Video Transcript
The uniform lamina 𝐴𝐵𝐶𝐷 is a rectangle, where 𝐴𝐵 equals 48 centimeters, 𝐵𝐶 equals 64 centimeters, and 𝐸 is an element of line segment 𝐴𝐷 such that 𝐴𝐸 equals 48 centimeters. The corner 𝐴𝐵𝐸 is folded over along the line 𝐵𝐸 such that the side 𝐴𝐵 meets the side 𝐵𝐶 as shown in the figure. Find the coordinates of the center of mass of the lamina in this new shape.
Alright, so in our diagram, we see this lamina, which started out as a rectangle, rectangle 𝐴𝐵𝐶𝐷, but then had one of its corners folded in so that now that corner rests here. And this folded edge of the lamina rests along line segment 𝐵𝐶. If this lamina was made of paper, then it looks like we’re starting to make a paper airplane. But anyway, we want to solve for the coordinates of the center of mass of this lamina in its new folded shape.
Let’s first record a few of the distances we’re given. Segment 𝐴𝐵, we’re told, is 48 centimeters; side 𝐵𝐶 is 64 centimeters; and the length of the folded edge of our lamina is 48 centimeters. Knowing all this, here’s the approach we’ll take to solve for the overall center of mass of this shape. First, recognizing that we can see two shapes here, a rectangle and a triangle, we’re going to solve for the center-of-mass locations of those shapes individually. They’ll be located roughly here and roughly here.
Remembering that the center of mass of a shape is the location where all of its mass is effectively concentrated, we can then solve for the center of mass of our overall lamina by considering these two individual centers of mass as points that we combine. The process will be clearer as we move through it, so let’s start on that now.
First, we’ll clear some space, and we’ll begin by solving for the coordinates of the center of mass of our rectangle. As we saw, that rectangle is here. And since point 𝐶 is located at the origin of our 𝑥𝑦-coordinate frame and the width of our rectangle is equal to 64 centimeters minus 48 centimeters, or 16 centimeters, we can say that its center-of-mass 𝑥-coordinate is eight. And we’ll leave off the units. And then its center-of-mass 𝑦-coordinate must be 24. That’s half the distance between points 𝐴 and 𝐵. And we’re relying on the fact that the lamina that makes up this shape is uniform.
So we now know the coordinates of this point, the center of mass of our rectangle. Let’s now solve for the coordinates of that point on our triangle. To start figuring this out, we can recall that the coordinates of the center of mass of any uniform triangular lamina are equal to the average coordinates of the vertices of that triangle. That is, if we know the coordinates of this vertex, this vertex, and this vertex, the three that describe our triangle, then when we average their 𝑥- and 𝑦-values, we’ll have solved for the 𝑥- and 𝑦-values of the triangle’s center of mass.
We see that two of these vertices already have labels, point 𝐸 and point 𝐵. Let’s call our third unlabeled vertex point 𝐹. So what are the coordinates of point 𝐹? We can see it has an 𝑥-value of 16 and a 𝑦-value of zero. Moving on then to vertex 𝐵, we can see that this has an 𝑥-value of 64 and a 𝑦-value of zero again. Lastly, vertex 𝐸 has an 𝑥-value of 16 and a 𝑦-value of 48. This tells us that the average 𝑥-value of our three vertices equals 16 plus 64 plus 16 divided by three. This comes out to 96 over three, or 32. And likewise, the average 𝑦-value is equal to zero plus zero plus 48 divided by three, which comes out to 16. Having calculated these values, we now know the coordinates of the center of mass of our triangular shape.
Now that we know the coordinates of these two pink dots, representing the centers of mass of our rectangle and triangle, we’ll want to calculate how much mass is effectively concentrated at each one of these points, in other words, how much mass is contained in the rectangle and how much in the triangle. In our problem statement, we’re not told anything about mass. But we are told that the lamina that makes up these shapes is uniform. That tells us that the mass contained in each one of these shapes scales with their area. Put another way, the ratio of the mass of the rectangle to that of the triangle is equal to the ratio of the area of the rectangle to that of the triangle.
If we solve then for the areas of these two shapes, we’ve effectively solved for their masses. The area of the rectangle is equal to its base times its height, 16 centimeters by 48 centimeters. That’s 768 centimeters squared. The area of the triangle is equal to one-half its base of 48 centimeters times its height of 48 centimeters. This equals 1152. If we imagine that each one of our units of area is equal to one unit of mass, whatever that unit is, then we can say that our rectangle has 768 mass units, while our triangle has 1152.
Knowing all this about our scenario, we now have the information needed to calculate the overall center of mass of this folded lamina. We can recall that, for a collection of masses, 𝑚 one, 𝑚 two, and so on, the 𝑥-coordinate of their overall center of mass is equal to the sum of the product of each mass with its average 𝑥-position all divided by the sum of the individual masses. As we apply this equation to our situation, we’ll want to be careful to recall that we have a folded lamina. It’s not only that we have a rectangle and a triangle in our lamina, but this triangle is doubled over on itself. Essentially, there’s two triangles of this lamina, only one of which we can see. This means that when we calculate the 𝑥-coordinate of our lamina’s center of mass, we’ll multiply the effective mass of our rectangle by the 𝑥-coordinate of its center of mass, add to that the effective mass of our triangle multiplied by its average 𝑥-position.
But then note that we multiply this by two, and we do that as we mentioned because there are really two triangles here. Likewise, in the denominator, when we add masses, we add one times the mass of our rectangle but two times the mass of our triangle. This comes out to 79872 over 3072, which reduces to 26. So the overall 𝑥-coordinate of the center of mass of our folded lamina is 26 centimeters to the right of point 𝐶.
And now let’s continue on to calculating the 𝑦-coordinate of the center of mass. We’ll use the same formula that we used to calculate the 𝑥-coordinate, except now we use the average 𝑦-values of each mass. For our lamina then, the 𝑦-coordinate of its center of mass equals the effective mass of our first shape, the rectangle, times its average 𝑦-position plus the effective mass of our triangle times its average 𝑦-position. And once again, we double this factor up top because there are two triangles. And this is all divided by the mass of the rectangle plus two times the mass of the triangle. This equals 55296 divided by 3072, which reduces to 18.
So now we have our answer. The center of mass of this folded lamina is located at the point 26, 18.
