Calculating the Center of Mass for a Dog on a Flatboat

[Punchlinegirl]

A dog, with a mass of 12.0 kg, is standing on a flatboat so that he is 25.5 m from the shore. He walks 6.0 m on the boat toward the shore and then stops. The boat has a mass of 46.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?
i used $$x_cm = m_1 x_1 + m_2 x_2 / m_1 + m_2$$
x= (12)(25.5) + (46)(6) / 12+46
and got 10.0 m
Which wasn't right.. can someone please help me?

 

[Doc Al]

The key is that the center of mass of the "boat + dog" system cannot change. So, if the dog ends up X meters closer to shore, how much further from the shore must the boat end up (in terms of X)? Now make use of the fact that you know how far the dog moved with respect to the boat to solve for X. (To help visualize what's going on, pretend the dog was standing on a line painted on the boat. The dog moves toward the shore; the boat--and line--moves away from the shore. But you know how far apart the dog moved from the line.)

 

[Punchlinegirl]

Ok I get it now. Thank you!