Finding the Center of Mass for Stacked Cubes of Different Masses

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SUMMARY

The center of mass for three stacked cubic boxes, with the lowest box containing 25 kg of gold bars, the middle box containing 10 kg of aluminum bars, and the top box containing 2 kg of balsa wood, is calculated to be 0.137 meters above the floor. The formula used for this calculation is \(\Sigma mx / \Sigma m\), where the origin is set at the bottom of the gold bar. The total mass of the stacked boxes is 37 kg, and the heights of the center of mass for each box are factored into the calculation. The discussion highlights the importance of recognizing the center of mass in symmetrical objects.

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  • Understanding of center of mass calculations
  • Familiarity with basic physics concepts related to mass and weight
  • Knowledge of cubic geometry and dimensions
  • Ability to perform weighted averages
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PeachBanana
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Homework Statement



9) Three cubic boxes, each with side 0.1m, are stacked on top of one another. The lowest one is filled with gold bars and has mass 25kg; the middle one contains aluminum bars and has mass 10kg. The top one contains balsa wood and has mass 2kg. The height above the floor of the center of mass of the three boxes is _______.

Homework Equations



\Sigmamx / \Sigmam

The Attempt at a Solution



I put the origin at the bottom of the gold bar.

(25 kg)(0.1m) + (10 kg)(0.2 m) + (2 kg)(0.3 m) / 37 kg = 0.137 m
 
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Hi PeachBanana! :smile:
PeachBanana said:
I put the origin at the bottom of the gold bar.

(25 kg)(0.1m) + (10 kg)(0.2 m) + (2 kg)(0.3 m) / 37 kg = 0.137 m

Are these those special boxes with the centre of mass at the top? :wink:
 
Oh. It looks like I forgot the rule about symmetrical objects having their CM's in the center.
 

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