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Homework Help: A Dog on a Raft

  1. Nov 14, 2008 #1
    This is problem 17 in Chapter 9 of Halliday and Resnick, 8th Edition (p.231). This is one of their "three star" (= hard) problems. Just thinking about it makes my head spin. I've managed to eventually get the right answer, but I'd appreciate someone describing the situation in their own words so I have another perspective.

    A raft is floating, stationary, offshore, with mass= 18kg. On the raft is a dog, mass 4.5kg. The dog is 6.1m from the shore. The water is considered a frictionless surface for the raft. The dog walks 2.4m down the raft towards the shore. How far is the dog now from the shore?

    My Approach:
    Okay, the point is the dog moves one way, while the raft moves the other way under the dog. The center of mass (com) does not move. I took the location of the com as the center of a one-dimensional co-ordinate axis. I imagined the dog-raft system to be like two point masses with the com inbetween somewhere. But here we don't know how long the raft is. But we do know that the com is not moving.

    The key equation is (using the x-axis as a our one-dimensional coordinate):

    (total mass of system)( x coordinate of center of mass)= (m1)(x1)+(m2)(x2), with m1=dog mass, ,m2=raft mass, and x1,x2 the location of those particles.

    If the com is at the origin of the axis (=0) then in the above 0=(m1)(x1)+(m2)(x2). Now the dog moves 2.4m down the raft. But the raft is moving away from the shore, under the dog's feet. So 2.4m of raft passes under the dog without the dog actually moving 2.4m closer to the shore. So you can't say 0=(m1)(2.4)+(m2)(x2) to get x2 (if you do this, you get the wrong answer).

    If the dog is walking in the +x direction, then he is actually moving a distance 2.4m - d, where we don't know d. But this d is also the distance that the raft moves towards the com. So for our equation we can use (2.4m - d) for the dog, and -d for the raft, since the raft is moving in the -x direction: So the correct equation is 0=(m1)(2.4 - d)+(m2)(-d). Putting in the masses gives 4.5(2.4+d)+18(d)=0 and solving this for d gives d=.48m. So the dog moves 2.4 - d=1.92m as seen from the sea/shore. And the dog starts 6.1m from shore so now he's so 6.1m-1.92m=4.2m from shore (= correct answer).

    Is there any simpler way to think about this? What throws me is the dog moving one way while the raft is moving the other way (as seen from the shore).
  2. jcsd
  3. Nov 14, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I'll tell you how I would do solve this. It's pretty much what you did, but it might help to see it from a slightly different view.

    All I care about is the change in the position of the dog and raft with respect to the shore. Say the raft moves a distance x away from the shore, so its change in position is -x. The dog moves 2.4m with respect to the raft, so his change in position with respect to the shore is -x+2.4.

    Since the center of mass cannot change, we have: (-x)Mr + (-x + 2.4)Md = 0. Solve for x, then for (-x + 2.4), then for the dog's distance from the shore and you're done.
  4. Nov 14, 2008 #3
    Thanks. Just hearing another person describe it helps. This problem is in a class of problem I occasionally come across where everything is moving and its hard to pin things down to get started. I've discussed these sorts of problems with my brother who's a statistician, and its not just physics-- its a particular sort of mathematical situation. For example, when most people think about the change from standard to daylight savings time, this sort of problem comes up. They can't figure if it should be darker or lighter at a given hour, after the switch. I have to walk myself through it, e.g. "Okay, if its this dark now, after setting the clocks back, it will be this dark, but when the clock says one hour earlier." It's just interesting to me there are some conceptual situations where its hard to get traction.
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