Calculating Raft Movement with Walking Boy

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In summary, the problem involves a 32 kg boy standing on a 59 kg raft, both 8 m wide. The boy walks 6 m from the left edge to the right edge of the raft, and the question asks how far the raft moves if water resistance is ignored. By finding the center of mass of the system, it is determined that the final position of the boy is 2.33 m from the left edge and the final position of the raft is 5.055 m from the left edge. The distance the raft moved during the procedure is 2.33 m.
  • #1
Smartguy94
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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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  • #2
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?
 
  • #3
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:
 
  • #4
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?
 
  • #5
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:
  • #6
nvm, it was just a miscalculation, i got it now, thank you :)
 

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