Question Video: Finding the Position of the Centre of Mass of a Uniform Equilateral Triangular Lamina Mathematics

Find the position of the center of mass of the uniform lamina 𝐴𝐡𝐢, which is in the shape of an equilateral triangle.

03:42

Video Transcript

Find the position of the center of mass of the uniform lamina 𝐴𝐡𝐢, which is in the shape of an equilateral triangle.

Okay, in our sketch, we see this equilateral triangle oriented on an π‘₯𝑦-plane. Vertex 𝐴 is at the origin, 𝐡 is up here, and 𝐢 is here along the π‘₯-axis. This shape is filled up by a uniform lamina. We can think of it as a very thin sheet of mass of material. Mass is evenly distributed throughout this lamina, and we want to solve for its center of mass. This is the location where all of this lamina’s mass is effectively concentrated.

Now, since the lamina 𝐴𝐡𝐢 is uniform, that means if we can find the geometric center of this shape, then we’ll also have found its center of mass. For any triangle, whether equilateral or not, its geometric center can be found from the coordinates of its three vertices. That is, for our triangle, if we know the coordinates of vertex 𝐴, 𝐡, and 𝐢, then we can use that information to solve for the geometric center of our triangle. And as we said, that’s located at the same point as its center of mass.

Our first task then is to solve for the coordinates of these three vertices. Considering first vertex 𝐴, since it’s located at the origin, we know that its π‘₯- and 𝑦-coordinates are zero, zero. Then let’s consider vertex 𝐡. Its π‘₯- and 𝑦-coordinates are indicated here, and we can see that its π‘₯-coordinate will be one-half the base of our triangle, where that base is seven π‘Ž. But what about the 𝑦-coordinate of this vertex? At this point, we can recall that this is an equilateral triangle. That means that all of the interior angles are the same. And so they must be 60 degrees.

Knowing that, we can say that this height of the triangle, what we’re really trying to solve for here, is equal to the hypotenuse of this right triangle multiplied by the sin of 60 degrees. That hypotenuse length is the length of all the sides of the triangle, seven times π‘Ž. So therefore, the 𝑦-coordinate of vertex 𝐡 is seven π‘Ž times the sin of 60 degrees. We can then recall that the sin of 60 degrees equals exactly the square root of three over two. So we now have the π‘₯- and 𝑦-coordinates of vertex 𝐡. And if we then consider those coordinates of vertex 𝐢, we see that the π‘₯-coordinate of this point is seven times π‘Ž, while the 𝑦-coordinate is zero.

Let’s now remind ourselves of what we noted earlier that the π‘₯- and 𝑦-coordinates of the center of mass of our triangle are equal, respectively, to the average π‘₯- and 𝑦-coordinates of our vertices. In other words, if we average out these three values, then we’ll have found the π‘₯-coordinate of the center of mass of our triangle. And the same is true for the 𝑦-coordinate of our triangle center of mass. Plugging in the π‘₯-coordinates of our three vertices, their average will be zero plus seven π‘Ž over two plus seven π‘Ž all divided by three. This equals three-halves times seven π‘Ž divided by three or in simplified form seven π‘Ž divided by two.

Now let’s move on to calculate the 𝑦-coordinate of our triangle center of mass. Similarly to the π‘₯-coordinate, the 𝑦-coordinate of the center of mass equals the average 𝑦-value of the three vertices. This simplifies to the square root of three over six times seven π‘Ž. We can now write out the coordinates of our center of mass. For this equilateral triangle, its center of mass is located at seven π‘Ž over two, root three over six times seven π‘Ž or seven π‘Ž over two, seven root three π‘Ž over six.

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