Question Video: Finding Unknown Masses from a Group of Three Discrete Masses on the Same Axis given the Coordinates of Their Centre of Mass Mathematics

Suppose three masses of 6 kg, 9 kg, and π‘š kg are at points (5, 9), (0, 6), and (βˆ’4, 3) respectively. If the center of mass of the three masses is (1, 𝑦), what are the values of π‘š and 𝑦?

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Video Transcript

Suppose three masses of six kilograms, nine kilograms, and π‘š kilograms are at points five, nine; zero, six; and negative four, three, respectively. If the center of mass of the three masses is one, 𝑦, what are the values of π‘š and 𝑦?

If we wish, we can begin this question with a coordinate grid representing the three different masses at the given coordinates. With the circles representing the masses, here we have a mass of six kilograms at the point five, nine; a mass of nine kilograms at the point zero, six; and a mass of π‘š kilograms at negative four, three. We are also given that the coordinates of the center of mass of these three masses is one, 𝑦. We need to work out both the unknown mass of π‘š kilograms and the value of 𝑦.

The center of mass of a set of masses isn’t simply just the geometric center. The center of mass needs to take into account the mass of the system. And there is a formula which we can use to help us. This formula tells us that the position vector of the center of mass, represented as vector 𝐑, is equal to one over 𝑀 times the sum from 𝑖 equals one to 𝑛 of π‘š sub 𝑖 times vector 𝐫 sub 𝑖. 𝑀 represents the total mass of the system of masses. π‘š sub 𝑖 represents the mass of object 𝑖. And vector 𝐫 sub 𝑖 is the position vector of object 𝑖.

Although this formula can look complicated, what we’re really doing is multiplying every object’s mass by its position vector then adding up all these products and multiplying by one over the total mass. Because this formula uses position vectors and we have coordinates, then the first thing we’ll do is work out these four coordinates in terms of their position vectors.

Let’s take these in the order we were given with the coordinate five, nine. It can be written in terms of 𝐒 and 𝐣 as five 𝐒 plus nine 𝐣. Then the second coordinate of zero, six can be written as zero 𝐒 plus six 𝐣. Since zero 𝐒 is simply zero, then this position vector can be given as six 𝐣. The third coordinate of negative four, three can be written as negative four 𝐒 plus three 𝐣. We will also convert the coordinates one, 𝑦 into a position vector as one 𝐒 plus 𝑦𝐣. This is more simply written as just 𝐒 plus 𝑦𝐣.

So now we have enough information to substitute these values into the given formula, remembering that we already have the position vector of the center of mass. When we begin our substitution, 𝐒 plus 𝑦𝐣 will be on the left-hand side of the equation. On the right-hand side, we have one over six plus nine plus π‘š, remembering that that’s simply the total mass of the system. This is then multiplied by six times five 𝐒 plus nine 𝐣, which is the mass times the position vector of this first object. We add this to the mass times the position vector of the second mass, so we have nine times six 𝐣. Finally, we add this to the mass of the third object, which is π‘š times its position vector of negative four 𝐒 plus three 𝐣.

Once we have this, the next stage is to simplify on the right-hand side. And so on the right-hand side, we begin with the fraction of one over 15 plus π‘š. We have then distributed six across the parentheses, which gives us 30𝐒 plus 54𝐣 plus 54𝐣 minus four π‘šπ’ plus three π‘šπ£. The next stage of simplifying will really be like collecting like terms in terms of the horizontal and vertical components 𝐒 and 𝐣.

The completed equation is then 𝐒 plus 𝑦𝐣 equals one over 15 plus π‘š times 30 minus four π‘š 𝐒 plus 108 plus three π‘š 𝐣. Notice in this equation we still have two unknowns; we have the unknown of 𝑦 and the unknown of π‘š, which appears twice. What we will do next is to recognize that the horizontal and vertical components on each side of the equation must be equal. In order to help us do this, let’s distribute one over 15 plus π‘š across the parentheses.

When we have done this, on the right-hand side we have 30 minus four π‘š over 15 plus π‘š 𝐒 plus 108 plus three π‘š over 15 plus π‘š 𝐣. Now we can equate the coefficients of 𝐒 and the coefficients of 𝐣, remembering that on the left-hand side that 𝐒 will simply be the same as one 𝐒. We can clear some space above this working to equate the coefficients. We now have two equations: one is equal to 30 minus four π‘š over 15 plus π‘š and a second equation that 𝑦 is equal to 108 plus three π‘š over 15 plus π‘š.

Let’s see if we can solve equation one first. Multiplying both sides by 15 plus π‘š would give us 15 plus π‘š is equal to 30 minus four π‘š. Then, in either one step or two steps, we add four π‘š and subtract 15 from both sides, leaving us with five π‘š equals 15. Dividing both sides by five, we get that π‘š is equal to three. This is the first answer that we are required to find. And it means that the unknown mass of π‘š is three kilograms.

We can also use this to help us solve the second equation by substituting π‘š equals three into this equation. So we have 𝑦 is equal to 108 plus three times three over 15 plus three. Simplifying the numerator, we have 108 plus nine, which is 117. And on the denominator, we have 18. As a decimal, this means that 𝑦 is equal to 6.5. Let’s remember that the value of 𝑦 was the value in the coordinates for the center of mass. If we added this onto the diagram, the center of mass at one, 6.5 would appear approximately here.

If we have already drawn a diagram, then adding the center of mass to it can be a useful way to check the value is sensible. Since we also worked out that the unknown mass is three kilograms, then we might expect to see the center of mass slightly to the right of this nine-kilogram mass. And so this is a good visual confirmation that the center of mass is correct. And so we can give the answer that π‘š is equal to three and 𝑦 is equal to 6.5.

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