Question Video: Finding the Coordinates of the Centre of Gravity of a Uniform Lamina Composed of a Triangle and Rectangle Mathematics

Two uniformed laminas made of the same material are joined together to make one body. The first is a rectangle 𝐴𝐵𝐶𝐷 where 𝐴𝐵 = 16 cm and 𝐵𝐶 = 7 cm. And the second is an isosceles triangle 𝐶𝐸𝐷 where 𝐷𝐸 = 𝐶𝐸 = 17 cm and vertex 𝐸 lies outside of the rectangle. Find the coordinates of the center of gravity of the laminas, given that rectangle 𝐴𝐵𝐶𝐷 is in the first quadrant, 𝐵 is at the origin, and 𝐶 is on the 𝑥-axis.

08:47

Video Transcript

Two uniformed laminas made of the same material are joined together to make one body. The first is a rectangle 𝐴𝐵𝐶𝐷, where 𝐴𝐵 equals 16 centimeters and 𝐵𝐶 equals seven centimeters. And the second is an isosceles triangle 𝐶𝐸𝐷 where 𝐷𝐸 equals 𝐶𝐸 equals 17 centimeters and vertex 𝐸 lies outside of the rectangle. Find the coordinates of the center of gravity of the laminas, given that rectangle 𝐴𝐵𝐶𝐷 is in the first quadrant, 𝐵 is at the origin, and 𝐶 is on the 𝑥-axis.

Okay, we see in our sketch these two laminas. Here we have our rectangle and here’s our triangle. And we’re told some of the dimensions involved in the sketch. 𝐴𝐵, we know, is 16 centimeters. 𝐵𝐶 is seven centimeters. And segment 𝐷𝐸 on our triangle is 17 centimeters, which we’re told is identical in length to the line segment 𝐶𝐸. Knowing all this, we want to find the coordinates of the center of gravity of these laminas. And the first thing we can say about that is, in this case, the center of gravity is located at the same point as the center of mass of these laminas.

Because we assume a uniform gravitational field for these laminas, in solving for one quantity, we solve for the other. Since we have these two laminas, the rectangle and the triangle, being joined together, here’s the approach we’ll take. First, treating these shapes separately will solve for their individual centers of mass. Once we know those quantities, we’ll combine them to solve for our system’s overall center of mass. And as we mentioned, that will be the same as the center of gravity of these laminas. As we get started, let’s clear some space on screen. And we can begin by solving for the center of mass of our rectangular lamina.

This point will be located at the average position of mass in a rectangle. Because the lamina that makes up this rectangle is uniform, that tells us that its center of mass will be halfway across its width and halfway up its total length. That is, the 𝑥-coordinate of a rectangle center of mass will be one-half of seven centimeters, while the 𝑦-coordinate will be one-half of 16 centimeters. We have then the 𝑥- and 𝑦-coordinates of the center of mass of a rectangular lamina. Now let’s solve for the position of the center of mass of our triangular lamina.

What we’re going to do here is use the fact that, for any triangle, whether isosceles like we have here or not, when that triangle is uniform, its center of mass is located at the average position of the 𝑥- and 𝑦-coordinates of its vertices. In other words, if we solve for the 𝑥- and 𝑦-coordinates of vertices 𝐷, 𝐸, and 𝐶, then this triangle center of mass in the 𝑥-direction will be the average of those three 𝑥-values. And likewise, its center of mass in the 𝑦-direction will be the average of those 𝑦-values.

So let’s solve for the coordinates of these three vertices, and we’ll start with vertex 𝐷. And as we do this for brevity’s sake, we’ll leave out our units of centimeters. So considering the coordinates of vertex 𝐷, we know that point 𝐵 in our rectangle is located at the origin of our coordinate frame. We can say then that the coordinates of vertex 𝐷 are seven in the 𝑥-direction and 16 in the 𝑦-direction. If we next consider the coordinates of vertex 𝐶, the 𝑥-coordinate here is once again seven, while the 𝑦-coordinate is now zero.

