For the purpose of calculation we will take an idealised scale of the following dimensions and calculate the rotational forces operating on the beam

The beam will have a length of two meters and height of four centimetre's and will have a suspension point in the dead centre for the fulcrum and similar suspension points precisely in line at each end for the pans .

The beam mass will be zero and the mass of the pans will be 1/9.82 Kg.

The beam will be suspended at the fulcrum point so that the pans are 10cm above the working surface.

Let us next push the left hand pan down to the surface, the right hand pan will rise and exert a torque of

(1-(.1^2))^.5 = 0.99498 newton meters tending to rotate he beam clockwise while the left hand pan will exert a similar torque tending to rotate the beam anti clockwise hence the system will be stable and the left hand pan will remain down.

Now let us calculate the effect of gravity, taking the radius of the Earth as 6,366,197.8 meters the left hand pan will be this distance from the centre of the Earth while the right pan will be 6,366,197.6 meters away.

Applying Newton's inverse square law the gravitational attraction on the right pan will be reduced by one part in (6,366,197.8/6,366,197.6)^2 =1.0000000631 hence there will be a net force of 0.99498*0.0000000631=0.0000000628 Newton's tending to hold the left hand pan down.

Now we come to the effect of raising the fulcrum point as it would normally be on any practical set of scales, If the fulcrum point is raised by one cm relative to the line of the pan suspension points when the left hand pan is pushed down the effective length of the left hand side the beam is reduced by one part in a ten thousand while that of the right hand beam is increased by the same amount hence a restoring force tending to move the beam to a horizontal position of 0.0002 Newton meters is generated vastly more than any gravitational effects.