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Physics Lesson 6.2.4 - Centre of Mass of Three non-Collinear Objects
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Welcome to our Physics lesson on Centre of Mass of Three non-Collinear Objects, this is the fourth lesson of our suite of physics lessons covering the topic of Determining the Centre of Mass in Objects and Systems of Objects, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Centre of Mass of Three non-Collinear Objects
In this case, we must consider a triangle in which objects' centres of gravity act as vertices as shown in the figure. 
Again, we think the sides of this triangle as very light sticks with negligible weight. They are drawn in the figure only for calculation purpose. Also, we can think the inner part of the triangle as made by a very thin and light layer, such as a transparent vinyl sheet or anything similar.
It is obvious that the centre of gravity of the system is somewhere inside the internal area of triangle. It acts as an equilibrium point of the system, i.e. if we put the system on a pivot, which is exactly below the centre of mass, it stays in equilibrium as shown below. 
We must consider two dimensions in order to work out the centre of mass if the system. Therefore, the equation (1) is applied for both x and y coordinates. This means the centre of mass of the system is written in the format (xC, yC). Hence, we write
and
A comprehensive approach in this regard involves dealing with some mathematical concepts such as the equation of a line that passes through two known points, the angles of a triangle when the side lengths are known, etc. These concepts are not difficult; a student learns all them in high school math. However, dealing with such concepts goes beyond the scope of this tutorial. Here we will discuss only situations involving objects whose centre of mass have known coordinates, so that we can apply directly the equation (1) for two dimensions.
Let's illustrate this point through an example.
Example 4
Three objects are connected with very light sticks to form a single system. The first object has a mass of 4 kg and it is at (0m, 3m). The second object has a mass of 1 kg and it is at (-1m, 0m), and the third object has a mass of 2 kg and it is at (5m, 4m). Where is the centre of mass of the system?
Solution 4
In this example, it is better to draw a coordinate system and show all objects visually.
Now, let's write the clues. We have:
m2 = 1 kg
m3 = 2 kg
x1 = 0 m
x2 = -1 m
x3 = 5 m
y1 = 3 m
y2 = 0 m
y3 = 4 m
xC = ?
yC = ?
Applying the equation (1) for the x and y-direction separately, we obtain
= 0 × 4 + (-1) × 1 + 5 × 2/4 + 1 + 2
= -1 + 10/7
= 9/7 m
≈ 1.29 m
and
= 3 × 4 + 0 × 1 + 4 × 2/4 + 1 + 2
= 12 + 8/7
= 20/7
≈ 2.86 m
Therefore, the centre of gravity of the system is at (1.29 m, 2.86 m). Look at the figure:
Even for three dimensions, we use the same method. The only difference is that we have another dimension, the z-dimension. Therefore, we must apply three times the equation (1).
You have reached the end of Physics lesson 6.2.4 Centre of Mass of Three non-Collinear Objects. There are 4 lessons in this physics tutorial covering Determining the Centre of Mass in Objects and Systems of Objects, you can access all the lessons from this tutorial below.
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