Lastly, we want to know the coordinates of vertex 𝐸. Regarding the 𝑥-coordinate of this point, we can see that that will be seven centimeters plus whatever distance this is here. That distance turns out to be what we could call the height of this triangle. That is, if this side length from 𝐷 to 𝐶 is the base, then the dashed line represents the height. If we think of that height as one side in a right triangle, we see that the other two side lengths are eight centimeters and 17 centimeters. By the Pythagorean theorem then, this height we want to solve for equals the square root of 17 squared minus eight squared. That comes out to 15.

So if this dimension of our triangle is 15 centimeters, then the 𝑥-coordinate of vertex 𝐸 must be 15 centimeters plus seven centimeters. That’s 22 centimeters. And now we can write in the 𝑦-coordinate of this vertex. Because we’re working with an isosceles triangle, that 𝑦-value of vertex 𝐸 is equal to half the height of our rectangle, that is, eight centimeters. Now that we know the coordinates of our three triangle vertices, we’re ready to solve for the average 𝑥- and 𝑦-values of these vertices.

The average 𝑥-value equals seven plus seven plus 22 divided by three — this is equal to 36 over three or 12 — while the average 𝑦-value equals 16 plus zero plus eight divided by three. That’s 24 over three or eight. Therefore, the center of mass of our triangle has coordinates 12, eight. Now that we know this value as well as the center of mass coordinates of a rectangle, we can say that the entire mass of our two laminas is effectively concentrated at these two points. The mass of the rectangle is effectively located here and that of the triangle here.

Recall now that our goal is to solve for the center of mass coordinates of our entire system, the two laminas together. Mathematically, we can now treat our system as these two points, effectively containing all the mass of the rectangle and the triangle, respectively. To calculate the overall center of mass of our system, we’ll need to know the relative mass of the one shape to the other. To do this, we’ll rely on the fact that the laminas that make up these shapes are uniform. That means that the ratio of the areas of these two shapes to one another is the same as the ratio of their masses to one another.

If, for example, our triangle had twice as much area as our rectangle, then that would mean it would also have twice as much mass. Due to this correspondence between area and mass, our next step will be to calculate the areas of these two shapes. Starting with the rectangle, we know it has dimensions of seven centimeters by 16 centimeters. Seven times 16 equals 112 which is the area of the rectangle in square centimeters. For the area of our triangle, in general, this is equal to one-half a triangle’s base times its height. Here, we’ve referred to this distance as the base of our triangle and this distance as its height. We can therefore say that our triangle has an area of one-half times 16 centimeters multiplied by a height of 15 centimeters. And multiplying these values together, we get a result of 120.

We see then that our triangle has a greater area than our rectangle. And therefore, by the same ratio, it has a greater mass. We could say that if the triangle has 120 units of mass, then the rectangle has 112 of those same units. All this is important for calculating the overall center of mass of our system because, in general, the 𝑥-coordinate of a collection of masses is equal to the sum of the product of each mass with its average 𝑥-coordinate all divided by the sum of the masses individually. In our scenario, we have two masses, that of the rectangle and that of the triangle. And each one, as we’ve seen, has an 𝑥-coordinate that we’ve solved for earlier.

Applying this relationship then to solve for the 𝑥-coordinate of our system’s overall center of mass, it would be equal to the effective mass of our rectangle, 112, multiplied by the 𝑥-coordinate of its center of mass plus the effective mass of our triangle times its center of mass 𝑥-coordinate all divided by the sum of our two effective masses. This is equal to 1832 over 232 or, fully reduced, 229 over 29. We can now move on to calculating the overall 𝑦 center of mass coordinate. The formula for doing this is the same as for calculating the 𝑥-coordinate, except now we use the average 𝑦-value of each of the masses in our system.

Applying this relationship to our scenario, we use the effective mass of our rectangle multiplied by the 𝑦-coordinate of this shape’s center of mass and add to it the effective mass of our triangle times its center of mass 𝑦-coordinate. All this gets divided by the sum of our two masses, and this is equal to 232 times eight divided by 232. We see then that this simplifies to eight. So bringing it all together, we can say that the center of mass coordinates of our system overall are 229 over 29 centimeters in the 𝑥-direction and eight centimeters in the 𝑦-direction. This is the location of our overall system’s center of mass.

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