# Woman walking along a canoe

• ph123

#### ph123

A 45.0-kg woman stands up in a 60.0-kg canoe of length 5.00 m. She walks from a point 1.00 m from one end to a point 1.00 m from the other end. If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

At first I thought that the canoe must move 3m in the opposit direction she walked along it since there was so fluid resistance. However, I decided the center of mass shifts so that the canoe should move the distance that center of mass shifted from one end to the other.

My center of mass equation is:

x_cm = (60kg * 5m/2)/(60kg + 45kg)
= 1.429

So the center of mass in the canoe is 1.429 m from the tip of the canoe on the side that the woman is standing. Since she walks 3 m and ends up on the opposite side of the canoe in the exact same spot relative to the tip of the canoe, the center of mass has also shifted to 1.429m from that end.

I added the two center of masses, and got 2.857m. I figured that subtracting that amount from the length of the canoe (5m) should give me the total distance moved by the canoe while she was walking, or 2.143m. This isn't correct, however.

Does anyone know if I may have overlooked something?

Since the system canoe+woman experiences NO EXTERNAL FORCES, how much can the center of mass of this system move when it begun at rest?

well, i suppose it can't. So this is a trick question, and the canoe doesn't move at all? According to Newton, however, equal and opposite and all that jazz. Shouldnt friction between the woman and the canoe cause the canoe to move opposite to her motion since the water is essentially frictionless?

well, i suppose it can't. So this is a trick question, and the canoe doesn't move at all? According to Newton, however, equal and opposite and all that jazz. Shouldnt friction between the woman and the canoe cause the canoe to move opposite to her motion since the water is essentially frictionless?

the canoe does move. The center of mass of the combined woman+canoe system does not move. Since the woman moves, the canoe must move to keep the CM of the combined system at the same point.

Be very careful to understand the difference between the CM of the canoe alone, the CM of the woman alone and the CM of the combined woman+canoe system.

By the way, when you calculated your x_CM, you did it wrong. If you want the CM of the combined woman plus canoe system, you must also include the woman in the calculation of the numerator!

The key is to find the Initial and Final centers of mass.

Initial: Woman's CoM = 1m , Canoe's CoM = 2.5...What's the CoM of the entire system?

Final: Woman's CoM = 4m , canoe's CoM = 2.5...CoM of System?

Final CoM - Initial CoM = your answer

thanks everyone. i did indeed forget the woman in the numberator. the final answer was 1.286m.