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Horse Trailer and Center of Mass

  1. Oct 26, 2011 #1


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    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    [itex]x_{cm} = \frac{1}{m}\sum_{i=0}^{n} m_ix_i[/itex]

    3. 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?
  2. jcsd
  3. Oct 26, 2011 #2


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    Homework Helper

    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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