Question Video: Finding the Center of Gravity of Three Equal Discrete Masses Placed on the Sides of a Triangle Mathematics • Third Year of Secondary School

Video Transcript

A triangle 𝐴𝐵𝐶, where 𝐴𝐵 equals 33 centimeters, 𝐵𝐶 equals 44 centimeters, 𝐶𝐴 equals 55 centimeters, and 𝐷 and 𝐸 are on the midpoints of line segment 𝐴𝐵 and line segment 𝐴𝐶, respectively, is located in the first quadrant of a Cartesian plane such that 𝐵 is at the origin and point 𝐶 is on the 𝑥-axis. Three equal masses are placed at points 𝐵, 𝐷, and 𝐸. Determine the coordinates of the center of gravity of the system given that the scales of the axes are defined such that each unit represents one centimeter of distance.

In the triangle 𝐴𝐵𝐶 below we can see that there are three masses labeled with the letter 𝑚. In order to work out the center of gravity or center of mass of this system, we’ll need to find the coordinates or position vectors of each of these three masses. We can use the given lengths of the line segments to help us along with the fact that point 𝐵 is at the origin. It will have the coordinates zero, zero. And we can also give this as a position vector. The position vector of point 𝐵 is zero 𝐢 plus zero 𝐣.

We are also given that 𝐴𝐵 is equal to 33 centimeters. And the point 𝐷, we are told, is at the midpoint Of line segment 𝐴𝐵. Point 𝐷 will therefore be at a distance of 33 over two centimeters away from 𝐵. We could write the coordinates of point 𝐷 as zero, 33 over two, which as a position vector is zero 𝐢 plus 33 over two 𝐣.

We also need to find the position vector of this mass at point 𝐸. This vector can be written as vector 𝚩𝐄, and it will be equal to vector 𝚩𝐂 plus a half of vector 𝐂𝚨. The vector 𝐂𝚨 is also equivalent to vector 𝐂𝚩 plus vector 𝚩𝚨. We can then substitute this into the equation. Then since vector 𝐂𝚩 is equal to negative vector 𝚩𝐂, the right-hand side simplifies to vector 𝚩𝐂 plus one-half times negative vector 𝚩𝐂 plus vector 𝚩𝚨.

Distributing one-half across the parentheses, the right-hand side of this equation becomes vector 𝚩𝐂 minus one-half vector 𝚩𝐂 plus a half vector 𝚩𝚨. This gives us that vector 𝚩𝐄 is equal to one-half of vector 𝚩𝐂 plus vector 𝚩𝚨. We can then work out both of these vectors 𝚩𝐂 and 𝚩𝚨 in terms of their horizontal and vertical components.

Remember, we were told that 𝚨𝚩 is 33 centimeters, and so point 𝐴 is at a position vector of zero 𝐢 plus 33𝐣. This will be the vector 𝚩𝚨. In the same way, 𝐵𝐶 is 44 centimeters, so vector 𝚩𝐂 can be given as 44𝐢 plus zero 𝐣. Now when we add 44𝐢 plus zero 𝐣 and zero 𝐢 plus 33𝐣, we get 44𝐢 plus 33𝐣. And so we have done all this working to find that this vector 𝚩𝐄 is equal to 44 over two 𝐢 plus 33 over two 𝐣. We can fill that onto the diagram and clear some space for the next part.

In order to find the center of gravity or center of mass of this system, we can apply this formula that vector 𝐑 is equal to one over 𝑚 times the sum from 𝑖 equals one to 𝑛 of 𝑚 sub 𝑖 times vector 𝐫 sub 𝑖, where 𝑚 is the total mass. 𝑚 sub 𝑖 is the mass of object 𝑖. And vector 𝐫 sub 𝑖 is the position vector of object 𝑖.

Notice that this is different than simply finding the geometric center of the masses at points 𝐵, 𝐷, and 𝐸. This is because we need to take account of the mass of each object. And so although this formula can look quite complicated, what we’re really doing is multiplying each mass by its position vector, adding up those products, and then multiplying by one over the total mass.

So let’s see how we can apply this formula in this context. The first thing we can calculate here is 𝑚, the total mass. And since we have three equal weights of 𝑚ℎ, then the total mass can be written as three 𝑚. The first mass times position vector we can take is that at point 𝐵, which will have a mass of 𝑚 multiplied by a position vector of zero 𝐢 plus zero 𝐣. We can then add a second mass of mass 𝑚 times zero 𝐢 plus 33 over two 𝐣. Finally, we add on the third mass times the position vector, which is 𝑚 times 44 over two 𝐢 plus 33 over two 𝐣.

Notice how when we simplify this by expanding the parentheses, the terms that have zero 𝐢 or zero 𝐣 will simply be equal to zero. Before we carry out the next stage of simplifying, which is distributing one over three 𝑚 across the parentheses, we might notice that each of the terms within the parentheses also has a variable 𝑚. So these will cancel.

And so one-third multiplied by 33 over two 𝐣 is 33 over six 𝐣 or 11 over two 𝐣. We then add one-third times 44 over two 𝐢, which is 44 over six 𝐢 or 22 over three 𝐢. Finally, we have one-third times 33 over two 𝐣, which is 33 over six simplifying to 11 over two 𝐣. We can write this as 22 over three 𝐢 plus 11𝐣. And so we have find the position vector of the center of gravity, but we need to give this as a set of coordinates. And so we can give the answer that the center of gravity of this system of masses is 22 over three, 11.

Finally, it’s always worth checking if the answer seems sensible. And if we’ve already drawn a diagram, this can be an easy check to make. If we draw on approximately where the coordinates 22 over three, 11 would be, they would be here. And this does seem like a sensible position for the center of mass. A point far outside the system of masses, particularly when all the masses are equal, is unlikely to be correct. So here we can simply give the answer as the coordinates 22 over three, 11.