# Dumb question?
**URL:** https://boards.straightdope.com/t/dumb-question/42062
**Category:** Factual Questions
**Created:** [November 21, 2000, 1:38am UTC](https://boards.straightdope.com/t/dumb-question/42062 "2000-11-21T01:38:05Z")
**Posts on this page:** 6
**Page:** 1
### Author:  [@insider](https://boards.straightdope.com/u/insider)
#### Post date: [November 21, 2000, 1:38am UTC](https://boards.straightdope.com/t/dumb-question/42062/1 "2000-11-21T01:38:05Z")
If you put five pounds of pressure on a thumb tack, what is the pressure in pounds per square inch at the tip of the tack? Thanks in advance for any help.
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### Author:  [@Gunslinger](https://boards.straightdope.com/u/Gunslinger)
#### Post date: [November 21, 2000, 4:23am UTC](https://boards.straightdope.com/t/dumb-question/42062/2 "2000-11-21T04:23:18Z")
Find the area of the top of the tack and the area at the point, then go to WalMart and buy a $2.00 calculator and do the math.
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### Author:  [@Arken](https://boards.straightdope.com/u/Arken)
#### Post date: [November 21, 2000, 1:29pm UTC](https://boards.straightdope.com/t/dumb-question/42062/3 "2000-11-21T13:29:24Z")
Wow. Thanks for the kind and courteous reply to insider’s question.
Now how about telling us what math you’re supposed to do for the rest of us?
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### Author:  [@steeljaw](https://boards.straightdope.com/u/steeljaw)
#### Post date: [November 21, 2000, 1:33pm UTC](https://boards.straightdope.com/t/dumb-question/42062/4 "2000-11-21T13:33:27Z")
all you have to do is divide the pounds of pressure by the area of the point on the tack.
5pounds/area of tack point.
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### Author:  [@tomndebb](https://boards.straightdope.com/u/tomndebb)
#### Post date: [November 21, 2000, 2:08pm UTC](https://boards.straightdope.com/t/dumb-question/42062/5 "2000-11-21T14:08:56Z")
And how many angels can dance under the pressure of five pounds?
While it occurs to me that **insider** ’s OP _might_ have been a smart-ass question, it does raise some interesting problems. What is the surface area at the point of the pin? Is there some minimum thickness that we use arbitrarily? Does anyone have a microscopic photo of a pin point, showing that it actually stops at a bluff or rounded shape that is x ten-thousands (or x hundred-thousands, x millionths, x whatever) of a square inch in surface area?
Given a “point” of .0001 of an inch, (which could be too large or too small by orders of magnitude), we could translate that to 0.000000694 square feet and divide 5 lbs. by that giving around 7,196,315.5 lbs/foot.
However, if the point is substantially larger or smaller, that 7M number changes rather drastically.
Once the pin has entered the wood (or your finger or whatever), we have more surface area to play with as the wedge shape of the pin presses apart the material into which it is being pressed. However, the _initial_ pressure gets very large, depending on how/where we define what the “point” is.
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### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [November 21, 2000, 10:17pm UTC](https://boards.straightdope.com/t/dumb-question/42062/6 "2000-11-21T22:17:26Z")
> [@](#):
>
> \*Originally posted by tomndebb \*
> \*\*
>
> Given a “point” of .0001 of an inch, (which could be too large or too small by orders of magnitude), we could translate that to 0.000000694 square feet and divide 5 lbs. by that giving around 7,196,315.5 lbs/foot.
> \*\*
You estimated the area to be 0.0001 in^2, which is O.K., though I think it may be a little high. (It’s equivalent to a round tip diameter of 0.01 in).
Let’s assume, for the sake of argument, that the tip is a flat circle with a radius of 0.001 in (which sounds reasonable):
diameter (inches) = 0.001
diameter (ft) = diameter (inches) / 12 = 8.33E-5
radius (ft) = diameter (ft) / 2 = 4.17E-5
area (ft^2) = (pi)_(r^2) = (3.1416)_((4.17E-5)^2) = 5.45E-9
If we apply 5 lbs, the pressure would be:
pressure (lbs./ft^2) = 5 / 5.45E-9 = 9.167E8 = 916,700,000
Or roughly **1 billion pounds per square foot**.
Now let’s say the tip is a flat circle with a radius of 0.0001 in (this value might be a good model for a very sharp tip):
diameter (inches) = 0.0001
diameter (ft) = diameter (inches) / 12 = 8.33E-6
radius (ft) = diameter (ft) / 2 = 4.17E-6
area (ft^2) = (pi)_(r^2) = (3.1416)_((4.17E-6)^2) = 5.45E-11
If we apply 5 lbs, the pressure would be:
pressure (lbs./ft^2) = 5 / 5.45E-11 = 9.167E10 = 91,670,000,000
Or roughly **100 billion pounds per square foot**.
We shouldn’t be very impressed with the big pressures; it’s just a numbers game. I mean, just think of what the pressure would be if we changed the units to “lbs./mile^2” or “micro-Newtons / ft^2”.