Balancing Forces on a Pivot: Q&A

[jendrix]

Hello, I've recently being learning about balancing forces on a pivot.I understand the part about calculating force x distance(from pivot) so that both clockwise and anti-clockwise balance but I'm stuck as to what effect it has when you move the pivot point.

Say for example you have a 10m beam with a central pivot, and a 200n weight 3m to the left of the pivot and a 300n weight 2m to the right of the pivot.This would balance as the forces on either side are 600n.

What effect would moving the pivot 2m metres to the left make if the beam weighed 1000n?

Just to add this isn't a homework question I just thought an example(I hope it's clear) might make it easier to understand, a lot of the articles I've found only cover the effect when the pivot is in the centre.

Thanks

 

[jendrix]

Am I right in thinking that I can disregard the weight of the beam and still work solely on the distance to the pivot on either side?

So the new forces would be:

Left side 1m from pivot x 200n =200n/m

Right side 4m from pivot x 300n = 1200n/m

This would result in a clockwise effect?

 

[sophiecentaur]

The problem with Moments questions (and it's sometimes hard to believe) is that it doesn't matter where you take moments - if you do it carefully, then you get the right answer. The secret is usually to choose an appropriate point to take the moments about and then believe in what you are doing. It will all come out in the wash but it may involve solving simple simultaneous equations. An 'appropriate point' would be one through which one of the unknown forces is acting - thus (effectively) making the simultaneous equations easier to solve.

In case you were worrying about it: The weight of the beam all acts through the centre of mass. Don't worry about what bits are on one side or the other of the pivot just use the total weight and a single distance from the pivot. It will come out ok without that added complication.

 

[jendrix]

Thanks I think I'm slowly getting there.

So when a question gives you the beam length and 2 weights to go either side with the aim of balancing the beam, say a 3m beam and 2 weights like 400n and 600n then you'd use an equation like

400n x Distance 1 = 600n x Distance 2 (Distance being from pivot)

I hope that's clear

 

[sophiecentaur]

That's what the principle of moments says. Go for it.

 

[jendrix]

Ok so now I've been told something different and was hoping one of you could confirm the right answer.


2m 3m
I-------P-----------I




In the above diagram where the beam is 5m long and the pivot in 2m in from the right, I've been told I should factor on the weight of the beam by taking it's position as being in the centre

So if beam was 500N I would multiply by 0.5 (As pivot is 0.5m from centre)

Is this correct?

Thanks again

 

[sophiecentaur]

That's what I said. Weight of beam at its cm is what counts. You COULD work out the weights of both sections and then say that they each act half way along that section - but you get the same answer.

If the pivot happens to be at the cm, there is zero turning effect. (Confirmation?)

 

[jendrix]

Thanks Sophie you've been a big help