Animated Solution for Physics - System of Particles: Three particles of masses $50\text{ g}$, $100\text{ g}$ and $150\text{ g}$ are placed at the vertices of an equilateral triangle of side $1\text{ m}$ (as shown in the figure). The $(x,y)$ coordinates of the centre of mass will be

Imagine three distinct masses resting at the corners of a perfect equilateral triangle. To find the "balance point" or the Center of Mass (CM) of this system, we first need to establish a coordinate system. Let's place the first mass, $m_1=50\text{ g}$, right at the origin $(0,0)$.

Since the triangle has a side length of $1\text{ m}$, the second mass, $m_2=100\text{ g}$, sits comfortably on the X-axis at $(1,0)$.

But what about the third mass, $m_3=150\text{ g}$, perched at the top vertex? Because it's an equilateral triangle, the top vertex lies exactly halfway along the base horizontally. So, its x-coordinate is $0.5\text{ m}$. To find its y-coordinate, we calculate the height of the triangle using the Pythagorean theorem:

So, the coordinates of $m_3$ are $\left(0.5,\frac{\sqrt{3}}{2}\right)$.

The center of mass for a system of discrete particles is the weighted average of their positions. The formula for the x-coordinate is:

Let's plug in our known values. The total mass of the system is $50+100+150=300\text{ g}$.

Dividing the numerator and denominator by $25$, we get:

We apply the exact same logic for the y-coordinate:

Notice that $m_1$ and $m_2$ both lie flat on the X-axis, meaning their y-coordinates are zero. This makes our calculation much simpler!

Dividing the numerator and denominator by $75$, we arrive at:

Combining our results, the exact coordinates of the center of mass for this system are $\left(\frac{7}{12}\text{ m},\frac{\sqrt{3}}{4}\text{ m}\right)$. Because the masses are unequal, the center of mass is pulled away from the geometric centroid, leaning closer to the heavier $150\text{ g}$ mass at the top.