# Video: Finding the Centre of Mass of a System of Three Masses Placed on the Sides of a Square

A square π΄π΅πΆπ· of side length πΏ. Three masses of 610 g are placed at π΄, π΅, and π·. Find the coordinates of the centre of mass of the system.

03:39

### Video Transcript

A square π΄π΅πΆπ· of side length πΏ. Three masses of 610 grams are placed at π΄, π΅, and π·. Find the coordinates of the centre of mass of the system.

In our diagram, we see the square π΄π΅πΆπ· as well as the masses that are placed at three of those four corners. We want to solve for this system of massesβ centre of mass. And we can describe that centre of mass as a pair of coordinates: the centre of mass in the π₯-direction and the centre of mass along the π¦-axis.

To begin solving for these two coordinates, letβs recall the mathematical relationship for centre of mass. If our system is composed of a collection of masses which weβve indexed with π and called π sub π, then if we multiply each of those individual masses by the distance π₯ sub π to that point mass or mass element is from the axis of rotation then sum up all those products and finally divide by the sum of all the individual masses, then weβve calculated the centre of mass of our system.

In our case, our system exists in two dimensions: both the π₯-dimension and the π¦-dimension. And there is a centre of mass of the system along each of these two dimensions. First, weβll solve for ππ sub π₯ β the centre of mass of our system along the π₯-direction. Since weβre oriented along that way, that means our axis of rotation is about the π¦-axis, where π₯ equals zero.

And as we calculate this value, we recall that the length of each side of our square is given as πΏ. Starting at corner π΄, the mass there has a distance of zero from our axis of rotation. Then, we move to point π΅, where the mass there has a distance of πΏ from our axis of rotation and finally to corner π·, where the mass there once again is on the π¦-axis and so has a distance of zero from the axis of rotation.

This means our numerator simplifies to 610 times πΏ and our denominator simplifies to three times 610. So we see the factor of 610 cancels out leaving us with a centre of mass in the π₯-direction of πΏ over three. Thatβs the π₯-coordinate of this systemβs centre of mass. Now, we move on to calculating the π¦-coordinate of the system centre of mass. This time rather than rotating about the π¦-axis, our axis of rotation is the π₯-axis, where π¦ equals zero.

When we again write out our equation for the centre of mass along the π¦-axis, starting at corner π΄, we see the mass there is on the axis of rotation. So its distance from that axis is zero. Likewise, for the mass at corner π΅, its distance is also zero. At π·, the mass has a distance of πΏ from our axis of rotation.

So like before, our numerator simplifies to 610 times πΏ and our denominator simplifies to three times 610. Once again, we see 610 cancelling out. This leaves us with a value of πΏ over three for ππ sub π¦. So we see that both the π₯- and π¦-coordinates of the systemβs centre of mass are πΏ over three.