in diameter.

Calculate the force necessary to insert the first object (A) THROUGH the hole in the second object (B).

I'd like to see the calculations to arrive at the solution as well.

I drew this picture about it.

See I am looking for a formula with variables for substituting

a) materials
b) diameters and thicknesses

So that one could have various cuboids of various materials and simply drop them into the equation.

Hope that makes sense.

 

[silverpig]

You'd need to know something about the material's shear resistance as the projectile piece will have to gouge out a hole in the hole piece...

 

[jagec]

yeah, seriously, it wouldn't fit without forcing it, and it gets REALLY complicated once you get to that point. You might be able to model it with a computer, but forget about solving it analytically.

 

[beansbaxter]

I know you'll have to force it, and I also understand that one or the other or both of the objects would experience some deflection. If you need to substitute a material you have more data about, that's cool; I just picked Iron just because.

 

[wkwong]

the calculations would totally depend on the material. it's too dyanamic to calculate manually. if A and B were made of the same materials, lets say iron. both A and B will start to bend. no matter what its made of, the shape of both objects will change if you are using the smae materials for both.

 

[beansbaxter]

I dont believe both objects would "bend".... lets say object B was 200 cm thick, but keep the rest of the dimensions for both the same. Object A would perhaps deform uniformly and the take on the shape of the round hole? In otherwords it would become a cyclinder, displacing it's corner mass to the sides? Hmmm... I suppose the first thing to do is calculate the volume of the hole in object B and the volume of A.

What would happen if object A was left as it is, but object B was only 2 mm thick? Would object A remain "square" and the hole would become a square?

The remifications of this are interesting, both in physics, and in interpersonal relations!

 

[jagec]

Originally posted by: beansbaxter
I know you'll have to force it, and I also understand that one or the other or both of the objects would experience some deflection. If you need to substitute a material you have more data about, that's cool; I just picked Iron just because.

it's more than just that, the calculations are literally too complicated to solve analytically, even if you DID have all the information. Differential equations can get mighty tricky, mighty fast.

This sort of experiment is best performed rather than modeled.

 

[DrPizza]

Okay, I'll be the first person to actually solve this

The amount of force necessary is all but negligible. (let's call it a net force)

You stand there with your round hole, and wait a while, mmmkay?
I'm going to head to the other side of the galaxy and exert an unimaginably small net force on the square peg, causing it to accelerate ever so slightly in the direction of you. A few thousand light years later, I'll have that sucker moving at 90% the speed of light, using a VERY VERY small force.

Hmmmm, on 2nd thought, that might be too fast. I'd be creating blackholes and all sorts of the other things that quantum physicists dream about. (Note, when the new particle accelerator is completed in (2007?), one of the expectations is that by colliding protons and anti-protons, they're hoping to create little tiny black holes - - this would be the first analytical evidence in support of string theory.)

 

[DrPizza]

Actually, upon thinking about it for a bit... if you're working with more brittle materials, example: a solid piece of chalk, or maybe some ceramic materials, I think you'd only need to know the compressive and sheer strengths of the material.

Then again, I remember one experiment from glass science. We tested the sheer strength of cylindrical glass rods, about 4 centimeters long, with a diameter of perhaps 1/2 a centimeter. (it's been a long while) I recall that there was a HUGE variation in the strengths of the rods. Those who held them in their hands longer had introduced a lot of surface scratches and etching from the pH of their sweat in their hands. So, not only the material is going to come into play, but also the manufacturing process for that material, the microstructure of the material, and surface defects. That ought to make it wayy to difficult right there. I think the best you could do, even experimentally, is come up with a reasonable range of values, rather than an actual value. And, that range may be small or large, depending on the material used.

 

[silverpig]

Summary of thread: Too hard to answer directly. Perform experiment to find out.

 

[AbsolutDealage]

Originally posted by: DrPizza
You stand there with your round hole, and wait a while, mmmkay?
I'm going to head to the other side of the galaxy and exert an unimaginably small net force on the square peg, causing it to accelerate ever so slightly in the direction of you. A few thousand light years later, I'll have that sucker moving at 90% the speed of light, using a VERY VERY small force.

The problem with this is, you need to exert your force on the object continually, otherwise there will not be constant acceleration. Have fun lugging that propellant out to the other side of the galaxy.

 

[DrPizza]

Originally posted by: AbsolutDealage

Originally posted by: DrPizza
You stand there with your round hole, and wait a while, mmmkay?
I'm going to head to the other side of the galaxy and exert an unimaginably small net force on the square peg, causing it to accelerate ever so slightly in the direction of you. A few thousand light years later, I'll have that sucker moving at 90% the speed of light, using a VERY VERY small force.

The problem with this is, you need to exert your force on the object continually, otherwise there will not be constant acceleration. Have fun lugging that propellant out to the other side of the galaxy.

Yeah, you kinda hit the nail on the head there. I was just trying to be a smartass.

Umm, I mean, I wasn't trying to be a smartass... businesses don't like that. I was "thinking outside the box." It's okay when you call it that.

 

[Smilin]

Originally posted by: silverpig
Summary of thread: Too hard to answer directly. Perform experiment to find out.

Sounds like the best answer yet.

You have sooo many variables:
How much does the hole material bend? How much force does that take? When does it break or tear? How much force does that take? How does temperature affect the properties in this material? How much heat is generated while the material bends? When it breaks? When it compresses? How are you applying the force? Evenly over time? Quickly? How long does the impact last? What is the friction coefficient between the two materials? What is the shape of the material look like as it passes through? What is the friction like with this given shape? Why does beans keep pulling this stuff out his @ss?

 

[gsellis]

You guys are slooooowwwww....

