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- Thread starter dBrandon/dC
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russ_watters

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How can you generate a pressure inside greater than the pressure at which you squeeze?

[edit] Err, wait, I see what you're saying and what the problem is: you've done 2d math on a 3d phenomena. If you cut the radius of a sphere in half, you decrease the volume and increase the pressure by a factor of 8, whereas the surface area decreases only by a factor of 4. So the force on the wall (and thus what you must apply to keep compressing it) has increased by a factor of two.

[edit] Err, wait, I see what you're saying and what the problem is: you've done 2d math on a 3d phenomena. If you cut the radius of a sphere in half, you decrease the volume and increase the pressure by a factor of 8, whereas the surface area decreases only by a factor of 4. So the force on the wall (and thus what you must apply to keep compressing it) has increased by a factor of two.

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pressure build up

Pressure is not something you can build up like money in the bank.

It disappears as soon as the applying agent is removed.

Nor does it increase just because you are considering a smaller area.

Pressure = Force/Area

One way to think of pressure is to think of it as a sort of distributed force, ie distributed over an area.

If you cut off (reduce) some of the area) you also remove the force that was distributed over it so the Force term in the equation is also reduced (in the same proportion) and the pressure remains constant.

Having said all that, your idea is worth something.

The rocks deep in the earth are under tremendous pressure in three dimensions. This is called triaxial stress. It has great significance for the properties and physical behaviour of these rocks that geological science is just beginning to appreciate.

go well

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One way to think of pressure is to think of it as a sort of distributed force, ie distributed over an area.

If you cut off (reduce) some of the area) you also remove the force that was distrubuted over it so the Force term in the equation is also reduced (in the same proportion) and the pressure remains constant.

Force can change when you're applying a constant pressure. An example of this is the hydraulic lift. But you can also apply a constant force, which would mean pressure

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Think of a hammer and nail. You apply, let's say, 10 pounds of force to the head of the nail, which has an area of, let's say, 1/10 of a square inch. That works out to a pressure of 100 pounds per square inch. The 10 pounds of force gets transferred to the tip of the nail, which has an area of, let's say, 1/1000 of a square inch. Thus, 10 pounds of force now generates a pressure of 10,000 pounds per square inch.

Mechanics just doesn't work like this.

You hit the nailhead with an impact of 10 lbs

It is this 10 lbs, not the force/area, that is transmitted to the point.

The timber offers a resistance of, say 9.5 lbs, so the resultant that move the nail forward is 0.5 lbs.

If this were not so the nail would not be driven in.

So the actual pressure at the point will be 0.5/point area.

Do not think that the pressure increases, just because the area decreases, you alwys have to consider the forces as well as the area.

If you have a tapered peg and hold the narrow end firmly against a solid plate and hit the wide end the pressure on the narrow end will indeed be greater than the impact pressure. If you hit it hard enough you will break either the plate or the peg.

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You hit the nailhead with an impact of 10 lbsforce.

It is this 10 lbs, not the force/area, that is transmitted to the point.

That's what I said. "The 10 pounds of force gets transferred to the tip of the nail ..."

Do not think that the pressure increases, just because the area decreases, you alwys have to consider the forces as well as the area.

Sometimes it does. Again, consider your feet. If I weigh 75 kg, and gravity accelerates me at 9.8 m/s^2, I apply a force against the floor of 735 N. If each shoe has an area of 0.015 m^2, the pressure on the floor if I stand on two feet is 24,500 Pascals. If I sand on one foot, the force is the same, but since the area is cut in half, the pressure doubles to 49,000 Pascals.

If you have a tapered peg and hold the narrow end firmly against a solid plate and hit the wide end the pressure on the narrow end will indeed be greater than the impact pressure. If you hit it hard enough you will break either the plate or the peg.

So we're in agreement.

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OK I finally understand your query and its resolution.

It is important to realise that the situation with the nail and my peg is not the same as the ever diminishing test circle you originally described.

The stress (pressure) in tapered peg as shown in my second sketch does indeed increase from section A to section B to section C.

This is because the whole of the applied force is transmitted across each section.

In the first sketch I have shown a limiting process taking ever smaller squares (squares are easier to work with than circles).

I have only shown two dimensions but the same principles apply in 3D.

The important point is that as we move from section A to B to C and select the smaller inner square, less force is transmitted across the section - or rather part section - which comprises the inner square.

I have shown this only on the y axis.

F_{y} is the total applied force to the outer square.

To calculate the stress we distribute it across the whole of section A, as shown by the line of small straight arrows.

At section B we select the middle square.

F_{y} is still distributed across the whole of section B, only part is now across our square.

I have shown the missing part as squiggly arrows and the active part as straight arrows as before.

At section C we are missing even more of the distribution of of F_{y}

At all sections the stress is the same.

Hope this helps

go well

It is important to realise that the situation with the nail and my peg is not the same as the ever diminishing test circle you originally described.

The stress (pressure) in tapered peg as shown in my second sketch does indeed increase from section A to section B to section C.

This is because the whole of the applied force is transmitted across each section.

In the first sketch I have shown a limiting process taking ever smaller squares (squares are easier to work with than circles).

I have only shown two dimensions but the same principles apply in 3D.

The important point is that as we move from section A to B to C and select the smaller inner square, less force is transmitted across the section - or rather part section - which comprises the inner square.

I have shown this only on the y axis.

F

To calculate the stress we distribute it across the whole of section A, as shown by the line of small straight arrows.

At section B we select the middle square.

F

I have shown the missing part as squiggly arrows and the active part as straight arrows as before.

At section C we are missing even more of the distribution of of F

At all sections the stress is the same.

Hope this helps

go well

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