Background: I'm on the pledge drive committee for a non-profit group, and there is an issue that is always contentious within this group; thankfully the group is amicable, although competitive, about the issue. So, we want to develop a computer-based equal-arm, weighing scale (e.g., the kind Lady Justice holds). During our pledge event, individuals will be able to donate to one side of the issue or the other and see the scale change accordingly. None of us on the pledge committee are well-versed in physics, so we aren't familiar with, and haven't been able to find, a formula for coding the angle of the scale in relation to the current donations. As a non-physicist, this seems like an equation that would have been identified, but we just don't know where to find it.

Is there a standard formula for the angle of the beam, in relation to the fulcrum, based on the relative 'weights' on each side? If so, what is it or where can we learn about it? If not, are there formulas that we could use as a basis for our simulation?

  • $\begingroup$ It is important for the fulcrum to be above the center of mass of the system in order to act as a pendulum. In your case realism is not as important, so just make the angle proportional to the difference. $\endgroup$ Nov 25, 2013 at 3:27

3 Answers 3


Mark is right: If your fulcrum lies on the straight line between the hanging points for weight 1 (w1) and weight 2 (w2); then the angle can be anything if w1=w2, and will be straight up and down if w1 does not equal w2 (the heavier weight being at the bottom).

Georg is correct, that the way to change this is to offset the fulcrum (make is higher than the straight line between w1 and w2 (or, equivalently, make it higher than the hanging points for the weighing baskets for w1 and w2).

I tried to get a simple closed form which described the balance angle as a function of w1/w2, but could not in the short time I have to devote to this problem. (Maybe someone can pick up where I left off). What I do have is an equation which describes that relationship, as follows:

Let B be the half-length of a balance beam between w1 (on the left) and w2 (on the right). Let H be the height of the fulcrum point (F) above this straight line. (So, lines drawn between points w1, w2 and F make a short, wide triangle.) Let L = the slant distance from F to w1 (and also from F to W2). L = SQRT(B^2 + H^2). If you draw a line through F and parallel to line w1w2 (the base of the triangle) then lines w1F and w2F both make a small angle 'A' (radians) with it. Use the convention that if w1 is heavier, it will cause a positive rotation of the balance beam by an angle of 'D' (radians). By writing and solving a moments eqn around point F, you get the eqn:

w1/w2 = (cos(D-A)) / (cos(D+A))

D must (and in the real world will) be limited to the range:

-((Pi/2)-A) radians< D < ((Pi/2)-A) radians.

  • 1
    $\begingroup$ This can be solved for D, to get tan(D) = tan(A)*(w1-w2)/(w1+w2). It is interesting that if one weight is not much bigger than the other this is not very different from the (simpler but not physically correct) formula I would've used in the pledge just to avoid solving that: simply D = (w1-w2)/(w1+w2). $\endgroup$
    – Arnoques
    Dec 5, 2011 at 14:00
  • $\begingroup$ The balance angle will not be up/down for any other load. 2:1 is 45 degrees. i.ibb.co/kHHzQC5/image.png and the angle is 90*(1-(A-B)/(A or B depending on which is larger)). $\endgroup$
    – J Decker
    Jun 7, 2022 at 9:39

To a first approximation, the scale is either balanced or not balanced, and remains so at all angles, so there is no formula that does what you're looking for. A simple model of a scale says that the angle should be either

  1. whatever angle you want if the weights on both sides are equal
  2. tipped all the way over to the heavier side if the weights are unequal

This assumes that the pans are equally-spaced from the fulcrum and the fulcrum is an idealized point and the scale is rigid and the masses are the same shape.

You can try this for yourself easily with an experiment I did just now. Balance a spoon on your finger. Put a penny in the bowl of the spoon. The spoon will tip all the way over, the penny will fall off, and the spoon falls off your finger. (It did for me at least - your results may vary depending on how ideal spoons and fingers are where you live.)

One way to modify this using a physics concept would be to move the fulcrum towards the heavier side. A scale is balanced when

$$L_1w_1 = L_2w_2$$

with $L_1$ and $L_2$ the length of the sides of the scale from the fulcrum and $w_1$ and $w_2$ the two weights. For example, if you put a 2 kg weight 10cm from the fulcrum to the right, it can be balanced by a 1 kg weight on the left if you put it 20cm away. This is the same principle a parent can use to play see-saw with their child. The physics term is a lever.

One way to see where this equation comes from is by balancing the torques exerted by the weights.

Another is to consider energy. The energy of the weights comes from the formula

$$U = w h$$

$U$ is the potential energy, $w$ is the weight, and $h$ is the height above ground. A basic principle of statics is that the scale will be balanced whenever this potential energy stays the same as you change the angle.

Suppose one arm is twice as long as the other. Then when you tilt the scale a little, that side will drop twice as far, so $h$ changes by twice as much for the long side as for the short side. That means that to have the scale balanced $w$ must be half as much for the long side. That way the change in $h$ is twice as much but $w$ is half at much, so when you multiply them to get the energy, you find the same thing for both sides.

For example, if with the 2kg and 1kg weights described above, the 1kg weight is on the long side. If it falls 1cm, then 2kg weight will rise 1/2 cm. Multiplying the change in height by the weight gives the same for both. (The weight is technically gravitational acceleration times the mass, so the weights are 10 Newtons and 20 Newtons because gravitational acceleration is 10 m/s^2.) This is true regardless of the angle, so the scale is either balanced or unbalanced at all angles.

To read some more, the scale is a simple example of a lever.

  • $\begingroup$ Such a scale would have the ultimate sensitivity, but lacking a way to read equilibrium, it would be useless. $\endgroup$
    – Georg
    Apr 30, 2011 at 21:03
  • $\begingroup$ @Georg Yes, I agree. As with all physics models, this is simply an idealization. $\endgroup$ Apr 30, 2011 at 21:22
  • $\begingroup$ balances do not tip all the way over. 2:1 is 45 degrees. i.ibb.co/kHHzQC5/image.png and the angle is 90*(1-(A-B)/(A or B depending on which is larger)). $\endgroup$
    – J Decker
    Jun 7, 2022 at 9:40

A scale has two options to work. a) It either acts as a pendulum with gravity counteracting the weight imbalance, or b) there is a return spring that tries to level the scale if it is not horizontal.

Either way the effect is the same. The angle is proportional to the imbalance. In your case the details do not matter, and so you get to pick the design (the constant of proportionality) in order to make it work reasonably well. Choose too low of a value and it will fail to show small differences (like trying to use a truck scale to weigh a fish). Choose to high and it will peg the scales with the slightest imbalance.

Have fun.


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