Walking on a canoe, finding its mass.

[PhyzicsOfHockey]

Homework Statement

A 61.7 kg canoeist stands in the middle of her canoe. The canoe is 3.9 m long, and the end that is closest to land is 2.5 m from the shore. The canoeist now walks toward the shore until she comes to the end of the canoe. Suppose the canoeist is 3.4 m from shore when she reaches the end of her canoe. What is the canoe's mass?

Homework Equations

I don't know

The Attempt at a Solution

1.95m/.9m * Mp

I tried taking a ratio of how far the canoeist moved to how far the canoe moved in the opposite direction and multiply that by the mass of the conoeist. I got 133.68 kg but that's wrong. How do I do this problem?

 

[mjsd]

several additional assumptions may be required to solve this one as there are so many unknowns here. eg. is the canoe moving? mass distribution on canoe uniform? how steady is the canoeist moving? is the canoesit moving at constant speed or accelerating? frictionless water?

from the wording, this looks to me like a "center of mass" problem...try that

 

[PhyzicsOfHockey]

I thought of that a realized I don't know how to go about it being a center of mass problem. The only thing I really undestand is its the sum of the mass*d/total mass of the system. In this case I don't know what distances to use. Distance from the shore? Distance from the center of the canoe? Just not sure how to go about it. Any help is appreciated.

 

[PhyzicsOfHockey]

I did the following...
Mc=Mass canoe
Mp=Mass person
4.45= distance from shore+half distance of canoe
Cm=Cm

(Mc+Mp)*4.45/(Mc+Mp)=Cm
Cm=4.45m
4.45=3.4Mp+5.35Mc/(Mp+Mc)
4.45Mp+4.45Mc= 3.4Mp+5.35Mc
1.05Mp=.9Mc
1.1667Mp=Mc
Mc= 71.983kg

Is this correct? or am I not suppose to assume the Cm is the same?

 

[mjsd]

ok, in order to do this as a "center of mass" problem, you do need to make quite a few assumptions. given the lack of info provided, I am guessing that we are to assume that (if it is indeed a center of mass problem) the person is "point mass" while the canoe is just a thin rod with uniform mass distribution. perhaps the person somehow rolls fritionlessly towards one end from the middle without doing anything silly like undergo large acceleration...etc..

ok, (if this is indeed a center of mass problem...of course I don't know) the key concept here is that the position of the centre of mass of a system remain at the same place IF no external force acts on the system. so in this case, taking in all our assumptions (basically to say that there is no external force present)...the change in the internal structure of the system gives rise to change in the location of the center of mass , but since location of it can't change (no external force) the entire system must be shifted relatively to the original coordinate grid.