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General => General Discussion => Topic started by: MarkVS on January 21, 2015, 06:46:58 PM

Title: An engineering question about weight distribution
Post by: MarkVS on January 21, 2015, 06:46:58 PM
The answer to this question may be simple or complex..I hope someone can confirm it for me.


Scenario:A bar is level with the ground, and runs between 2 end points.

If you place 100kg on top of the bar, is the weight split evenly between the 2 end points regardless of where along the bar you put the weight ?(except where it is put directly on top of an end point).
Or does something else apply depending on where along the bar you place the weight ?

I am trying to understand this in relation to weight placement along the drawbar and its effect on ball weight.


MarkVS
Title: Re: An engineering question about weight distribution
Post by: McGirr on January 21, 2015, 07:00:36 PM

I will attempt this.

If its in the middle the weight should be distributed evenly. If you move the weight either way the weight would be heavier towards that point.

What do I win  ;D


Mark 
Title: Re: An engineering question about weight distribution
Post by: Brij on January 21, 2015, 07:01:18 PM
The closer the weight is to the support the more weight will be carried by the support. Equal distance means equal weight distribution.

Example - gas bottle on drawbar 1/2 way between ball and axle means weight will be split evenly.

If the distance between the ball and axle is 4m, and the centre of a water tank is placed 1m in front of the axle the axle will see 3/4 of the water tank weight and the ball will see 1/4.

Peter H
Title: Re: An engineering question about weight distribution
Post by: Spada on January 21, 2015, 07:09:07 PM

If its in the middle the weight should be distributed evenly. If you move the weight either way the weight would be heavier towards that point.

Mark

I'm no engineer, but that is my understanding.

The weight at each fulcrum is proportional to where the weight is positioned.
IE- weight in the centre = 50/50 distribution, weight quarter way along the length = 25/75 distribution.
Title: Re: An engineering question about weight distribution
Post by: MarkVS on January 21, 2015, 07:13:11 PM
Thanks everyone.


What do I win  ;D

Mark 

Mark, you win our company on another trip at some point.... ;D ;D ;D
Title: Re: An engineering question about weight distribution
Post by: S.W on January 21, 2015, 07:18:25 PM
Hi Mark,
Looks like others have beat me too it while I was typing but yes, it does matter where along the beam the mass is placed. The closer to one end, the greater the mass supported by the end support at that end of the beam.

For example, assume a beam that is 4m long, with a 100kg mass 1m from the end. You have two unknowns, so need two equations to solve.

The first equation comes from the sum of the moments about each end support. Moment is force (mass x gravity) x distance
So, F1 - 3F2 (the minus is because moments are opposite directions)

The second equation is the sum of the forces must be zero
So, F1 + F2 - (100x9.81) = 0
Which gives F1 + F2 = 981 or F2 = 981 - F1

Solving for F1 and F2 we get:
F1 - 3(981-F1)=0
F1-2943+3F1=0
4F1=2943
F1=735.75N
Thus mass 1 is 75kg

Mass 2 is therefore 25kg.

Hope that makes sense and is not too complicated. It would be easier to explain with a diagram.

P.S I am a mechanical engineer so have tried to use laymans terms if possible.

Title: Re: An engineering question about weight distribution
Post by: muzza01 on January 21, 2015, 07:20:10 PM
OK Mark, firstly not an engineer but previous profession was ensuring the weight and balance of aircraft or (centre of gravity) remains intact and taught it to pilots, Loadmasters and Flight Engineers for a few years.

Imagine this, you have a 30 cm ruler. You place the ruler on top of the long end of a pencil on a table at the 15cm mark.  Theoretically, the ruler should have both ends in the air as it should be balanced.

If you put a 10 gram eraser (or weight) at one end of the ruler say 15cm from the balance point, 15 x 10 grams = 150. Say for the other end I have a 20 gram eraser. Where can I put it to ensure the ruler remains balanced?

150 / 20 grams = 7.5. Therefore the 20 gram eraser would need to put 7.5 cm from the balance point to ensure the ruler is perfectly  balanced.

I know this example is not a CT, but if you give me some more detailed info, I am sure I can help.
 :cheers:
Title: Re: An engineering question about weight distribution
Post by: McGirr on January 21, 2015, 07:25:59 PM

And I made it simple  :cheers:

Mark
Title: Re: An engineering question about weight distribution
Post by: ATC on January 21, 2015, 07:39:54 PM
Or remove the engineering calculations and try something different:-

TOWSAFE: Tow Ball Weight Scales, Weights up to 350kg. (http://caravansplus.com.au/catalog/product_info.php?products_id=13779)

(http://caravansplus.com.au/images_tiff/2013/towsafel.jpg)
Title: Re: An engineering question about weight distribution
Post by: oldmate on January 21, 2015, 08:15:44 PM
And I made it simple  :cheers:

Mark

Yeah I'll say.

