# Question Video: Determining the Location of the Center of Mass of a Multibody System

Three point masses are placed at the corners of a triangle as shown in the accompanying figure. The origin of the system is defined to be at the position of the 150 g mass, with displacement to the right of the origin corresponding to positive π₯ values and displacement above the origin corresponding to positive π¦ values. Find the π₯ coordinate of the center of mass system. Find the π¦ coordinate of the center of mass system.

06:42

### Video Transcript

Three point masses are placed at the corners of a triangle as shown in the accompanying figure. The origin of the system is defined to be at the position of the 150 gram mass, with displacement to the right of the origin corresponding to positive π₯-values and displacement above the origin corresponding to positive π¦-values. Find the π₯-coordinate of the center of mass system. Find the π¦-coordinate of the center of mass system.

Letβs call the center of mass π₯-coordinate π₯ sub cm and the center of mass π¦-coordinate π¦ sub cm. With reference to the diagram, weβre told that movement to the right and up are defined as motion in the positive directions. With that as background, we now want to find the π₯ and π¦-coordinates of the center of mass of these three mass system.

The center of mass for a given coordinate is equal to the sum of that coordinate times the mass related to it divided by the sum of all the masses. This equation may look complicated and strange, but as we apply it to our scenario weβll see that it makes sense. Letβs focus first on the π₯, center of mass coordinate.

Our center of mass equation tells us that we find this coordinate by creating a fraction where the numerator of that fraction is the sum of the products of all the masses we have multiplied by their π₯-coordinate. And the denominator of that fraction is the sum of all the masses without their coordinate added in.

Since we have three masses, this equation will look like this: π₯ one π one plus π₯ two π two plus π₯ three π three, all divided by the sum of the three masses. Now weβve numbered our masses one, two and, three. Letβs assign those numbers to particular corners of the triangle on our diagram. We can assign them anyway we want. The π₯-center of mass coordinate will come out the same way. But letβs choose our 100 gram mass to be mass one, our 150 gram mass to be mass two, and our 75 gram mass to be mass three.

Now that weβve defined our masses, letβs move ahead and solve in for the π₯-center of mass coordinate by figuring out what are π₯ one, π₯ two, and π₯ three. Looking again at our diagram, weβre told in the problem statement that the location of π two, the 150 gram mass, is our origin, meaning that is where our coordinates are zero, zero. The location of mass one relative to that origin is 4.0 centimeters to the left.

Since weβve defined positive motion to be motion to the right and up, the π₯-coordinate of mass one, also referred to as π₯ one, is negative 4.0 centimeters. Moving onto mass two, we know the π₯-coordinate of this mass is zero, because it sits at the origin. So we can write π₯ two equals zero centimeters. π three, our third mass, is located directly above π two, which has an π₯ coordinate of zero. Therefore, mass three also has an π₯-coordinate of zero or π₯ three equals zero.

Weβre now ready to plug in to our equation for the π₯-center of mass coordinate. We know π one, π two, π three and π₯ one, π₯ two, and π₯ three. Letβs plug those values in now. We see you that negative 4.0 centimeters times 100 grams, all divided by the sum of our three masses, is equal to the π₯-center of mass coordinate. When we calculate these numbers out, we find that that value is negative 1.2 centimeters. That is the π₯-coordinate of the center of mass of these three masses.

Now letβs turn our attention to the π¦-coordinate of the center of mass. Weβll use the same process and equation for π¦ as we did for π₯. Indeed, looking at the center of mass equation weβve written, that variable π₯ could be any variable. Itβs just the name of the position coordinate we use. So we can write out an equation for the center of mass π¦-coordinate with respect to our three masses.

That center of mass coordinate equals π¦ one π one plus π¦ two π two plus π¦ three π three, all divided by the sum of our three masses. We already know the values for π one, π two, and π three, but letβs solve now for π¦ one, π¦ two, and π¦ three: the π¦-values of our three masses.

Looking at mass one on our diagram, we see that itβs on the same horizontal line as the origin, which means that the π¦-value of π one is zero; likewise, for π two which sits at the origin where π¦ is zero, For π three, the 75 gram mass, we know that the π¦-value is not zero but is given as 3.0 centimeters.

Since weβve defined displacement above the origin to be motion in the positive direction, π¦ sub three is equal to positive 3.0 centimeters. Weβre now ready to plug in our π¦ and π-values into the equation for the center of mass π¦-coordinate.

When we enter these values in, we see that the π¦-coordinate of the center of mass equals 3.0 centimeters times 75 grams, all divided by the sum of the three masses. Entering these values into our calculator, we find that the center of mass π¦-coordinate is 0.69 centimeters. That is the center of mass of this system of masses in the π¦-direction.