Missing or mistaken door slam force formula?

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The discussion revolves around calculating the impact force of a slamming door, with a focus on a simplified approach. The user estimates the door's weight and dimensions to derive a velocity of 1.13 m/s, leading to a calculated force of approximately 10.24 Newtons. Key points include the importance of measuring the final velocity at the point of impact and the distinction between force and kinetic energy, which are not the same. The impact force is influenced by the area of contact, with smaller contact areas resulting in higher forces. Overall, accurately calculating impact force is complex due to variables like object deformation upon impact.
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Correct formula to work out impact force of a slamming door?
Hi Guys,
I am trying to keep this as simple as possible as I only need a ballpark figure. I am sure there are thousands of other variables which need/could be taken into account, but for this purpose, as simple as could be.

A door in my house slams when other doors/windows open.
The door is 72mm(radius for future calculation) wide. I approximate it weighs 16kg(average door weight).
The door is at 90 degrees to it's closing position when it starts to slam.

Using Pi formula, I work out that the diameter would be :
72mm x 2 = 144mm(Diameter)
144 x 3.14 = 452.16mm(Circumference)
452.16 / 4 = 113.04mm(gives me quadrant size)

So distance traveled = 1.13 metre

The time the door took to slam, from start position, was roughly 1 sec.
Therefore I presume velocity= 1.13 m/s

From what I have read...Force/kinetic energy equals "..half the mass x velocity squared... "

So...
Half the mass is... 16kg/2 = 8
Velocity squared of 1.13 = 1.28

So 8 x 1.28 = 10.24
I think the unit is Newtons in this case...

Again, so...in a simplistic formula...would the Impact force of the door closing be 10.24 Newtons?

My second question is probably my main question...

If I put my hand to stop the door(in roughly the same position as where the door frame would stop the door) would the impact force be the same OR is the force dependent on the area(size) of my hand it hits...I'm sure it does.
If I put my finger out(obviously smaller in size) to stop the door I am sure the impact force would be more.
Therefore...what would be the formula to work out impact force on a particular object it hits(hand or finger or anything)..presuming my initial formula is correct at 10.24N ?
I apologize for the simple way I have put this but it's years since I done math and formula.
Please let me know if I have done the formula correct and point out any mistakes(which I am sure there are!).
Thanks in advance.
BTW...I have looked at these related questions but they give indefinite answers or, are far too complex for my understanding...hence why i needed to keep it simple!

https://www.physicsforums.com/threa...mpact-energy-of-a-door-hitting-a-door.821759/

https://www.physicsforums.com/threa...ocity-momentum-of-a-door.840398/#post-5273155
 
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Several things to look at:

1) A 72 mm door is a mouse or squirrel door. A 720 mm (0.72 m) door is a small people door.

2) The velocity of concern is the velocity at the point where the door suddenly stops, not the average velocity. If a breeze is closing the door, it is accelerating the entire time it is moving. The final velocity will be higher than the average velocity. There are other factors (non-constant acceleration for one) that complicate things, so I suggest measuring the final velocity. Make a video recording of the door slamming, then measure the distance traveled between frames for the last two or three frames.

Duffy said:
If I put my hand to stop the door(in roughly the same position as where the door frame would stop the door) would the impact force be the same OR is the force dependent on the area(size) of my hand it hits...I'm sure it does.
If I put my finger out(obviously smaller in size) to stop the door I am sure the impact force would be more.

3) The door slamming force, as you noticed, depends on both the closing velocity and the hardness of the door stop. A soft hand will stop the door more gradually than a hard door stop, so the force will be lower. For now, focus on finding the closing velocity.

4) When doing physics calculations, always include units. Always.

5) Since the door is rotating about its hinges, the velocity of interest is the angular velocity in radians per second. The inertia of interest is the moment of inertia, AKA rotational inertia (search those terms). Don't try to study the Wikipedia article, it's way too complicated for what you want.

That is plenty to keep you busy for now. Work on it for a while, then come back.
 
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Hi jrmichler,
Many thanks for the reply.
In response to your pointers...
1) Yep...no mice or squirrels here at the moment so it is indeed for a human door! I have tried to edit the post to change the unit but can't seem to find an edit button(it was there yesterday?)
2) Good point about acceleration and closing speed and good idea to record final few frames.
If i was to record and just track the last 10cm of the doors journey...do you think this would give me an accurate enough time for the velocity..or do you think the distance would need to be less...ie...5cm etc?
5) I shall look at those two terms today and take in what i can.

Once again, thanks for taking the time to respond.
Have a great day! 👍
 
Duffy said:
From what I have read...Force/kinetic energy equals "..half the mass x velocity squared... "

So...
Half the mass is... 16kg/2 = 8
Velocity squared of 1.13 = 1.28

So 8 x 1.28 = 10.24
I think the unit is Newtons in this case...
No, kinetic energy and force are not the same thing. Kinetic energy has units of Newton-meters.
 
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russ_watters said:
No, kinetic energy and force are not the same thing. Kinetic energy has units of Newton-meters.
Hi Russ,
Thanks for clarifying...sort of...!
As it's been many years since I studied this stuff...i was just quoting what I had read on the web the previous day. I had seen it called a few different things and was slightly confused.
In this particular instance, what is the unit of measurement to use Newton or Newton-meters?
 
Duffy said:
In this particular instance, what is the unit of measurement to use Newton or Newton-meters?
A Newton-meter is Newtons times meters. I think you want force, which is just Newtons.

The larger problem here is impact force is very difficult to calculate because it requires analysis of the deformation of the object at impact. It is essentially impossible to find analytically, and isn't constant. But maybe that's not a big problem: Why do you think you need to calculate force? What do you want to do with that information?
 
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