Everyone knows that you cannot fit a square peg in a round hole!

And the real answer involves a lathe, and a micrometer and gauge to measure OD and ID. Lathe the diagonal of the peg to 2.94mc +-.01 and it will require minimal force to insert it.

You guys are making this too hard. If you don't want to lathe it, a**2 + b**2 = c**2 ~= 4.24. Add 6 thousands (it does not have to be mechanically tight). So, bore the hole to 4.3cm diameter.

Beans, you sure are bored.

 

[joelT33]

i think that gsellis just got it.. when i was looking at it i see that the square peg has a length of 1.412 cm from corner to corner and so it doesn't fit through a whole with a diameter of 1cm. So as gsellis says your going to need a lathe or a boring machine if your working with iron. Really no calculations that i can see cept 1**2=1**2=1.412**2

 

[DrPizza]

Originally posted by: gsellis
You guys are slooooowwwww....

Everyone knows that you cannot fit a square peg in a round hole!

And the real answer involves a lathe, and a micrometer and gauge to measure OD and ID. Lathe the diagonal of the peg to 2.94mc +-.01 and it will require minimal force to insert it.

You guys are making this too hard. If you don't want to lathe it, a**2 + b**2 = c**2 ~= 4.24. Add 6 thousands (it does not have to be mechanically tight). So, bore the hole to 4.3cm diameter.

Beans, you sure are bored.

There's an original response
Your math is a bit off.. :Q (How do you bore a hole with a diameter of 4.3 cm in an object that's only 3 cm across?!) Maybe you read the question wrong.

But, to make it really simple, yeah, lathe the "peg" to a diameter of 1.0 cm, or .998 cm if you want it to slide easier. Or, bore the hole to 1.42 cm. (diagonal across the peg = sqrt(2) )

 

[jagec]

Originally posted by: gsellis
You guys are slooooowwwww....

Everyone knows that you cannot fit a square peg in a round hole!

And the real answer involves a lathe, and a micrometer and gauge to measure OD and ID. Lathe the diagonal of the peg to 2.94mc +-.01 and it will require minimal force to insert it.

You guys are making this too hard. If you don't want to lathe it, a**2 + b**2 = c**2 ~= 4.24. Add 6 thousands (it does not have to be mechanically tight). So, bore the hole to 4.3cm diameter.

Beans, you sure are bored.

Lathes require energy, you know

I think most of us were going by the assumption that the peg would be FORCED through the "hole", deforming the metal as it went through.

 

[gsellis]

Originally posted by: DrPizza

Originally posted by: gsellis
You guys are slooooowwwww....

Everyone knows that you cannot fit a square peg in a round hole!

And the real answer involves a lathe, and a micrometer and gauge to measure OD and ID. Lathe the diagonal of the peg to 2.94mc +-.01 and it will require minimal force to insert it.

You guys are making this too hard. If you don't want to lathe it, a**2 + b**2 = c**2 ~= 4.24. Add 6 thousands (it does not have to be mechanically tight). So, bore the hole to 4.3cm diameter.

Beans, you sure are bored.

There's an original response
Your math is a bit off.. :Q (How do you bore a hole with a diameter of 4.3 cm in an object that's only 3 cm across?!) Maybe you read the question wrong.

But, to make it really simple, yeah, lathe the "peg" to a diameter of 1.0 cm, or .998 cm if you want it to slide easier. Or, bore the hole to 1.42 cm. (diagonal across the peg = sqrt(2) )

Oops, my bad... I looked back up at the top and saw it was 3cm "LONG" and thought that was the w and h. So, 1.42 would be the correct hole size. Worse still, I knew that 5 to 6 thousands is getting to a loose fit, but then I wrote it as hundredths. I am starting to lose it.

 

[HokieESM]

There's not an "analytical" solution to this problem--as you're looking at yielding (going past the yield strength) of both the plate and the penetrator. My area of research is computational simulations of plastic deformations at high strain rates (things more along the lines of machining, high speed impact, etc). Its an extremely complicated problem--both in theory AND in practice (the set of numerical equations is very very stiff).

Here's an estimate, though--calculate the area of "overlap" (the corners of the square) and multiply this by the yield stress of the material. You will have to supply AT LEAST this much force--just to get the penetrator to begin to plastically deform (actually, you'll have local plastic deformation before this amount, but you'll most likely see this as a lower bound estimate).

 

[azndelite6983]

Do you think I can fit the rectangular cuboid in your butt? How much force would THAT take?

And would you like it?

Let's try and answer some important questions here people.

 

[Shummy007]

Do you think I can fit the rectangular cuboid in your butt? How much force would THAT take?

Depends if the butt has ever been yielded before

 

[bonkers325]

Originally posted by: beansbaxter
Some friends and I were talking during the Lakers game last night, and we are having no luck with this, curious to see if and how this can be done.

Let's say there are two SOLID objects, both made of the same material.... (let's say iron).

The first object (A) is a cuboid that is 3 cm long with sides of 1 cm each.

The second object (B) is a cuboid with sides 3 cm x 3cm and is 1 cm thick.

The latter object (B) has a round hole in the center of it's 3x3 side of 1 cm in diameter.

Calculate the force necessary to insert the first object (A) THROUGH the hole in the second object (B).

I'd like to see the calculations to arrive at the solution as well.

I drew this picture about it.

See I am looking for a formula with variables for substituting

a) materials
b) diameters and thicknesses

So that one could have various cuboids of various materials and simply drop them into the equation.

Hope that makes sense.

If the object is solid, the only way it'll fit through a 1cm hole is if u apply enough force such that the edges of the cuboid will shear off. The object will shear before any type of bending occurs.

You'll have to apply a shear stress equation to find the max. shear.