Well, the cg did not move.
The cg is (1.5+2.6) from shore at the start.
cg*totalmass= 3.3*63+masscanoe*(3.3+1.5)
(1.5+2.6)(63+masscanoe)= 3.3*63+masscanoe(4.8)
masscanoe(4.1-4.8)=3.3*63-4.1**63
solve for mass canoe
A 63 canoeist stands in the middle of her canoe. The canoe is 3.0 long, and the end that is closest to land is 2.6 from the shore. The canoeist now walks toward the shore until she comes to the end of the canoe. Suppose the canoeist is 3.3 from shore when she reaches the end of her canoe.
What is the canoe's mass?
2 answers
better to sum torques about the center of mass.
from standing in the middle of the canoe we know that the CM is at x = 1.5+2.6 = 4.1
when the person is 3.3m from the shore that makes them (4.1m-3.3m)=0.8m from the CM and because we know the canoe is 3m long the CM of the canoe is (3.3+1.5)=4.8m from the shore and consequently (4.8m-4.1m)=0.7m from the CM of the system.
now sum the torques about the CM and set equal to zero with CCW rotation as being positive we get
0=(63kg)*(0.8m)-(Mcanoe)*0.7m
Mcanoe=(0.8/0.7)*63kg
from standing in the middle of the canoe we know that the CM is at x = 1.5+2.6 = 4.1
when the person is 3.3m from the shore that makes them (4.1m-3.3m)=0.8m from the CM and because we know the canoe is 3m long the CM of the canoe is (3.3+1.5)=4.8m from the shore and consequently (4.8m-4.1m)=0.7m from the CM of the system.
now sum the torques about the CM and set equal to zero with CCW rotation as being positive we get
0=(63kg)*(0.8m)-(Mcanoe)*0.7m
Mcanoe=(0.8/0.7)*63kg
2 answers