Typical engineers though, type 17 pages of calculation to tell you that 1l of water weighs a kilo  ;D ;D ;D.  :angel: :angel:
Title: Re: An engineering question about weight distribution
Post by: Bad Scott on January 21, 2015, 08:35:59 PM
Yeah I'll say.

Typical engineers though, type 17 pages of calculation to tell you that 1l of water weighs a kilo  ;D ;D ;D.  :angel: :angel:

SW has nailed it on the head ;D ;D

That's pure distilled water. Tap water will weight slightly more due to the minerals in it. Now that's 18 pages for calcs >:D >:D >:D >:D

Title: Re: An engineering question about weight distribution
Post by: xcvator on January 21, 2015, 08:57:40 PM
SW has nailed it on the head ;D ;D

That's pure distilled water. Tap water will weight slightly more due to the minerals in it. Now that's 18 pages for calcs >:D >:D >:D >:D


What temperature did you measure the water  :-*
Title: Re: An engineering question about weight distribution
Post by: Bad Scott on January 21, 2015, 09:15:58 PM
What temperature did you measure the water  :-*
sorry forgot to add @20 deg c
Title: Re: An engineering question about weight distribution
Post by: nab on January 21, 2015, 09:24:59 PM
Its like when you help a mate carry a desk and he quickly grabs the end without the drawers. You end up sweating like a pig and he is using 1 hand haha
Title: Re: An engineering question about weight distribution
Post by: Footy Shorts Shane on January 22, 2015, 04:54:40 AM
Reading this thread is like trying to have a 'simple' conversation with my Dad.

Shane
Title: Re: An engineering question about weight distribution
Post by: oldmate on January 22, 2015, 06:27:15 AM
sorry forgot to add @20 deg c

And what was the ambient temp? And the humidity?  :angel:
Title: Re: An engineering question about weight distribution
Post by: Bad Scott on January 22, 2015, 06:36:10 AM
And what was the ambient temp? And the humidity?  :angel:
I only went to a tech school. >:D
Title: Re: An engineering question about weight distribution
Post by: oldmate on January 22, 2015, 06:37:05 AM
 ;D ;D
Title: Re: An engineering question about weight distribution
Post by: Symon on January 22, 2015, 06:40:23 AM
What temperature did you measure the water  :-*

I didn't realise that temperature came into it.
Title: Re: An engineering question about weight distribution
Post by: Bird on January 22, 2015, 06:40:59 AM
Hi Mark,
Looks like others have beat me too it while I was typing but yes, it does matter where along the beam the mass is placed. The closer to one end, the greater the mass supported by the end support at that end of the beam.

For example, assume a beam that is 4m long, with a 100kg mass 1m from the end. You have two unknowns, so need two equations to solve.

The first equation comes from the sum of the moments about each end support. Moment is force (mass x gravity) x distance
So, F1 - 3F2 (the minus is because moments are opposite directions)

The second equation is the sum of the forces must be zero
So, F1 + F2 - (100x9.81) = 0
Which gives F1 + F2 = 981 or F2 = 981 - F1

Solving for F1 and F2 we get:
F1 - 3(981-F1)=0
F1-2943+3F1=0
4F1=2943
F1=735.75N
Thus mass 1 is 75kg

Mass 2 is therefore 25kg.

Hope that makes sense and is not too complicated. It would be easier to explain with a diagram.

P.S I am a mechanical engineer so have tried to use laymans terms if possible.
yea, exactly what I was going to say....
Title: Re: An engineering question about weight distribution
Post by: Bad Scott on January 22, 2015, 06:45:26 AM
I didn't realise that temperature came into it.
Density comes into play. All measuring equipment that needs to be calibrated have temp and humidity controlled rooms.
Title: Re: An engineering question about weight distribution
Post by: DannyG on January 22, 2015, 09:41:12 AM
This is the simple equation you need to do, its easy......

(http://ej.iop.org/images/1475-7516/2004/02/004/Full/img49.gif)
Title: Re: An engineering question about weight distribution
Post by: Symon on January 22, 2015, 10:34:01 AM
Density comes into play. All measuring equipment that needs to be calibrated have temp and humidity controlled rooms.

Yes and?  We are talking mass here,  not density.  Since when does the mass,  or weight,  of water change with temperature?

Edit: I'm talking in typical terms here, I'm not going into high energy states like plasma for instance.  ;D

Sent from my SM-N9005 using Tapatalk

Title: Re: An engineering question about weight distribution
Post by: robbo1172 on January 22, 2015, 10:52:20 AM
Interesting reading here, you learn something new every day...

http://en.wikipedia.org/wiki/Properties_of_water (http://en.wikipedia.org/wiki/Properties_of_water)

I turns out water is most dense at 4°C, getting less dense as it gets hotter or colder.

