Question Video: Finding the Centre of Mass of a Uniform Lamina Composed of an Isosceles Triangle and a Square Knowing the Relation between Their Lengths Mathematics

A uniform square lamina 𝐴𝐡𝐢𝐷 has a side length 𝑙. Another uniform lamina 𝐡𝐢𝐸 of the same density, shaped as an isosceles triangle, is attached to the square such that 𝐸 lies outside the square and 𝐡𝐸 = 𝐢𝐸. Given that the square’s side length is 5/3 times the length of the triangle’s height, find the center of mass of the system.

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Video Transcript

A uniform square lamina 𝐴𝐡𝐢𝐷 has a side length 𝑙. Another uniform lamina 𝐡𝐢𝐸 of the same density, shaped as an isosceles triangle, is attached to the square such that 𝐸 lies outside the square and 𝐡𝐸 is equal to 𝐢𝐸. Given that the square’s side length is five-thirds times the length of the triangle’s height, find the center of mass of the system.

Taking a look at our diagram, we see our square 𝐴𝐡𝐢𝐷 and the isosceles triangle attached to one side of the square. In this exercise, we want to find the center of mass of the square and triangle combined. To help us do that, we’re told that the square has a side length 𝑙 and that the length of the triangle’s height, this distance here, is related to the square side length by this relationship: five-thirds times β„Ž equals 𝑙. Having written this down, let’s clear some space on screen and begin working on our solution of what is the center of mass π‘₯- and 𝑦-coordinate for these combined shapes.

For our overall strategy, we can begin by solving for the center of mass of the shapes individually, the square and the triangle by themselves. Once we’ve done that, we can use the fact that a two-lamina center of mass, that is, the center of mass of two individual laminas combined together, is given by these relationships for the π‘₯- and 𝑦-coordinates of that center of mass. Here, 𝐴 one and 𝐴 two are the respective areas of our two laminas involved. π‘₯ one and π‘₯ two are the center of mass π‘₯-coordinates of these respective shapes. And 𝑦 one and 𝑦 two are their corresponding 𝑦 center of mass coordinates. In order to apply these relationships then, we’ll need to solve for the centers of mass of our square and triangle as well as their areas.

Let’s start by considering our square. And if we consider the fact that this square’s geometric center is located at its center of mass, we know that that is at a point here with both π‘₯- and 𝑦-coordinates of 𝑙 over two. So let’s write those down as 𝐢𝑂𝑀 π‘₯ square and 𝐢𝑂𝑀 𝑦 square coordinates. And then let’s consider the area of our square. Since it has side length 𝑙, that area is simply 𝑙 squared. Now let’s move on to our triangle, calculating the center of mass π‘₯- and 𝑦-coordinates for it. We can recall that for such a shape, the center of mass π‘₯- and 𝑦-coordinates are equal to the average π‘₯- and 𝑦-coordinates, respectively, of the triangle’s vertices. Considering the coordinates at vertex 𝐡, we can see that these are 𝑙, 𝑙. At vertex 𝐢, they are 𝑙 and zero.

But then what about at vertex 𝐸? We were told in our problem statement that this is an isosceles triangle. This means that the 𝑦- coordinate of point 𝐸 is midway between that of points 𝐢 and 𝐡. This tells us that it’s 𝑙 over two. To solve for the π‘₯-coordinate though, we’ll need to add the height β„Ž to the length 𝑙 that we already have as the side length of our square. Considering this relationship between β„Ž and 𝑙, if we multiply both sides of the equation by three divided by five, then we find that β„Ž, the height of this isosceles triangle, is three-fifths 𝑙, which means that the π‘₯-coordinate of vertex 𝐸 will be 𝑙 plus three-fifths 𝑙. Combining these gives us eight 𝑙 divided by five.

Now that we know the coordinates of each of the three vertices of our triangle, we can use the fact that the center of mass π‘₯-coordinate is equal to the average of these three π‘₯-values and the center of mass 𝑦-coordinate is equal to the average of these three 𝑦-values. And note that this is true for any triangle whether isosceles or not. Beginning with our π‘₯-coordinate, this is equal to 𝑙 plus 𝑙 plus eight 𝑙 over five all divided by three. This is equal to 18𝑙 over five divided by three or 18 over 15𝑙, simplifying to six-fifths 𝑙. So, we’ll record this value as our center of mass π‘₯-coordinate for the triangle and then clear this away so we can start to work on the center of mass 𝑦-coordinate.

This is equal to the average 𝑦-coordinate of the three vertices. That’s 𝑙 plus zero plus 𝑙 over two divided by three, which is three-halves 𝑙 over three or simply 𝑙 over two. And we’ll then enter this value into our coordinate pair. The last thing we need to do before we can apply these two center of mass equations is calculate the area of our triangle. In general, triangle area is one-half base times height. Here we’ve said that our height is three-fifths times 𝑙. And with that orientation, we can see that our base here is 𝑙. The area then is one-half 𝑙 times three-fifths 𝑙 or three-tenths 𝑙 squared. We’re now completely prepared to apply these two relationships to solve for our overall center of mass.

The center of mass of our two laminas combined has an π‘₯-coordinate given by the area of the square multiplied by the center of mass π‘₯-coordinate of the square plus the area of the triangle times the center of mass π‘₯-coordinate of that shape all divided by the areas of these two shapes added together. When we plug in all these values, we can note that a factor of 𝑙 squared appears in both numerator and denominator in all terms. This means we can factor it out and then cancel it out, leaving us with this expression for our center of mass π‘₯-coordinate.

Note that in the numerator we can now factor out one power of 𝑙. And then if we add together one-half and three-tenths times six-fifths, that’s equal to 43 divided by 50. And then in our denominator, one plus three-tenths is equal to 13 divided by 10. If we then multiply top and bottom by 10 divided by 13, our denominator simplifies to one, and we have 𝑙 times forty-three fiftieths times ten thirteenths. This simplifies to forty-three sixty-fifths 𝑙. That’s the center of mass π‘₯-coordinate of our combined two-lamina shape. So we enter this value into that coordinate pair, and now we clear space to calculate our overall center of mass 𝑦-coordinate.

Similar to our π‘₯-coordinate, this equals the area of the square times the center of mass 𝑦-coordinate of that shape plus the area of the triangle times the center of mass 𝑦-coordinate of the triangle, and all this is divided by the sum of the areas of those two shapes. With those values substituted in, once again, we see that 𝑙 squared can be factored out of numerator and denominator. Just like before then, that factor cancels out, and we see that, once more, one power of 𝑙 can be factored out of the numerator.

When we go to add one-half to three-tenths times one-half, we get 13 over 20. And just like before, one plus three-tenths simplifies to thirteen-tenths. Multiplying numerator and denominator once more by 10 over 13, we see that our denominator simplifies to one while our numerator becomes 𝑙 times one-half. This then is the center of mass 𝑦-coordinate of our two laminas together, which means that our final answer is that the center of mass coordinates of these two laminas is forty-three sixty-fifths 𝑙 and 𝑙 over two.

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