Question Video: Finding Unknown Discrete Masses given the Coordinates of Their Centre of Mass | Nagwa Question Video: Finding Unknown Discrete Masses given the Coordinates of Their Centre of Mass | Nagwa

Question Video: Finding Unknown Discrete Masses given the Coordinates of Their Centre of Mass Mathematics • Third Year of Secondary School

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Three points (0, 6), (0, 9), and (0, 4) on the 𝑦-axis are occupied by three solids of masses 9 kg, 6 kg, and π‘š kg, respectively. Determine the value of π‘š given the center of mass of the system is at the point (0, 7).

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Video Transcript

Three points zero, six; zero, nine; and zero, four on the 𝑦-axis are occupied by three solids of masses nine kilograms, six kilograms, and π‘š kilograms, respectively. Determine the value of π‘š given the center of mass of the system is at the point zero, seven.

The masses of the components of a system as well as their locations are precisely the pieces of information we need to calculate the center of mass of that system. For example, we calculate the 𝑦-coordinate of the center of mass by multiplying each mass by the 𝑦-coordinate of its location, adding all of those products together, and dividing by the total mass of the system. Here though, instead of being given the mass and location of every component and being asked to find the center of mass, we are instead given the center of mass and asked to find the mass of one of the components.

So because we know the 𝑦-coordinate of the center of mass, all we need to do is rearrange this formula to solve for the unknown mass. We’re focused on the 𝑦-coordinate of the center of mass because each mass in our system is located at a point with the same π‘₯-coordinate, which means that the actual value of π‘š doesn’t affect the π‘₯-coordinate of the center of mass. Okay, so let’s set up our equation. The 𝑦-coordinate of the center of mass is given as seven. The numerator of the fraction is the sum of each mass times the 𝑦-coordinate of its location. So we have nine times six plus six times nine plus π‘š times four. And the denominator is just the sum of all the masses, which is nine plus six plus π‘š.

Now we just need to solve for π‘š. Let’s start by simplifying the numerical terms in the fraction. Nine plus six is 15, and nine times six is equal to six times nine, which is equal to 54. Replacing 54 plus 54 with 108, we have seven equals 108 plus four π‘š divided by 15 plus π‘š. Now we multiply both sides by 15 plus π‘š. On the right-hand side, 15 plus π‘š divided by 15 plus π‘š is just one. And on the left-hand side, 15 plus π‘š times seven is 105 plus seven π‘š. So 105 plus seven π‘š equals 108 plus four π‘š. Subtracting 105 and four π‘š from both sides, on the left-hand side, we have 105 minus 105 is zero and seven π‘š minus four π‘š is three π‘š. On the right-hand side, 108 minus 105 is three and four π‘š minus four π‘š is zero. So three π‘š equals three and π‘š equals one. And this is the answer that we’re looking for.

It’s worth making a brief note about the difference between the number one and one kilogram. The correct answer is one and not one kilogram because the unit of kilogram is already given to us in the question. So when we determine the value of π‘š, what we’re determining is a number, not a number with units. In fact, whenever we answer questions that involve numbers with units, it’s always important to pay attention to how the question is worded to see if our answer is really supposed to be just a number or if it’s supposed to be a physical quantity that also includes units.

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