Question

A 45.0-kg woman stands up in a 69.0-kg canoe that is 5.00 m long. She walks from a point 1.00m from one end to a point 1.00 m from the other end. If you ignore resistance to the motion of the canoe in the water, how far does the center of the canoe move during this process? (or the left end, or the right end, whichever point in the canoer is easier for you to think of)

Answer 1

Answer:

The center of the canoe is displaced 1.18m in the opposite direction to the movement of the woman.

Explanation:

Let's assume the woman is on the left of the canoe and that the origin of our system is in the middle of the canoe.

The position of the center of gravity of the whole system remains in the same position, so:

$m_{c}*r_{oc}+m_{w}*r_{ow}=m_{c}*r_{fc}+m_{w}*r_{fw}$

Since our origin is in the middle of the canoe, all of the positions become:

$r_{oc}=0$

$r_{ow}=-(2.5-1)=-1.5$

$r_{fc}=-X$  (It is negative, assuming that the canoe will move to the left).

$r_{fw}=-X+(2.5-1)=-X+1.5$

Being X the displacement we want to find.

Replacing all of these values, we get:

$69*0+45*(-1.5)=69*(-X)+45*(-X+1.5)$

Solving for X:

X=1.18m  Since the result is positive, we conclude that our assumption was correct.