Question Video: Finding the Center of Mass of a System of Three Masses Placed on the Vertices of a Square Mathematics

Video Transcript

A square 𝐴𝐵𝐶𝐷 has side length 𝐿. Three masses of 610 grams are placed at 𝐴, 𝐵, and 𝐷. Find the coordinates of the center of mass of the system.

In our diagram, we see that masses of 610 grams are placed at three of the vertices of our square. We are told that the square has side length 𝐿. As point 𝐴 lies at the origin, it has coordinates zero, zero. Point 𝐵 has coordinates 𝐿, zero; point 𝐶 has coordinates 𝐿, 𝐿; and point 𝐷 has coordinates zero, 𝐿.

We are asked to find the coordinates of the center of mass of the system. This can be written as the ordered pair 𝐶𝑀 sub 𝑥 and 𝐶𝑀 sub 𝑦, where the center of mass in the 𝑥-direction is equal to the sum of the products of the masses multiplied by the corresponding 𝑥-coordinates divided by the sum of the masses. The center of mass in the 𝑦-direction can be calculated in the same way, but this time using the 𝑦-coordinates of each mass.

Starting with the 𝑥-coordinates of points 𝐴, 𝐵, and 𝐷, we see that the center of mass in the 𝑥-direction is equal to 610 multiplied by zero plus 610 multiplied by 𝐿 plus 610 multiplied by zero all divided by 610 plus 610 plus 610. This simplifies to 610 multiplied by 𝐿 divided by 610 multiplied by three. Dividing the numerator and denominator by 610, this in turn simplifies to 𝐿 over three.

We can then repeat this process for the 𝑦-coordinates at points 𝐴, 𝐵, and 𝐷. Once again, our equation simplifies to 610 multiplied by 𝐿 divided by 610 multiplied by three. Dividing the numerator and denominator by 610 once again, we see that the center of mass in the 𝑦-direction is also equal to 𝐿 over three.

We can therefore conclude that the coordinates of the center of mass of the system are 𝐿 over three, 𝐿 over three.