Math Problem Statement

Three masses of 5 kg, 9 kg, and 12 kg are placed at (-4, 3), (2, 2) , and (5, 0) respectively. Find the center of mass.

Solution

To find the center of mass of the system of three masses, we use the formula for the coordinates (xcm,ycm)(x_{cm}, y_{cm}) of the center of mass:

xcm=(mixi)mix_{cm} = \frac{\sum (m_i x_i)}{\sum m_i} ycm=(miyi)miy_{cm} = \frac{\sum (m_i y_i)}{\sum m_i}

where mim_i is the mass and (xi,yi)(x_i, y_i) are the coordinates of the masses.

Given:

  • Mass m1=5kgm_1 = 5 \, \text{kg} at coordinates (4,3)(-4, 3)
  • Mass m2=9kgm_2 = 9 \, \text{kg} at coordinates (2,2)(2, 2)
  • Mass m3=12kgm_3 = 12 \, \text{kg} at coordinates (5,0)(5, 0)

Step 1: Find xcmx_{cm}

xcm=(5)(4)+(9)(2)+(12)(5)5+9+12x_{cm} = \frac{(5)(-4) + (9)(2) + (12)(5)}{5 + 9 + 12} xcm=20+18+6026x_{cm} = \frac{-20 + 18 + 60}{26} xcm=58262.23x_{cm} = \frac{58}{26} \approx 2.23

Step 2: Find ycmy_{cm}

ycm=(5)(3)+(9)(2)+(12)(0)5+9+12y_{cm} = \frac{(5)(3) + (9)(2) + (12)(0)}{5 + 9 + 12} ycm=15+18+026y_{cm} = \frac{15 + 18 + 0}{26} ycm=33261.27y_{cm} = \frac{33}{26} \approx 1.27

Thus, the center of mass is approximately at (2.23,1.27)(2.23, 1.27).

Would you like more details or have any questions?

Related Questions:

  1. How does changing the mass at a specific point affect the center of mass?
  2. What if the mass distribution were in 3D? How would we calculate the center of mass?
  3. How do we calculate the center of mass for continuous distributions?
  4. What happens if one mass is at the origin?
  5. Can the center of mass lie outside the system of particles?

Tip:

The center of mass is a balance point, so it can be useful in analyzing the motion of the entire system in physics problems.

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Math Problem Analysis

Mathematical Concepts

Physics
Center of Mass
Weighted Averages
Coordinate Geometry

Formulas

x_cm = (Σ(m_i * x_i)) / Σ(m_i)
y_cm = (Σ(m_i * y_i)) / Σ(m_i)

Theorems

Center of Mass Formula

Suitable Grade Level

Grades 10-12