The figure shows three weights arranged in an equilateral triangle of side length 12 centimeters. Find the coordinates of the center of gravity of the system.
Knowing that the side lengths of each of the legs of the triangle is 12 centimeters, we can sketch those distances in on our diagram. We want to solve for the coordinates of the center of gravity of the system. We’ll call these coordinates 𝑐𝑔 sub 𝑥 and 𝑐𝑔 sub 𝑦, respectively. We’ve been asked to solve for the coordinates of the center of gravity of this shape. We’ll assume a uniform gravitational field such that the center of gravity is equal to the center of mass.
Working under this assumption, we can recall the relationship for the center of mass of an object. If an object is made up of a collection of discrete masses, then if we add up the product of each one of those masses with the distance of that mass element from the axis of rotation and then divide that sum by the sum of all the masses of our system, then that fraction is equal to the center of mass. The center of mass is something we’ll calculate for each dimension in our problem. In our case, we have an 𝑥- and a 𝑦-dimension. So we’ll calculate a center of mass for each one of those two.
We can begin by calculating the center of gravity or center of mass in the 𝑥-direction. Looking at our diagram, we see that there are three total masses that make up our object, applying the mathematical relationship for calculating center of mass along a certain dimension. When we calculate the center of mass in the 𝑥-direction, the axis of rotation we assume is the 𝑦-axis. Therefore, each of the distances we write down, 𝑥 sub 𝑖 in our center of mass equation, is the distance of a given mass from that 𝑦-axis. This means that for our first mass, a mass with a value of eight, we have a distance of zero since that mass sits on the 𝑦-axis. For the second mass, the mass of value nine, we have a distance of 12 since that’s the length of each leg of our equilateral triangle. And since the triangle is equilateral, that means the distance from the axis of rotation that our third mass exists is equal to 12 times the cosine of 60 degrees.
We’ve now accounted for the numerator of our center of mass formula. We’ve added up the products of each of our mass elements with the distance that mass element is from the axis of rotation. In the denominator of our fraction, we simply add up each mass by itself, eight plus nine plus 13. Calculating this value, eight times zero is zero. Nine times 12 is 108. And 13 times 12 times the cosine of 60 degrees is equal to 78. So all told, our numerator simplifies to 186. The numbers in our denominator add to 30. And when we reduce this fraction, we find it’s equal to 31 over five. That’s the center of gravity of our shape in the 𝑥-dimension.
Now we’ll move on to solving for the center of gravity in the 𝑦-direction. In this case, our axis of rotation will be the 𝑥-axis. The distance from each elemental mass from that axis is what we’ll use in our center of mass calculation. When we write out the distance of our first mass from the axis of rotation, we see that that distance is zero. And likewise with our second mass of value nine, the distance of that mass from our axis of rotation, the 𝑥-axis, is also zero. Our third mass has a distance of 12 times the sine of 60 degrees from the axis of rotation. And in our denominator, we again have the sum of the three elemental masses. Since the sine of 60 degrees is the square root of three over two, our overall fraction simplified to 156 root three over 60. This fraction itself further simplifies down to 13 times the square root of three divided by five. This then is the overall coordinate of our center of gravity in the 𝑦-dimension.
So our overall center of gravity is at 31 over five in the 𝑥-direction and 13 root three over five in the 𝑦.