Does leverage apply when working with mass times distance equals mass times distance equations?

Question

My question deals with a balence beam.
Lets say I have a huge cement slab that hangs down verticaly 8ft exactly from the pivot point. The balence arm is 25ft long and has a weight which can slide from tip to pivot point. This cement slab weighs lets say 5,000lbs and the sliding weight is unknown, "X".

Does the fact that the cement slab hangs well below the pivot point acting as a type on lever the moveable weight now has to work against effect the equation?
MxD=MxD is now MxD=MxDxL
M=mass D=distance L=leverage

I have uploaded a picture of what I mean to my forum...
It can be viewed here...
http://www.secrets2flight.dfuzed.org/forum/viewtopic.php?p=9#9

What kind of equation would I use if my guess was wronge???
I am guessing leverage has something to do with it, but I have not gotten the chance to take highschool physics yet.

Answer 1

EDITED 9/27: THIS ANSWER HAS BEEN CHANGED

You must consider all the masses as concentrated as a point at the center of gravity of each component as defined by the pivot point. The weight vector always points straight down and the torque exerted by that weight is W times the perpendicular distance from the weight to the pivot, see http://en.wikipedia.org/wiki/Torque That distance will vary with the angle the slab makes with the horizontal. For balance, the sum of all the torques must be zero.

For a given angular postion A, the weight of the slab produces a torque equal to M*.5*L*cos(A). (M=weight of slab, L= length of slab.) It looks from your drawing like the slab and arm are at right angles, therefore the torque effect of the slider's weight is W*s*sin(A), where W = weight of the slider, and "s" is its position along the arm from the pivot. [If the weight of the arm is significant, you must also include that as concentrated in its center, also multiplied by sin(A).] Ignoring the arm weight (which you did not mention), the torques are balanced when W*s*sin(A) = .5*M*L*cos(A).

Therefore the relation between slider positon s and angle of rotation A is

s = (.5*M*L/W)*cot(A). If you adjust the slider position so that A=45 degrees, cot(A)=1, so the the weight of the slab is then

M = 2*s*W/L

Answer 2

Yes you have to deal with leverage. But it is not so simple. First you have to determine the center of mass for both "arms", then you must determine the component of gravitational force which is responsible for the leverage for both centers of mass. Third you must set them equal (equal in magnitude, but opposite signs). Your renderings are nice, but you will have a better chance of getting a formula if you draw it in 2D and introduce an angle representing the position of the arms, indicate the length of the arms etc...

Answer 3

No, the force on the lever is at the point of attachment, not where the weight hangs.