# Center of mass of dog

1. Feb 23, 2008

### NAkid

1. The problem statement, all variables and given/known data
A dog, weighing 11.1lb, is standing on a flatboat so that he is 20.0ft from the shore (to the left in Figure (a)). He walks 8.5ft on the boat toward shore and then halts. The boat weighs 40.4lb, and one can assume there is no friction between it and the water. How far is the dog then from the shore?

2. Relevant equations
x(cm) = (m1x1 + m2x2) / (m1 + m2)

3. The attempt at a solution
The center of mass of the system has to be the same before and after the dog moves, so taking the shoreline as the origin and the center of mass of the boat(m1) at 20ft, I said that initially,
x(cm) = [(m1*20) + (m2*20)] / (m1 + m2) -- m2 is the dog

and after the dog moves

x*(cm) = [(20-x)m1) + (11.5-x)m2] / (m1+m2)

where x is the displacement. but i'm pretty sure this isn't right.. help!

2. Feb 23, 2008

### Staff: Mentor

Try this: Figure out how the center of mass of the "dog + boat" (measured with respect to the edge of the boat) changes as the dog moves. That will tell you how the boat moves.

3. Feb 23, 2008

### NAkid

well as the dog moves to the left closer to shore the center of mass of dog+boat shifts back, right?

Last edited: Feb 23, 2008
4. Feb 23, 2008

### Staff: Mentor

Here's what I mean: Before the dog moves the center of mass of dog+boat is somewhere, let's say X1 feet from the edge. As the dog moves towards that edge, the center of mass (with respect to the boat edge, not the shore) must shift because the dog's mass shifts. The new center of mass will be X2 feet from the edge. Find X1 - X2.

5. Feb 23, 2008

### NAkid

i'm still a bit confused, but this is what i have:

x(cm,1) = (x1m1 + x2m2)/(m1+m2) --where m1 is the boat and m2 is the dog

x(cm, 2) = [(x1-x2)m1 + (8.5-x2)]/(m1+m2) --the dog is displaced less than 8.5ft because the boat shifts in the opposite direction

6. Feb 23, 2008

### Staff: Mentor

OK. x1 is the initial distance from dog to edge of boat; x2 is the distance of center of boat to edge of boat; x(cm,1) is the initial distance of the cm of dog+boat to the edge of the boat.

For x(cm, 2), the only thing that changes is the dog's position, since it walks 8.5m closer to the edge of the boat:
x(cm,2) = [(x1-8.5)m1 + x2m2]/(m1+m2)

So figure out how much the cm of dog+boat shifted towards the edge of the boat.

7. Feb 23, 2008

### NAkid

ok, and subtract that from 20 to find the final distance of the dog from the shore?

8. Feb 23, 2008

### Staff: Mentor

No. For example: If the center of mass shifts one meter closer to the edge, that means that the boat must shift one meter away from the shore. (Since the cm must remain the same distance from the shore.) Which you can then use to figure out how far the dog moved toward the shore.

From what I gave in #6 you can calculate exactly how far the center of mass shifted.

Last edited: Feb 23, 2008