Why does the beam in a weighing balance get tilted proportional to the weights added to each pan?

I'm talking about a beam balance(a simple weighing balance with a beam and two pans hung on either side)

As answered in a previous question, the beam comes back to the original position when one side is pushed down and released because the centre of gravity of the beam is a little below the axis of rotation.

practically, when unequal weights are put to the pans, the beam gets tilted proportional to the weight added to each side. Yet, according to my knowledge, this cannot happen theoretically.

for example, if a weight of M is added to the left pan , and a weight of m is added to the right pan, and M>m, and the length of the beam is 2l, the torques acting around the point of rotation would be Ml- anticlockwise and ml-clockwise. Then the resultant torque would be (Ml-ml) anticlockwise. hence theoretically, the beam must rotate anticlockwise by nearly 90 degrees, which means the left pan (the pan with weight M) must move down as much as possible.(Please note the small torque due to the weight of the beam has been ignored)

But what really happens is the beam gets tilted proportional to the weight added to each side. How can this be?

Suppose you have a balance that looks like this: where $m_1$ and $m_2$ are the weights you're comparing and $M$ is some large weight fixed to the balance. For simplicity let's $m_1$ is zero, so one one end of your scales you have some weight $m_2$ and there is nothing else on the other end.

You are quite correct that with simple lever scales the lever would tip over until it was vertical with $m_2$ at the bottom. However the scales pictured above would rotate only a small distance because the restoring force from the large weight $M$ would balance out the tilting force from $m_2$. The result is that the angle increases as you increase $m_2$, just as you observe in practice.

You can work out the angle of rotation just by taking moments around the pivot point, and you get:

$$m_2gl\cos\theta = Mgh\sin\theta$$

or

$$\tan\theta = \frac{m_2 l}{Mh}$$

up to about 25$^\circ$ $tan(\theta)$ is approximately proportional to $\theta$ so you get:

$$\theta \propto m_2$$

i.e. for angles less then 25$^\circ$ the angle of tilt is proprotional to the weight on the scales.

protected by Qmechanic♦Jun 22 '16 at 8:41

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?