Calculating Raft Movement with Walking Boy

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The discussion revolves around calculating the movement of a raft as a boy walks on it, emphasizing the principle of center of mass. The boy weighs 32 kg and the raft 59 kg, with the boy walking 6 m across the raft. Participants clarify that the center of mass of the system remains stationary, which helps determine the final positions of both the boy and the raft. Initial calculations led to confusion, but through corrections, the proper movement of the raft was established. Ultimately, the focus is on understanding the relationship between mass distribution and movement in a closed system.
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Homework Statement


A 32 kg boy is standing on a 59 kg raft that is 8 m wide. He walks 6 m from a point 1.0 m from the left edge of the raft to a point 1.0 m from the right edge of the raft. If you ignore resistance of the water to the motion of the raft, how far does the raft move during this procedure


Homework Equations



F=ma
p=mv

No idea

The Attempt at a Solution



i draw a force diagram for it and looks like the one that will effect the movement of the boat is

F = friction force
but I have no idea what the friction constant is. and also will the mass affect the movement of the boat?

and I also don't have any velocity to use the equation of

m1v1 + m2v2 = m3v3
32(v1) + 0 = 91(v3)
 
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Smartguy94 said:

Homework Statement


A 32 kg boy is standing on a 59 kg raft that is 8 m wide. He walks 6 m from a point 1.0 m from the left edge of the raft to a point 1.0 m from the right edge of the raft. If you ignore resistance of the water to the motion of the raft, how far does the raft move during this procedure


Homework Equations



F=ma
p=mv

No idea

The Attempt at a Solution



i draw a force diagram for it and looks like the one that will effect the movement of the boat is

F = friction force
but I have no idea what the friction constant is. and also will the mass affect the movement of the boat?

and I also don't have any velocity to use the equation of

m1v1 + m2v2 = m3v3
32(v1) + 0 = 91(v3)


This is just a centre of mass question. The centre of mass of the system will remain stationary. The c of m of the two components moves. Where will they each finally be?
 
PeterO said:
This is just a centre of mass question. The centre of mass of the system will remain stationary. The c of m of the two components moves. Where will they each finally be?

x1 = (m1x1 + m2x2) / (m1+m2)
91(1) + 91(7) / (182)
x1 = 4m

uhmm i believe i did something wrong :confused:
 
Smartguy94 said:
x1 = (m1x1 + m2x2) / (m1+m2)
91(1) + 91(7) / (182)
x1 = 4m

uhmm i believe i did something wrong :confused:

suddenly the boy and raft have become 91 kg each?
 
PeterO said:
suddenly the boy and raft have become 91 kg each?

x1 = (m1x1 + m2x2) / (m1+m2)
x1 = (32(1) + 59(4)) / (91)
x1 = 2.725

x1 = (m1x1 + m2x2) / (m1+m2)
x1 = (32(7) + 59(4)) / (91)
x1 = 5.055

x = 5.055 - 2.725
x = 2.33m

but it's wrong, I'm confused where my error is
 
Last edited:
nvm, it was just a miscalculation, i got it now, thank you :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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