Video: Finding the Center of Mass of a System of Point Masses Placed on the Vertices of a Right Triangle

The figure shows a system of point masses placed at the vertices of a triangle. The mass placed at each point is detailed in the table. Determine the coordinates of the center of gravity of the system.

02:18

Video Transcript

The figure shows a system of point masses placed at the vertices of a triangle. The mass placed at each point is detailed in the table. Determine the coordinates of the centre of gravity of the system.

Now, because the shape we’ve been given isn’t symmetrical, we’re going to need to be a little bit careful about how we determine the centre of mass of our object. Now, if we’re given a system of 𝑛 particles of masses π‘š one, π‘š two, and so on up to π‘šπ‘›, whose position vectors are π‘Ÿ one, π‘Ÿ two all the way up to π‘Ÿ 𝑛, respectively. The centre of mass of the system is the point with position vector π‘Ÿ and is equal to the sum from 𝑖 equals one to 𝑛 over π‘š sub 𝑖 times π‘Ÿ sub 𝑖 all over capital 𝑀, where that’s the total mass of the system.

In reality though, it can be much easier, especially in two dimensions, to just fit this into the π‘₯-direction and the 𝑦-direction. The π‘₯-coordinate of the centre of mass of a system of 𝑛 particles π‘₯ sub 𝐢 is π‘š one π‘₯ one plus π‘š two π‘₯ two all the way up to π‘š 𝑛 π‘₯ 𝑛 over π‘š one plus π‘š two all the way up to π‘š 𝑛. And the 𝑦-coordinate of the centre of mass 𝑦 sub 𝐢 is π‘š one 𝑦 one plus π‘š two 𝑦 two all the way up to π‘š 𝑛 𝑦 𝑛 all over the total sum of the masses. Now here, π‘₯ one, π‘₯ two, and π‘₯ 𝑛, 𝑦 one, 𝑦 two, and 𝑦 𝑛 are the π‘₯- and 𝑦-coordinates, respectively, of each of the individual particles.

We’ll begin by dealing with the π‘₯-coordinate of the centre of mass. π‘š one π‘₯ one is the mass of 𝐴 times the distance that 𝐴 is from the origin. So that’s 13 times zero. 𝐡 is six times zero. Again, the horizontal distance from the origin is zero. And for 𝐢, it is 15 times six. This is all over the sum of their masses. On the top of our fraction, this simplifies to 90 and, on the bottom, we get 34. This simplifies to 47 over 17.

We’re going to repeat this process for the 𝑦-coordinate. π‘š one 𝑦 one is 13 times zero. Remember, 𝐴 is at the origin, so it’s zero units away in the vertical direction. This time, 𝐡 is eight centimetres away from the origin in the vertical direction, and 𝐢 is zero units away in the vertical direction. Once again, this is all over the sum of their masses. And this gives us 48 over 34. 48 over 34 simplifies to 24 over 17. And so we found the coordinates of the centre of mass or centre of gravity of our system. They are 47 17ths and 24 17ths.

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