Solving the Tug of War Problem: Man Moves 240m

[rent981]

1. Well this might not be exactly a tug of war problem but its pretty close.
A 100 kg fisherman and a 500 kg supply crate are on a frozen pond that is virtually frictionless. The man and the crate are initially 600 m apart. The fisherman uses a very light rope to pull the crate closer to him. How far has the man moved when the crate reaches him?



2. This is Newtons Third law of Mechanics.



3. TI know that if the man and the crate were the same mass, they would meet in the middle. So the answer would be 300m. So I think that since the crate is 5 times the mass it will only move one fifth of that distance. Therefor the man would have moved 240 meter. Does this sound right?

 

[PhanthomJay]

1. Well this might not be exactly a tug of war problem but its pretty close.
A 100 kg fisherman and a 500 kg supply crate are on a frozen pond that is virtually frictionless. The man and the crate are initially 600 m apart. The fisherman uses a very light rope to pull the crate closer to him. How far has the man moved when the crate reaches him?



2. This is Newtons Third law of Mechanics.



3. TI know that if the man and the crate were the same mass, they would meet in the middle. So the answer would be 300m. So I think that since the crate is 5 times the mass it will only move one fifth of that distance. Therefor the man would have moved 240 meter. Does this sound right?

No. The cm of the system doesn't move. Where is the center of mass?

 

[rent981]

The center of mass would be (100kg)(1m)+(500kg)(600m)/(100+500kg)=500.167 m

 

[rent981]

well actually 1m should be zero, but it still works out to be 500.

 

[rent981]

so if friction is ignored, in a game of tug of war, the opposing sides of the rope would meet at the center of mass? Provided there are no other contributing factors.

 

[denverdoc]

yea in order to conserve momenta. They would have to.

YOU= mv1
the crate=mv2

Neither has any velocity, therefore no momenta before or after the tug of war.There is only way to get there, at the center of mass with opposing velocities, proportionate to mass.

 

[PhanthomJay]

so if friction is ignored, in a game of tug of war, the opposing sides of the rope would meet at the center of mass? Provided there are no other contributing factors.

No, that would be true only if the weights of the opposing team were the same. If one of the teams has a mass of 500 kg (say 5 football players) and the the opposing team has a mass of 100 kg (say 5 small children), and the rope is 6 m long, then they meet, as you noted, at 1 m away from the football players' original position (5 m away from the kids' original position). This assumes a frictionless surface.

 

[PhanthomJay]

so if friction is ignored, in a game of tug of war, the opposing sides of the rope would meet at the center of mass? Provided there are no other contributing factors.

yes, correct, sorry, i misread your response earlier.