To find the coordinates of the center of mass of this system of four masses, we will use the following formula:

$x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i} \quad \text{and} \quad y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i}$

Where:

• $m_i$ represents the mass of the $i$-th object,
• $x_i, y_i$ represent the coordinates of the $i$-th mass,
• $x_{\text{cm}}, y_{\text{cm}}$ are the coordinates of the center of mass.

Given:

• Mass $A$ is at the origin, so its coordinates are $(0, 0)$,
• Mass $B$ is at $(0, l)$,
• Mass $C$ is at $(l, l)$,
• Mass $D$ is at $(l, 0)$,
• $l = 43.3 \, \text{cm} = 0.433 \, \text{m}$,
• The masses are:
  • $m_A = 467.0 \, \text{g}$,
  • $m_B = 362.0 \, \text{g}$,
  • $m_C = 349.0 \, \text{g}$,
  • $m_D = 341.0 \, \text{g}$.

Step 1: Calculate the total mass

$M_{\text{total}} = m_A + m_B + m_C + m_D = 467.0 + 362.0 + 349.0 + 341.0 = 1519.0 \, \text{g}$

Step 2: Find the center of mass coordinates

$x_{\text{cm}}$:

The $x$-coordinates of the masses are:

• $A$ at $x_A = 0$,
• $B$ at $x_B = 0$,
• $C$ at $x_C = l = 0.433 \, \text{m}$,
• $D$ at $x_D = l = 0.433 \, \text{m}$.

Now calculate $x_{\text{cm}}$: $x_{\text{cm}} = \frac{(m_A \cdot x_A) + (m_B \cdot x_B) + (m_C \cdot x_C) + (m_D \cdot x_D)}{M_{\text{total}}}$ $x_{\text{cm}} = \frac{(467.0 \cdot 0) + (362.0 \cdot 0) + (349.0 \cdot 0.433) + (341.0 \cdot 0.433)}{1519.0}$ $x_{\text{cm}} = \frac{(349.0 \cdot 0.433) + (341.0 \cdot 0.433)}{1519.0}$ $x_{\text{cm}} = \frac{151.217 + 147.353}{1519.0} = \frac{298.57}{1519.0} = 0.1965 \, \text{m}$

$y_{\text{cm}}$:

The $y$-coordinates of the masses are:

• $A$ at $y_A = 0$,
• $B$ at $y_B = l = 0.433 \, \text{m}$,
• $C$ at $y_C = l = 0.433 \, \text{m}$,
• $D$ at $y_D = 0$.

Now calculate $y_{\text{cm}}$: $y_{\text{cm}} = \frac{(m_A \cdot y_A) + (m_B \cdot y_B) + (m_C \cdot y_C) + (m_D \cdot y_D)}{M_{\text{total}}}$ $y_{\text{cm}} = \frac{(467.0 \cdot 0) + (362.0 \cdot 0.433) + (349.0 \cdot 0.433) + (341.0 \cdot 0)}{1519.0}$ $y_{\text{cm}} = \frac{(362.0 \cdot 0.433) + (349.0 \cdot 0.433)}{1519.0}$ $y_{\text{cm}} = \frac{156.646 + 151.217}{1519.0} = \frac{307.863}{1519.0} = 0.2026 \, \text{m}$

Final Answer:

The coordinates of the center of mass are approximately: $(x_{\text{cm}}, y_{\text{cm}}) = (0.1965 \, \text{m}, 0.2026 \, \text{m})$

Would you like further details or have any questions?

Here are five related questions:

1. How does changing the mass at point A affect the center of mass?
2. What is the significance of placing the origin at point A?
3. How would the center of mass change if the rods were not massless?
4. How can you apply the center of mass concept to a continuous object instead of discrete masses?
5. How does the position of the center of mass affect the stability of a system?

Tip: When calculating center of mass, always ensure you correctly assign the coordinates relative to a chosen origin, as this affects your results.