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To solve this problem, we need to determine the \( x \)- and \( y \)-coordinates at which to place the third particle so that the center of mass of the three-particle system is at \((-0.750 \, \text{m}, -0.0807 \, \text{m})\).

### Step-by-Step Solution

The center of mass (COM) coordinates \((x_{\text{com}}, y_{\text{com}})\) for a system of particles are given by:

\[
x_{\text{com}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}
\]
\[
y_{\text{com}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}
\]

Where:
- \( m_1 = 2.63 \, \text{kg} \), \( x_1 = -1.67 \, \text{m} \), \( y_1 = 0.0713 \, \text{m} \)
- \( m_2 = 5.44 \, \text{kg} \), \( x_2 = 0.0117 \, \text{m} \), \( y_2 = -0.489 \, \text{m} \)
- \( m_3 = 4.66 \, \text{kg} \), and we need to find \( x_3 \) and \( y_3 \)

The center of mass is given as:
\[
x_{\text{com}} = -0.750 \, \text{m}
\]
\[
y_{\text{com}} = -0.0807 \, \text{m}
\]

### Finding \( x_3 \)

Using the equation for the \( x \)-coordinate of the center of mass:
\[
-0.750 = \frac{(2.63)(-1.67) + (5.44)(0.0117) + (4.66)(x_3)}{2.63 + 5.44 + 4.66}
\]
Simplifying the denominator:
\[
2.63 + 5.44 + 4.66 = 12.73
\]
Now, simplifying the numerator:
\[
(2.63)(-1.67) = -4.3921
\]
\[
(5.44)(0.0117) = 0.063648
\]
The equation becomes:
\[
-0.750 = \frac{-4.3921 + 0.063648 + 4.66x_3}{12.73}
\]
Multiply both sides by 12.73:
\[
-9.5475 = -4.3921 + 0.063648 + 4.66x_3
\]
Combine terms:
\[
-9.5475 = -4.328452 + 4.66x_3
\]
Solve for \( x_3 \):
\[
4.66x_3 = -9.5475 + 4.328452
\]
\[
4.66x_3 = -5.219048
\]
\[
x_3 = \frac{-5.219048}{4.66} \approx -1.12 \, \text{m}
\]

### Finding \( y_3 \)

Now, using the equation for the \( y \)-coordinate of the center of mass:
\[
-0.0807 = \frac{(2.63)(0.0713) + (5.44)(-0.489) + (4.66)(y_3)}{12.73}
\]
Simplify the numerator:
\[
(2.63)(0.0713) = 0.187519
\]
\[
(5.44)(-0.489) = -2.65968
\]
The equation becomes:
\[
-0.0807 = \frac{0.187519 - 2.65968 + 4.66y_3}{12.73}
\]
Multiply both sides by 12.73:
\[
-1.025471 = 0.187519 - 2.65968 + 4.66y_3
\]
Combine terms:
\[
-1.025471 = -2.472161 + 4.66y_3
\]
Solve for \( y_3 \):
\[
4.66y_3 = -1.025471 + 2.472161
\]
\[
4.66y_3 = 1.44669
\]
\[
y_3 = \frac{1.44669}{4.66} \approx 0.31 \, \text{m}
\]

### Final Answer
- \( x_3 \approx -1.12 \, \text{m} \)
- \( y_3 \approx 0.31 \, \text{m} \)

Would you like more details or have any questions?

Here are some related questions for deeper understanding:
1. How is the center of mass affected by the distribution of mass in the system?
2. How would the solution change if the mass of one of the particles increased?
3. Can the center of mass of a system be located outside the object itself?
4. How do we calculate the center of mass in three-dimensional systems?
5. What is the physical significance of the center of mass in mechanical systems?

**Tip:** Always ensure units are consistent (e.g., meters, kilograms) when calculating the center of mass.

A 2.63 kg particle has the xy coordinates (-1.67 m, 0.0713 m), and a 5.44 kg particle has the xy coordinates (0.0117 m, -0.489 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 4.66 kg particle such that the center of mass of the three-particle system has the coordinates (-0.750 m, -0.0807 m)?

Solve for the coordinates where a 4.66 kg particle must be placed to ensure the center of mass of the system lies at (-0.750 m, -0.0807 m). This problem involves calculating the center of mass using the weighted average of the mass and position of particles.

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