An equilateral triangle 𝐴𝐵𝐶 has a side length of 82 centimeters. When three equal masses are placed at the vertices of the triangle, the center of mass of the system is 𝐺. When the mass at vertex 𝐶 is removed, the center of mass of the system is 𝐺 prime. Find the coordinates of the center of mass of the two systems 𝐺 and 𝐺 prime.
In this question, we are given two figures that represent two systems involving the same equilateral triangle 𝐴𝐵𝐶. The triangle has a side length of 82 centimeters. In the first figure, we have equal masses placed at the vertices 𝐴, 𝐵, and 𝐶. And we are told the center of mass of the system is 𝐺. In the second figure, the mass at 𝐶 has been removed, so we only have masses at vertex 𝐴 and vertex 𝐵. This time, the center of mass is 𝐺 prime.
We have been asked to find the coordinates of both 𝐺 and 𝐺 prime. In order to find the coordinates of 𝐺, we will begin by completing the following table for figure one. We will let the equal masses at vertices 𝐴, 𝐵, and 𝐶 equal 𝑚 grams. As point 𝐵 lies at the origin, it has coordinates zero, zero. Point 𝐶 lies 82 centimeters along the 𝑥-axis, so it has coordinates 82, zero. Since our triangle is equilateral, the 𝑥-coordinate of 𝐴 will be equal to 82 divided by two, and this is equal to 41. The 𝑦-coordinate of point 𝐴 will be equal to the height of the right triangle drawn.
Using the Pythagorean theorem, we know that ℎ squared is equal to 82 squared minus 41 squared. This is equal to 5043. Next, we can take the square root of both sides of this equation. And since ℎ must be positive, we have ℎ is equal to 41 root three. The coordinates of vertex 𝐴 are 41, 41 root three.
We are now in a position to find the coordinates of the center of mass of the first system 𝐺. The 𝑥-coordinate of 𝐺, which we will call 𝑥 one bar, is equal to 𝑚 multiplied by 41 plus 𝑚 multiplied by zero plus 𝑚 multiplied by 82 all divided by 𝑚 plus 𝑚 plus 𝑚. We multiply the relative masses at each of the vertices 𝐴, 𝐵, and 𝐶 by their corresponding 𝑥-coordinate. We then find the sum of these values and divide by the sum of the masses. The expression for 𝑥 one bar simplifies to 41𝑚 plus 82𝑚 divided by three 𝑚. We can divide through by a common factor of 𝑚 as the mass cannot be zero. This leaves us with 123 divided by three, which is equal to 41. The 𝑥-coordinate of point 𝐺 is 41. Repeating this for the 𝑦-coordinates, we have 41 root three 𝑚 over three 𝑚. Once again, the 𝑚’s cancel leaving us with 41 root three over three. This is the 𝑦-coordinate of point 𝐺.
Let’s now move on to the second figure where we need to find the coordinates of 𝐺 prime. Vertices 𝐴, 𝐵, and 𝐶 have the same coordinates. However, this time, there’s no mass at vertex 𝐶. The 𝑥-coordinate of the center of mass of this system, which we will call 𝑥 two bar, is equal to 𝑚 multiplied by 41 plus 𝑚 multiplied by zero divided by 𝑚 plus 𝑚. We note that the total mass of this system is two 𝑚. Our expression for 𝑥 two bar simplifies to 41𝑚 over two 𝑚. We can once again divide through by 𝑚, giving us 41 over two. The 𝑥-coordinate of 𝐺 prime is 41 over two. Repeating this for the 𝑦-coordinate, we have 41 root three 𝑚 divided by two 𝑚. And this simplifies to 41 root three over two.
We now have coordinates for 𝐺 and 𝐺 prime as required, and it is worth noting that 𝐺 prime is the midpoint of 𝐴 and 𝐵. The center of mass of the first system is 41, 41 root three over three, and the center of mass of the second system is 41 over two, 41 root three over two.