Consider a triple-beam balance, like so:

enter image description here
(source: microscopesamerica.com)

An unknown mass is placed on the left pan, and the provided weights are moved on the right until the lever arm comes to rest at an exactly horizontal position, as indicated by a line on the rightmost tip of the lever arm. Often, when I'm using the device and I'm close to the equilibrium point, the lever arm will come to rest slightly higher or lower than horizontal, so I'll adjust the weights accordingly until the lines align perfectly.

However, I don't see why the arm must be perfectly horizontal to read an accurate value. Isn't a stationary lever -- no matter what the angle -- an indication that there is no net torque on the system? If so, wouldn't that demonstrate that the masses are balanced in an appropriate way?

In other words, why doesn't a lever which is stationary in a non-horizontal position remain stationary when placed horizontally (and vice versa), since the net torque is zero in both cases?

  • $\begingroup$ Try it: adjust the weights a bit and observe static equilibrium away from the zero point. Then take a look at the effective torque about the pivot point (due to gravity) as a function of the lever arm's angle. $\endgroup$ Commented Mar 27, 2014 at 21:38
  • $\begingroup$ This is a good question and the answer depend in detail on how the balance is constructed. $\endgroup$ Commented Mar 29, 2014 at 4:36

2 Answers 2


On a well designed balance scale, the center of gravity of the beam or lever arm will be just slightly below the center pivot point.

If the beam is not level, the center of gravity will be to one side or the other of the pivot, and will thus create torque as it tries to move directly below the pivot point.

The distance between the pivot point and the center of gravity of the beam will determine the scale's sensitivity, ie.. the closer the CG is to the pivot, the less torque it will apply when it is out of alignment, and the smaller the difference the scale will be able to detect.

There are no springs, or magnets that cause this effect, just gravity at work.

  • $\begingroup$ This is confirmed by I noticed that when I tilt the entire device, the arm remains horizontal (relative to the ground) as if it were springs, magnets etc. then the arm would rotate with the scale. $\endgroup$
    – Rick
    Commented Oct 2, 2015 at 15:56

(NOTE: This is not a definitive answer and is largely guesswork, but it might be useful for further answerers.)

At first I suspected it was some sort of weak spring or magnetic device which applied an opposing torque to the arm, so as to make the center reading unique. Searching Google, I found this explanation of the theory of a triple-beam balance offered by Ohaus, and it mentions that there is a magnetic dampener that couples to the free end of the arm that is used to stop the oscillation of the beam (presumably by the generation of opposing electrical eddy currents):

enter image description here

Unfortunately, the slide in question also specifically notes that "this resistance is to movement and not an attractive force thus no added torque is applied", so this is not the correct reason. I searched through the rest of the slides to see if there was any mention of an angle-dependent opposing force, but there didn't seem to be any.

However, it is obvious that there has to be an opposing torque that occurs when the arm moves up and down, as otherwise if the two arms were balanced, there would be a wide range of accessible angles which the arm would be stable in, whereas having worked with these sort of scales before I can say that there is a restoring force that depends on angle.

Possible leads of investigation: according to the same set of slides, the weighing platform makes contact with the balancing arm using a knife-edge bearing:

enter image description here

As a result, if the weighing arm angle is $\theta$, the torque $T_\theta$ provided by the weighing platform will "correctly" be $$T_\theta=T_0\cos(\theta).$$

Otherwise, using a different sort of bearing could cause the load to be applied closer or further down the arm, destroying the $\cos(\theta)$ dependence.

However, there is no mention of how the sliding weights make contact with the arm. So I'd try to find out how the sliding weights make contact with the arm, as this could potentially be used to ensure that there is a net angular difference on how the two sides react to changes in angle. For example, if the right edge of the sliding weight makes contact with the balancing arm when the sliding-weight section is raised higher, it would cause the torque to decrease slower than the normal $\cos(\theta)$ dependence that the left-hand side experiences, and thus there would be a restoring force that drives the sliding-weight section downwards.

Or maybe the post which is connected to the weighing platform and extends downwards into the hole on the top of the balance is actually connected to a spring:

enter image description here

Unfortunately I don't have access to a mechanical scale at the moment (all mine are digital).

Honestly the easiest way would be to ask one of their engineers, but that'd sort of be cheating :).

  • $\begingroup$ Thanks for your detailed response. I noticed that when I tilt the entire device, the arm remains horizontal (relative to the ground), lending weight to your sliding weights hypothesis. I'll do some more testing and see what I can determine. $\endgroup$ Commented Mar 27, 2014 at 23:34
  • $\begingroup$ I inserted strips of metal underneath the right side of each sliding weight, so the point of contact was forced to remain approximately steady. However, the restoring torque was still present to the same degree as before the change. There isn't any sort of spring in the device in the base; rather, there's a very light lever which seems to serve to prevent the pan from moving left-to-right. Do you think it is possible that the knife-edge bearing system on the pan causes the force to remain perpendicular to the arm somehow? If so, any ideas on how would that occur? $\endgroup$ Commented Mar 28, 2014 at 23:15
  • 1
    $\begingroup$ The damping magnets server to cause any swinging to settle out faster, I don't believe they impart any force on the bar when it is at rest. $\endgroup$ Commented Mar 29, 2014 at 4:35
  • $\begingroup$ A scale will typically be constructed so that if everything is perfectly balanced, moving the beam off the center point will cause mass to move up more than it moves down (if there were only one "knife-edge" support, the center of mass of the swinging section would be below it; the fact that there are additional supports to keep the pan level complicates things but does not change the principle). A scale maker can easily adjust the effective distance between the center of mass and the support point, and thereby adjust the sensitivity. $\endgroup$
    – supercat
    Commented Oct 30, 2014 at 22:26

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