Question

The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point ($p$) of normal bisector, as shown in the figure is _________

• A. $(\frac{\sqrt{3}}{4}, 1.25)$
• B. $(\frac{\sqrt{3}}{4}, 1.0)$
• C. (0,0)
• D. (1.25, 0)

Hint:

Define a coordinate system with point $p$ as origin. Calculate the altitude of the triangle to find the y-coordinates of the masses, and use the side lengths to find the x-coordinates.

Answer 1

The Correct Option is A

We can also determine the center of mass by first calculating its position relative to the base and then shifting the origin to point $p$.

Assume the midpoint of the base is the origin $M(0,0)$.
The base masses are at positions:
$m_1=2$ kg at $(-5\sqrt{3}, 0)$
$m_2=3$ kg at $(5\sqrt{3}, 0)$
The height of the triangle is $10 \times \cos(60^\circ) = 5$ m. So the top mass is at:
$m_3=15$ kg at $(0, 5)$

Calculate CM relative to $M$:

$$X_{cm,M} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$$

$$Y_{cm,M} = \frac{2(0) + 3(0) + 15(5)}{20} = \frac{75}{20} = 3.75$$

Point $p$ is the midpoint of the altitude from $(0,0)$ to $(0,5)$, so its coordinates relative to $M$ are $(0, 2.5)$.

Now, find the CM position relative to point $p$ by subtracting $p$'s coordinates from the system's CM:

$$X_{rel} = X_{cm,M} - X_p = \frac{\sqrt{3}}{4} - 0 = \frac{\sqrt{3}}{4}$$

$$Y_{rel} = Y_{cm,M} - Y_p = 3.75 - 2.5 = 1.25$$

The final result is $(\frac{\sqrt{3}}{4}, 1.25)$, matching option 1.