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The problem in the image involves finding the coordinates where you should place a 1.49 kg particle so that the center of mass of the three-particle system is located at \((-0.549\, \text{m}, -0.360\, \text{m})\).

The key to solving this problem is using the formula for the center of mass \((x_{\text{cm}}, y_{\text{cm}})\) of a system of particles:

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

Here are the known values:
- First particle: mass = 4.64 kg, coordinates = \((-1.76\, \text{m}, 0.373\, \text{m})\)
- Second particle: mass = 4.03 kg, coordinates = \((0.543\, \text{m}, -0.929\, \text{m})\)
- Third particle: mass = 1.49 kg, coordinates = \((x_3, y_3)\) (which we need to find)
- Desired center of mass: \((-0.549\, \text{m}, -0.360\, \text{m})\)

### Step 1: Solve for \(x_3\)
Using the center of mass equation for the x-coordinate:

\[
-0.549 = \frac{(4.64)(-1.76) + (4.03)(0.543) + (1.49)(x_3)}{4.64 + 4.03 + 1.49}
\]

Simplifying:

\[
-0.549 = \frac{-8.1664 + 2.18729 + 1.49 x_3}{10.16}
\]

\[
-0.549 \times 10.16 = -5.97911 + 1.49 x_3
\]

\[
-5.57484 = -5.97911 + 1.49 x_3
\]

\[
1.49 x_3 = 0.40427
\]

\[
x_3 = \frac{0.40427}{1.49} \approx 0.271 \, \text{m}
\]

### Step 2: Solve for \(y_3\)
Now, use the center of mass equation for the y-coordinate:

\[
-0.360 = \frac{(4.64)(0.373) + (4.03)(-0.929) + (1.49)(y_3)}{10.16}
\]

Simplifying:

\[
-0.360 = \frac{1.73072 - 3.74587 + 1.49 y_3}{10.16}
\]

\[
-0.360 \times 10.16 = -2.01515 + 1.49 y_3
\]

\[
-3.6576 = -2.01515 + 1.49 y_3
\]

\[
1.49 y_3 = -1.64245
\]

\[
y_3 = \frac{-1.64245}{1.49} \approx -1.103 \, \text{m}
\]

### Final Answer:
- \(x_3 = 0.271\, \text{m}\)
- \(y_3 = -1.103\, \text{m}\)

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. How do you calculate the center of mass for a continuous object, such as a rod or plate?
2. What happens to the center of mass when the third particle's mass is much larger than the others?
3. How does the center of mass relate to rotational motion?
4. Can the center of mass be located outside the physical body of a system?
5. How would the center of mass change if the plane were inclined instead of horizontal?

**Tip**: Always ensure the units are consistent when calculating physical quantities like center of mass.

A 4.64 kg particle has the xy coordinates (-1.76 m, 0.373 m), and a 4.03 kg particle has the xy coordinates (0.543 m, -0.929 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 1.49 kg particle such that the center of mass of the three-particle system has the coordinates (-0.549 m, -0.360 m)?

Learn how to calculate the coordinates where a 1.49 kg particle must be placed such that the center of mass of a three-particle system, with given masses and coordinates, lies at (-0.549 m, -0.360 m). This solution involves key concepts such as the center of mass and weighted averages, and uses the center of mass formula.

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