Video Transcript
A uniform square-shaped lamina 𝐴𝐵𝐶𝐷 of side length 222 centimeters has a mass of one kilogram. The midpoints of line segments 𝐴𝐷, 𝐴𝐵, and 𝐵𝐶 are denoted by 𝑇, 𝑁, and 𝐾, respectively. The corners 𝑇𝐴𝑁 and 𝑁𝐵𝐾 were folded over so that they lie flat on the surface of the lamina. Bodies of masses 365 grams and 294 grams were attached to the points 𝑇 and 𝐾, respectively. Find the coordinates of the center of mass of the system rounding your answer to two decimal places if necessary.
This question can be solved by modeling the system as consisting of only positive-mass bodies. These would be the rectangle 𝐶𝐷𝑇𝐾, the triangle 𝐾𝑇𝑁, and the added masses at 𝑇 and 𝐾. However, in this case, we will solve the problem using the negative mass method. We will model the system as containing positive-mass bodies consisting of the added masses at 𝑇 and 𝐾, the square 𝐴𝐵𝐶𝐷, and the triangle 𝐾𝑇𝑁, and also containing negative-mass bodies consisting of triangles 𝑇𝐴𝑁 and 𝐾𝐵𝑁. In order to use the negative mass method, we need to find the relative mass of each of these together with the 𝑥- and 𝑦-coordinates of their corresponding centers of gravity.
Let’s begin by considering the relative mass of each part of the system. We are told in the question that the square-shaped lamina 𝐴𝐵𝐶𝐷 has a mass of one kilogram. And converting to grams, this is equal to 1000 grams. At point 𝑇, we have a mass of 365 grams and at point 𝐾 a mass of 294 grams. Since the area of the triangle 𝐾𝑇𝑁 is a quarter of the area of the square, its mass is a quarter of the mass of the square lamina. And since one-quarter of 1000 is 250, the mass of the additional triangle 𝐾𝑇𝑁 is 250 grams.
Finally, the total mass of the two negative-mass triangles is negative one-quarter of the mass of the square lamina. Halving this value of negative 250, we see that the relative mass of triangles 𝑇𝐴𝑁 and 𝐾𝐵𝑁 are negative 125 grams. Our next step is to find the 𝑥- and 𝑦-coordinates of the centers of gravity of each part of the system. In order to do this, we will begin by identifying the coordinates of the points on the outside of the square. As point 𝐶 lies at the origin, it has coordinates zero, zero.
We are told that the square had side length 222 centimeters. This means that 𝐵 has coordinates 222, zero; point 𝐴 has coordinates 222, 222; and point 𝐷 has coordinates zero, 222. We were told in the question that the points 𝑇, 𝑁, and 𝐾 are the midpoints of line segments 𝐴𝐷, 𝐴𝐵, and 𝐵𝐶. Since one-half of 222 is 111, 𝐾 has coordinates 111, zero; 𝑇 has coordinates 111, 222; and 𝑁 has coordinates 222, 111. We know that the square 𝐴𝐵𝐶𝐷 is uniform, so the position of its center of gravity is its geometric center. This lies at the point 111, 111. So the 𝑥- and 𝑦-coordinates of the center of gravity of 𝐴𝐵𝐶𝐷 are both 111.
As already established, the additional mass of 365 grams at 𝑇 lies at the point 111, 222. And the additional mass of 294 grams at 𝐾 lies at the point 111, zero. Next, we consider the triangle 𝐾𝑇𝑁, recalling that there are several methods to find the center of gravity of a triangle. If we know the coordinates of the three vertices of the triangle, we can find the average of the 𝑥-coordinates and the average of the 𝑦-coordinates. The sum of the 𝑥-coordinates at vertices 𝐾, 𝑇, and 𝑁 is 111 plus 111 plus 222. And dividing this by three gives us 148. The 𝑥-coordinate of the center of gravity of triangle 𝐾𝑇𝑁 is 148. Repeating this for the 𝑦-coordinates of the three vertices, we have zero plus 222 plus 111 divided by three. As this is equal to 111, the 𝑦-coordinate of the center of gravity of triangle 𝐾𝑇𝑁 is 111.
We can repeat this process for triangles 𝑇𝐴𝑁 and 𝐾𝐵𝑁. The center of gravity of triangle 𝑇𝐴𝑁 lies at the point 185, 185. And the center of gravity of triangle 𝐾𝐵𝑁 lies at the point 185, 37. We are now in a position where we can find the 𝑥- and 𝑦-coordinates of the center of gravity of the system. To calculate the 𝑥-coordinate, which we will call 𝑥 bar, we multiply the mass of each body in the system by the 𝑥-coordinate of the relative body’s center of gravity and then divide by the sum of the masses. This gives us the following equation.
Clearing some space, we have 𝑥 bar is equal to 174899 divided by 1659, which, to two decimal places, is 105.42. We can then repeat this process to calculate the 𝑦-coordinate 𝑦 bar. This time, the equation simplifies to 192030 divided by 1659. And rounding to two decimal places, this is equal to 115.75. We can therefore conclude that the coordinates of the center of mass of the system are 105.42, 115.75.
