Horse Trailer and Center of Mass

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SUMMARY

The discussion centers on the physics of a 1000 kg horse trailer with a 500 kg horse moving within it. When the horse walks from one end of the 4 m trailer to the other, the trailer shifts to maintain the center of mass (CM) of the system. The initial calculation incorrectly assumed the trailer's CM remained stationary, leading to an erroneous conclusion of no movement. The correct calculation reveals that the trailer moves 4/3 m to keep the CM stationary, demonstrating the principles of conservation of momentum and center of mass in a closed system.

PREREQUISITES
  • Understanding of center of mass (CM) concepts
  • Knowledge of conservation of momentum principles
  • Familiarity with basic physics equations, particularly x_{cm} = \frac{1}{m}\sum_{i=0}^{n} m_ix_i
  • Ability to analyze systems with multiple masses
NEXT STEPS
  • Study the concept of center of mass in multi-body systems
  • Learn about conservation of momentum in closed systems
  • Explore practical applications of physics in real-world scenarios, such as vehicle dynamics
  • Investigate the effects of friction on motion and center of mass calculations
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of moving systems, particularly in relation to center of mass and momentum conservation.

TMO
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Homework Statement



A 1000 kg horse trailer with frictionless wheels is sitting in a level parking lot. The trailer is 4 m long, and its center of mass is at its center. Its passenger, a 500 kg horse, breaks free from its stall at one end of the trailer and walks to the other end. How far does the trailer move relative to the ground? Treat the horse as a point particle. The mass of the trailer above does not include the 500 kg horse.

Homework Equations



x_{cm} = \frac{1}{m}\sum_{i=0}^{n} m_ix_i

The Attempt at a Solution



I knew that xcm*m = (1000 kg)(2 m) + (500 kg)(0 m), which is 2000 kg*m, therefore the new position must be (2000 kg*m) = (1000 kg)x + (500 kg)(4 m) because no external forces are acting upon it. This yields x = 0m, but the correct answer is x = 4/3 m. Why is that?
 
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TMO said:

Homework Statement



A 1000 kg horse trailer with frictionless wheels is sitting in a level parking lot. The trailer is 4 m long, and its center of mass is at its center. Its passenger, a 500 kg horse, breaks free from its stall at one end of the trailer and walks to the other end. How far does the trailer move relative to the ground? Treat the horse as a point particle. The mass of the trailer above does not include the 500 kg horse.

Homework Equations



x_{cm} = \frac{1}{m}\sum_{i=0}^{n} m_ix_i

The Attempt at a Solution



I knew that xcm*m = (1000 kg)(2 m) + (500 kg)(0 m), which is 2000 kg*m, therefore the new position must be (2000 kg*m) = (1000 kg)x + (500 kg)(4 m) because no external forces are acting upon it. This yields x = 0m, but the correct answer is x = 4/3 m. Why is that?

The actual size of the horse is being ignored.
That answer is predicated on the c of m of the trailer being at the centre of the trailer, as defined, and the c of m of the horse being, at first at one end of the trailer, then later at the other end of the trailer.

If you work out the centre of mass of the trailer-horse system, you will find it is some way towards the end of the trailer where the horse is.
After the horse walks to the other end of the trailer, the c of m of the trailer-horse system will be someway towards the other end of the trailer.

The c of m of course will remain stationary, which can only be achieved if the trailer itself moves an appropriate amount.

For example if the c of m of the system was offset 80cm to the right to begin with, and so 80cm to the left at the end, the trailer must have moved 160 cm to the right in order for the c of m to actually remain stationary.
 

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