Don't forget there is a difference between mass and weight.  ;D

Robbo
Title: Re: An engineering question about weight distribution
Post by: Symon on January 22, 2015, 10:58:30 AM
Interesting reading here, you learn something new every day...

http://en.wikipedia.org/wiki/Properties_of_water (http://en.wikipedia.org/wiki/Properties_of_water)

I turns out water is most dense at 4°C, getting less dense as it gets hotter or colder.

Don't forget there is a difference between mass and weight.  ;D

Robbo


That is the reason why ice floats!

And its mass does not change, and unless you are taking your readings at different altitudes, the weight won't change either with changing temperature.
Title: Re: An engineering question about weight distribution
Post by: oldmate on January 22, 2015, 12:23:00 PM
I love how this started as a question of weight on a bar, and has turned into the weight and density of water at 40*c at a height of 30000 ft above GL.  ;D ;D :cheers:
Title: Re: An engineering question about weight distribution
Post by: Bad Scott on January 22, 2015, 01:21:47 PM
I love how this started as a question of weight on a bar, and has turned into the weight and density of water at 40*c at a height of 30000 ft above GL.  ;D ;D :cheers:
:cup: :cup: :cup:
Title: Re: An engineering question about weight distribution
Post by: sweetpea on January 22, 2015, 05:02:49 PM
I love how this started as a question of weight on a bar, and has turned into the weight and density of water at 40*c at a height of 30000 ft above GL.  ;D ;D :cheers:


Note that the topic is "An engineering question".

The weight and mass of water (pure or with specific impurities) on the moon should also be considered just in case.

Speaking of which; if I take my camper to the moon do I still need to air down the tyres when I go off-road?

http://www.myswag.org/Smileys/classic/grin.gif (http://www.myswag.org/Smileys/classic/grin.gif)
Title: Re: An engineering question about weight distribution
Post by: crappsy on January 22, 2015, 05:24:21 PM
[quote author=sweetpea link=topic=41355.msg683204#msg683204 date=1421910169
Speaking of which; if I take my camper to the moon do I still need to air down the tyres when I go off-road?

http://www.myswag.org/Smileys/classic/grin.gif (http://www.myswag.org/Smileys/classic/grin.gif)
[/quote]

I would think you should air up your tyers with Co2 or some othere heavy than air gas combination.
Title: Re: An engineering question about weight distribution
Post by: oldmate on January 22, 2015, 05:37:41 PM
Note that the topic is "An engineering question".

The weight and mass of water (pure or with specific impurities) on the moon should also be considered just in case.

Speaking of which; if I take my camper to the moon do I still need to air down the tyres when I go off-road?

http://www.myswag.org/Smileys/classic/grin.gif (http://www.myswag.org/Smileys/classic/grin.gif)


I would worry about letting air out, but you have to watch the wash outs down at moon beach. Hit them too fast and it will flip ya.
Title: Re: An engineering question about weight distribution
Post by: Barry G on January 22, 2015, 06:36:01 PM
Note that the topic is "An engineering question".

The weight and mass of water (pure or with specific impurities) on the moon should also be considered just in case.

Speaking of which; if I take my camper to the moon do I still need to air down the tyres when I go off-road?

http://www.myswag.org/Smileys/classic/grin.gif (http://www.myswag.org/Smileys/classic/grin.gif)

Having got your camper to the moon I wouldn't be worrying about this until you find yourself on a lunar road from which to go "off".   ????
Title: Re: An engineering question about weight distribution
Post by: MarkVS on January 22, 2015, 06:42:29 PM
I thought I asked an easy question.... ;D ;D ;D  Shows how much I know.....


I could ask the meaning of life...but I think at this rate, this thread is eventually going to get there anyway....
Title: Re: An engineering question about weight distribution
Post by: GeoffA on January 22, 2015, 06:43:39 PM
I could ask the meaning of life......

...42...at any temp, pressure, or altitude....
Title: Re: An engineering question about weight distribution
Post by: Barry G on January 22, 2015, 06:48:28 PM
...42...at any temp, pressure, or altitude....
Yeah, stupid bloody mice!
Title: Re: An engineering question about weight distribution
Post by: MarkVS on January 22, 2015, 06:57:55 PM
...42...at any temp, pressure, or altitude....

Geoff,
I'm impressed. 1 response only to get the 42 answer in.  :cup: :cup: :cup:
Title: Re: An engineering question about weight distribution
Post by: jclures on January 22, 2015, 09:38:27 PM
I had this very same question, ??? I asked my son in law, and he has made me a spread sheet to help give me the answers I wanted.
I had to give him some information so he could then work on the moments as he called them. (actual moments created rear and resisting moments with a moment check) I have no idea what this means but it sounds technical. :cup:
So I now have  spread sheet to help me work on my new project I would like to do. :cheers: