# Question #268cb

May 1, 2017 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:

$\sum F = m a = 0$

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 $18 - x$

${m}_{T} \cdot L + {m}_{F} \cdot \frac{L}{2} = {m}_{T} \cdot \left(L + x - 18\right) + {m}_{F} \cdot \left(\frac{L}{2} + x\right)$

$\implies 0 = {m}_{T} \cdot \left(x - 18\right) + {m}_{F} \cdot x$

Solves as: $x = \frac{54}{43} m$

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

${m}_{T} \cdot L + {m}_{F} \cdot {L}_{F} = {m}_{T} \cdot \left(L + x - 18\right) + {m}_{F} \cdot \left({L}_{F} + x\right)$