# Weight at a distance

1. May 11, 2017

### Rodney D

So I have a question that I figure there is a simple equation for but I don't know what it is and when I have done some tests I see that the equation may not be a simple as I thought. Any help with find the equation to this question is much appreciated.

Situation:
- I have a bar that is solid mounted to a structure and running at 90 degrees of the structure
- The bar length may be anything ie: 1 meter
- I have a weight sitting on that bar at the far end from the mounting point, maybe 50kg

Question:
- What is the calculation to find the resultant weight at the mounting point of the bar?

When I tested this with a 15kg weight on scales at 1 meter out from the mounting point the scales read 45kg so it has multiplied by 3 times but this was a crude test done on scales and in a garage so I can't trust it 100% and it doesn't give me a formula;)

Any help is appreciated!

2. May 11, 2017

### BvU

Hello Rodney,

So where did the additional 30 come from ? Could you add a sketch of the setup ?

3. May 12, 2017

### Rodney D

This is the idea, I was able to hold down the lever on a set of scales with my own weight. I weighed all the components together and the started sliding the 15kg weight down the lever until it got to about 1 meter out from the scales, this had added about another 30kg to the total weight. What's the equation for this?

4. May 12, 2017

### BvU

Scale should show 15 kg (if the mass of the lever can be ignored).
If it shows something else the scale (electronic?) algorithm is misled by the uneven distribution over the sensors.
A mechanical scale might show less than 15 kg, depending on how it's constructed.

 I had fun experimenting with a 5 kg suitcase holding it close to me and at arms length. No significant difference (< 0.3 %) Soehnle type 63749. Also shows no difference if I stand as much as possible to one side on the thing.

5. May 12, 2017

### Rodney D

Hmm, that's interesting because if you forget the scales and just hold a 15kg weight down by your side this is easy but if you try and hold it out at arms length it is difficult, why is that? I'm going to have to try my test again haha

6. May 12, 2017

### BvU

Has to do with the force needed to produce the torque needed to produce the lifting force at arm's-length distance . $\$ See the biceps figure here for an example. Shoulder muscles are even more complicated

7. May 12, 2017

### Mech_Engineer

The diagram you've drawn wouldn't cause any apparent increase in force at the scale, unless the scale isn't designed to control for a moment load at its surface.

The problem you're trying to solve might be better described as a "moment balance" problem. A moment (a.k.a. torque) is proportional to the length of a lever arm as well as the force applied at its end.

See here:
https://en.wikipedia.org/wiki/Moment_(physics)
https://en.wikipedia.org/wiki/Mechanical_equilibrium

8. May 12, 2017

### Dr.D

You need to investigate the mechanism by which the scale is resisting the moment developed by the applied weight and the effect that this moment has on the scale reading.