# Question Video: Finding the Coordinates of the Centre of Gravity of a Uniform Triangular Lamina with a Triangle Cut from It Mathematics

The diagram shows a uniform lamina 𝐴𝐵𝐶 from which a triangle 𝐺𝐵𝐶 has been cut out. 𝐴𝐵𝐶 was an equilateral triangle with a side length of 93 cm and center of mass 𝐺. Find the coordinates of the new center of mass of the resulting lamina. Round your answer to two decimal places if necessary.

08:36

### Video Transcript

The diagram shows a uniform lamina 𝐴𝐵𝐶 from which a triangle 𝐺𝐵𝐶 has been cut out. 𝐴𝐵𝐶 was an equilateral triangle with a side length of 93 centimeters and center of mass 𝐺. Find the coordinates of the new center of mass of the resulting lamina. Round your answer to two decimal places if necessary.

The statement refers to this arrow-shaped lamina formed from the equilateral triangle 𝐴𝐵𝐶 with the triangle 𝐺𝐵𝐶 cut out of it. The specific information we’re given is that 𝐴𝐵𝐶 has a uniform mass distribution, its side length is 93 centimeters, and 𝐺 is its center of mass. Since we’re looking for the center of mass of this resulting shape and we’re explicitly told that it was an equilateral triangle with 𝐺𝐵𝐶 cut out of it, this suggests that we should use the negative mass method. To use the negative mass method, we first start with this diagram here that shows the equilateral triangle 𝐴𝐵𝐶 shaded in orange and the triangle 𝐺𝐵𝐶 shaded over it in magenta.

Separating this out into its two pieces, we have the original lamina 𝐴𝐵𝐶 and the lamina 𝐵𝐺𝐶 that represents the part that was cut out. The uniform lamina 𝐴𝐵𝐶 is a regular object, so its mass is positive. However, since the combination of 𝐺𝐵𝐶 and 𝐴𝐵𝐶 results in a lamina with no mass at the bottom, where 𝐴𝐵𝐶 clearly does have mass, 𝐺𝐵𝐶 must have negative mass because negative plus positive is zero. We find the resulting center of mass the way we would find the center of mass of any two laminae. It’s just one of these laminae now has a negative number instead of a positive number for its mass. This is why we call it the negative mass method.

We now need to find the mass and the location of the center of mass of these two laminae. Since these laminae are uniform, we can express their total mass as their area times a constant, which is the mass per unit area of the material. For the lamina with positive mass, we need the area of the equilateral triangle 𝐴𝐵𝐶. From the diagram, we know the length of the base is 93 centimeters and the height is the length of the blue dashed line, which passes through the points 𝐴 and 𝐺. Because this dashed line is an altitude, it is perpendicular to the line 𝐵𝐶. Furthermore, because 𝐴𝐵𝐶 is equilateral, this angle here at point 𝐵 is 60 degrees. So the triangle formed by 𝐴𝐵 and the point where the altitude meets the side 𝐵𝐶 is a 30-60-90 right triangle.

Either directly using trigonometry or using facts about 30-60-90 right triangles, since the hypotenuse of this triangle is 93 centimeters, the height must be 93 times the square root of three divided by two. We’ll actually calculate a definite value for this later, but for now, let’s just call it 𝑎. So the triangle 𝐴𝐵𝐶 has a base of 93 and a height of 𝑎. So its area, one-half base times height, is one-half 93𝑎. If we call the constant uniform density 𝑚, then one-half times 93 times 𝑎 is the area of 𝐴𝐵𝐶, and one-half 93𝑎 times 𝑚 is the total mass of the lamina 𝐴𝐵𝐶.

What about 𝐺𝐵𝐶? Well, recall that 𝐺 is the center of mass of the equilateral triangle. For any uniform triangle, the center of mass is one-third of the way along any of the medians. In any collateral triangle, the median which connects any vertex to the midpoint of the opposite side is also the altitude of the triangle. So the distance from the midpoint of 𝐵𝐶 to the point 𝐺 is one-third of the total height of the triangle 𝐴𝐵𝐶. But we know what that value is; we called it 𝑎. So the height of 𝐵𝐺𝐶 is one-third of 𝑎. 𝐵𝐺𝐶 and 𝐴𝐵𝐶 have the same base. So the area of 𝐵𝐺𝐶 is one-half times 93 times one-third of 𝑎. Now remember, 𝐵𝐺𝐶 needs to have negative mass but have the same kind of mass distribution as 𝐴𝐵𝐶. So the density of 𝐵𝐺𝐶 will be negative 𝑚.

Let’s rearrange our expression for the mass of 𝐵𝐺𝐶 by factoring out the negative one and the factor of one-third. Written this way, we see that the mass of 𝐵𝐺𝐶 is negative one-third times the mass of 𝐴𝐵𝐶. The factor of negative one-third is present because 𝐴𝐵𝐶 has three times the area of 𝐵𝐺𝐶, and the mass of 𝐵𝐺𝐶 is negative. Just to make our life easier and clean up some space, let’s call one-half times 93 times 𝑎 times lowercase 𝑚 capital 𝑀 so that the mass of 𝐴𝐵𝐶 is capital 𝑀 and the mass of 𝐵𝐺𝐶 is negative one-third times capital 𝑀.

