Question #268cb

1 Answer
Eddie
May 1, 2017

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Because there is no external force on the truck/ferry system (eg we are told to disregard any effect of the water), Newton's 2nd says that:

#sum F = ma = 0#

And so the position of the combined centre of mass of the truck/ferry system is unmoved by this event. So we just need to calculate that position from the same point marked #O#, both before and after the event.

To be clear, the ferry boat moves left wrt the water by #x#, as the truck moves right wrt the ferry by #18 m#, and so wrt to the water by #18 - x#

# m_T cdot L + m_F cdot L/2 = m_T cdot (L + x - 18) + m_F cdot (L/2 + x) #

#implies 0 = m_T cdot (x - 18) + m_F cdot x #

Solves as: #x = 54/43 m#

Upon reflection, I didn't need to assume that the CoG of the ferry is at the midpoint, #L/2#. So if we re-write and solve our first equation with the CoG at generalised distance #L_F# from O, we get same solution:

# m_T cdot L + m_F cdot L_F = m_T cdot (L + x - 18) + m_F cdot (L_F + x) #

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