Author Topic: An engineering question about weight distribution  (Read 12154 times)

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Offline MarkVS

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An engineering question about weight distribution
« 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.


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Offline McGirr

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Re: An engineering question about weight distribution
« Reply #1 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.

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Offline Brij

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Re: An engineering question about weight distribution
« Reply #2 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.

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Offline Spada

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Re: An engineering question about weight distribution
« Reply #3 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.
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Offline MarkVS

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Re: An engineering question about weight distribution
« Reply #4 on: January 21, 2015, 07:13:11 PM »
Thanks everyone.


What do I win  ;D

Mark 

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Offline S.W

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Re: An engineering question about weight distribution
« Reply #5 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.


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Re: An engineering question about weight distribution
« Reply #6 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:

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Re: An engineering question about weight distribution
« Reply #7 on: January 21, 2015, 07:25:59 PM »

And I made it simple  :cheers:

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Re: An engineering question about weight distribution
« Reply #8 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.


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Re: An engineering question about weight distribution
« Reply #9 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:
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Re: An engineering question about weight distribution
« Reply #10 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

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Offline xcvator

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Re: An engineering question about weight distribution
« Reply #11 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  :-*
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Re: An engineering question about weight distribution
« Reply #12 on: January 21, 2015, 09:15:58 PM »
What temperature did you measure the water  :-*
sorry forgot to add @20 deg c
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Re: An engineering question about weight distribution
« Reply #13 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
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Re: An engineering question about weight distribution
« Reply #14 on: January 22, 2015, 04:54:40 AM »
Reading this thread is like trying to have a 'simple' conversation with my Dad.

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Offline oldmate

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Re: An engineering question about weight distribution
« Reply #15 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:
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Re: An engineering question about weight distribution
« Reply #16 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
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Re: An engineering question about weight distribution
« Reply #17 on: January 22, 2015, 06:37:05 AM »
 ;D ;D
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Re: An engineering question about weight distribution
« Reply #18 on: January 22, 2015, 06:40:23 AM »
What temperature did you measure the water  :-*

I didn't realise that temperature came into it.
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Re: An engineering question about weight distribution
« Reply #19 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....
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Re: An engineering question about weight distribution
« Reply #20 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.
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Re: An engineering question about weight distribution
« Reply #21 on: January 22, 2015, 09:41:12 AM »
This is the simple equation you need to do, its easy......

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Re: An engineering question about weight distribution
« Reply #22 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

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« Last Edit: January 22, 2015, 10:43:39 AM by Symon »
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Re: An engineering question about weight distribution
« Reply #23 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

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

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Re: An engineering question about weight distribution
« Reply #24 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

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.
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