Its really more Physics :) - Moments and Equalibrium!!

Ok, i have a big exam in january, and I am picking through stuf i need to brush up on :) I found moments :D and I was wondering if I could get a bit of help :)

Right, I pretty much know how to choose the points of force upon a thing, say a plank with a person and a mass and a pivot underneither has the pressure of the pivot, froce of gravity, the person and the mass right ? what I dont know EXACTLY how to do is check to see if they are in equalibrium :)

Could someone possibly explain it to me ?

thanks :)

You basically pick a point and calculate the clockwise and anticlockwise moments around that point.

Bit tricky to explain without a pic so I'll try to draw a system in 'code' below here....

Code:

---

            F1                                                          F2
            V                                                            V
---------------------------------------------------------------------
                              ^
                              F3
            |    d1          |                    d2                  |

---

I hope that comes out with the lineup correct.

Anyway the horizontal line is a beam, the 2 'V's are the points at which forces F1 and F2 act, and the '^' is the pivot, or fulcrum. There is an upward froce F3 at the fulcrum. The distances d1 and d2 are from F1 and F2 to the fulcrum, respectively.

Taking moments at the fulcrum we have...

clockwise: F2 x d2
anticlock: F1 x d1

and we have equilibrium if those 2 products are equal.

Lets put some values in. My world is metric, hope that's ok with you!

Let's say F1 is a 100kg man, then F1= 100x10 = 1000N
(Taking gravity as 10 m/s2... close enough)

Let's say F2 is a child, say 40kg, F2=400N.

Let's say d1 is 2m.

Let's say the question is to solve for d2 for equilibrium.

So,

F1d1 = F2d2,

or

d2 = F1d1/F2

d2 = 1000x2/400 = 5m

That makes intuitive sense... since the man is 2 1/2 times the mass of the kid, the kid needs to be 2 1/2 times away to "lever" the adult.

You can calc the moments anywhere you like, but the fulcrum makes sense here.

But for the heck of it, let's take moments around F1's point instead.

We need to introduce the force at the fulcrum first... let's call it F3 upwards.

For vertical equilibrium, F3= F1 + F2

Moments about F1:

clockwise: F2 x (d1 + d2)
anticlock: F3 x d1

Let's solve for d2 again but using F1 as our point of reference....

F2 x (d1+d2) = F3 x d1
F2d1 + F2d2= F3d1
or
d2 = (F3d1 - F2d1)/ F2
= (F3-F2)d1/F2
= (F1+F2-F2)d1/F2..... (F3=F1 + F2 remember)
= F1d1/F2
which is exactly what we had before.

Lastly, just in case you didn't know, or realise, the force acting at the point where you take the moments doesn't come into play, since it's arm is 0. Thus in the first method at the fulcrum, F3 is 'missing', and so is F1 when we take moments at its point of action.