Now that we know the masses, let’s find the centers of mass of 𝐴𝐵𝐶 and 𝐵𝐺𝐶. We know that the center of mass of 𝐴𝐵𝐶 is located at the point 𝐺. And we’ve already determined the 𝑦-coordinate of point 𝐺; it’s one-third of 𝑎. 𝐺 lies along the altitude of 𝐴𝐵𝐶. And we know that this altitude also being a median intersects the 𝑥-axis halfway between 𝐵 and 𝐶. Well, since the length of 𝐵𝐶 is 93, halfway between them is 46.5. So the coordinates of point 𝐺 are 46.5 comma one-third 𝑎. For triangle 𝐵𝐺𝐶, note that 𝐵𝐺𝐶 is an isosceles triangle. We can see this because the dashed line is the perpendicular bisector of 𝐵𝐶. So any point on this bisector including the point 𝐺 is equidistant from point 𝐵 and point 𝐶. So the sides 𝐵𝐺 and 𝐺𝐶 are congruent, and therefore 𝐵𝐺𝐶 is isosceles.

In an isosceles triangle, the altitude drawn from the vertex in between the two congruent sides is also a median, just like in the equilateral triangle. But that means that the altitude of 𝐵𝐺𝐶, which is the dotted line that we’ve already drawn, is also a median. Since the center of mass is exactly one-third of the way along the median, the center of mass of 𝐵𝐺𝐶 is right around here. Although we’ve only estimated the location of this point in the figure, we can still find its location exactly. It’s on the same vertical line as 𝐺, so its 𝑥-coordinate is also 46.5. The 𝑦-coordinate is one-third of the height of 𝐵𝐺𝐶, and the height of 𝐵𝐺𝐶 is one-third of 𝑎. So the 𝑦-coordinate is one-third of one-third or one-ninth of 𝑎.

Now we have the mass and center of mass of 𝐵𝐺𝐶 and 𝐴𝐵𝐶. To use this information to find what we’re looking for, we can use the center of mass formula to find each coordinate of the resulting center of mass of the combination of 𝐴𝐵𝐶 and 𝐵𝐺𝐶. For each coordinate, we plug in the mass and corresponding coordinate of the center of mass for each of our laminae into the numerator and the mass of each lamina into the denominator. Let’s start with the 𝑦-coordinate. The mass of 𝐴𝐵𝐶 is 𝑀, and the 𝑦-coordinate of the center of mass is one-third of 𝑎. So the first term of the sum of our numerator is 𝑀 times one-third 𝑎. The second term in the numerator is negative one-third 𝑀 times one-ninth 𝑎, which is the mass of 𝐵𝐺𝐶 times the 𝑦-coordinate of its center of mass.

The denominator is the sum of the two masses, 𝑀 plus negative one-third 𝑀. To start simplifying this fraction, let’s note that both terms in the numerator have an 𝑀, a one-third, and an 𝑎. So we’ll factor one-third times 𝑀 times 𝑎 out of both of these terms. In the denominator, 𝑀 plus negative one-third 𝑀 is just two-thirds 𝑀. So we have one-third times 𝑀 times 𝑎 times one minus one-ninth all divided by two-thirds 𝑀. 𝑀 divided by 𝑀 is one, and one-third in the numerator divided by two-thirds in the denominator leaves just two in the denominator. In the numerator, one minus one-ninth is eight-ninths. So we have 𝑎 times eight-ninths divided by two.

Note that the actual value of the mass of 𝐴𝐵𝐶 does not factor into this calculation because the only thing that’s important is the relative mass of 𝐴𝐵𝐶 and 𝐵𝐺𝐶 not their absolute masses. Eight-ninths divided by two is four-ninths. So 𝑎 times eight-ninths divided by two is four-ninths times 𝑎. Plugging back in our definition for 𝑎, we get that the 𝑦-coordinate of the center of mass of our new lamina is four-ninths times 93 times the square root of three divided by two. If we plug this into a calculator, we get 35.7957 and several more decimal places. Since we’re asked to round to two decimal places, the third decimal place is a five. So we add one to the second decimal place, which is nine. One plus nine is 10, so 35.7957 et cetera rounds to 35.80.

To find the 𝑥-coordinate of the center of mass, instead of going through this whole calculation again, we note that the 𝑥-coordinate of the center of mass for 𝐴𝐵𝐶 and 𝐵𝐺𝐶 is the same. This means that when we combine them, we won’t change the 𝑥-coordinate of the center of mass. So the 𝑥-coordinate of the center of mass of our resulting lamina will also be 46.5. So rounding as specified, we find that the center of mass of our new lamina is at 46.5 comma 35.80.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.