Balancing Ruler Problem – A Surprising Solution, Part II

This is the last part of a multi-part series on balancing rulers. If you’d like, you can check out the original problem or the first part of the solution.

So, I mentioned last time that I was rather surprised at the result of the Balancing Ruler Problem. I expected the balancing point to be further away from the center. In case you are having a hard time reading the ruler, the balancing point of the one penny-two pennies ruler was at 5 7/16″ inches. This means that the prism was moved 9/16 of an inch toward the two pennies end from the center.

One Penny - Two Pennies

One Penny – Two Pennies

So, I decided that I would see how the position of the prism changes if the proportion of the masses stays constant, but the amount of mass changes. So, I recreated the balancing act with dimes instead of pennies.

One dime - two dimes

One dime – two dimes

The balancing point appeared to be at 5 9/16″. The prism was located a full eighth-inch back toward the center. In my mind, this demonstrated that the mass at the end is a variable as much as the proportion of the two masses. Also, it seemed to suggest that the prism will move away from the center if the masses are greater. So, I decided to test this hypothesis by combining the coins so that there was one penny and one dime on one end and two of each at the other.

One Dime, One Penny - Two Dimes, Two Pennies

One Dime, One Penny – Two Dimes, Two Pennies

The balancing point is located a 5 3/8″ meaning that combining the masses moved the balancing point another eighth-inch away from the center. This supported my hypothesis.

Well, what about quarters? Balancing point: 5 3/16″

one quarter - two quarter

one quarter – two quarter

And quarters and pennies together? Balancing point: 4 15/16″.

photo 4

One quarter, one penny – two quarters, two pennies

So, it seems like there is more support for my hypothesis.

Then, a student and I decided to grow the project a bit.

One calculator - two calculators

One calculator – two calculators

With the number of variables that have changed, this photo doesn’t really mean a whole lot, but we had fun trying to balance calculators on a yardsticks.

The balancing point of the calculators

The balancing point of the calculators

So, how can we explain this phenomena? What causes the balance point to move predictably even though the ratio of the masses at the end is a constant?


One thought on “Balancing Ruler Problem – A Surprising Solution, Part II

  1. The force acting to push down on the ruler is equal to the mass multiplied by acceleration (in this case due to gravity), hence the magnitude of the turning moment is related to the mass and the distance from the pivot, (moment = force x perp distance). The ruler balances when the forces are equal (or in a state of equilibrium). If you use force in newtons exerted on the ruler by each pile, and measure distance from pivot in metres, you can calculate the clockwise and anticlockwise moments and see the maths behind the balancing act.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s