Solving Tug of War Physics: Who Wins?

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SUMMARY

The discussion centers on the physics of a tug of war, specifically analyzing the forces at play according to Newton's Third Law. It concludes that the winner is determined by the friction between the boys and the ground; the heavier boy will win on a frictionless surface. The key forces identified include the forces exerted by each boy on the rope and the corresponding reactions from the rope and ground. The tension in the rope remains constant, and the boy who pushes harder against the ground ultimately prevails.

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  • Understanding of Newton's Third Law of Motion
  • Basic knowledge of force and friction concepts
  • Familiarity with free body diagrams
  • Ability to analyze systems in equilibrium
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  • Explore the role of friction in force interactions
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Physics students, educators, and anyone interested in understanding the mechanics of force interactions in competitive scenarios like tug of war.

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Homework Statement


Two boys play tug of war.If according to Newton's third law, for every force a there is a reaction b with the same magnitude but opposite direction. What determines who wins?
Our professor gave a hint that it has something to do with the ground.

Homework Equations


f=ma and Newtons 3rd law

The Attempt at a Solution


I have identified these forces:
A= force exerted by boy1 to the rope
A'=reaction to A ,force exerted by the rope to boy1
B=force exerted by boy1 to the rope
B'=reaction to B ,force exerted by the rope to boy2
Fa=Friction exerted at boy1
Fb=Friction exerted at boy2
Our professor said that no one wins if the ground is frictionless but i can't see why.
thanks!

 
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iampaul said:
Our professor said that no one wins if the ground is frictionless but i can't see why.
thanks!

Actually the heavier person will win for frictionless ground.

Here's how to think about it: What is external force on the system? Where is the center of mass? Does it move? Where will the two players meet if they keep pulling?
 
Last edited:
iampaul said:

Homework Statement


Two boys play tug of war.If according to Newton's third law, for every force a there is a reaction b with the same magnitude but opposite direction. What determines who wins?
Our professor gave a hint that it has something to do with the ground.

Homework Equations


f=ma and Newtons 3rd law

The Attempt at a Solution


I have identified these forces:
A = force exerted by boy1 to the rope
A'=reaction to A ,force exerted by the rope to boy1
B=force exerted by boy1 to the rope
B'=reaction to B ,force exerted by the rope to boy2
Fa=Friction exerted at boy1
Fb=Friction exerted at boy2
Our professor said that no one wins if the ground is frictionless but i can't see why.
thanks!
You started out listing some great 3rd law couples:
Boy 1 pulls the rope, the rope pulls Boy 1
Boy 2 pulls the rope, the rope pulls Boy 2

You then degenerated into some general descriptions
Fa=Friction exerted at boy1
Fb=Friction exerted at boy2

You must express those forces, that you just called Friction, as 3rd law Couples.

Do that and I will get back to you if that doesn't already answer your problem.
 
boy2 pushes the ground,ground pushes boy2 ??
boy1 pushes the ground, ground pushes boy1 ??

The forces acting on boy1 are A' and the force from the ground.
The forces acting on boy2 are B' and the force from the ground.
The forces acting on the rope are A and B
 
iampaul said:
boy2 pushes on ground,ground pushes on boy2 ??
boy1 pushes on ground, ground pushes on boy1 ??

??

Great.
That is a total of 8 forces, drouped according to Newtons 3rd Law couple.

Now group the forces according to which body they act on: boy 1, boy 2, the rope and the ground

eg
Rope: Boy 1 pulls rope, Boy 2 pulls rope

Now You -

Boy 1:
Boy 2:
Ground:

Then look at the sizes of the pair of forces grouped this way.

NB: Newtons Third law says that the pair of forces : Boy 1 pulls Rope, Rope pulls Boy 1 are equal in size, but Newtons law does not compare the force of the Rope pulling Boy 1 to the force of Boy 2 pushing the Ground
 
Boy1: Ground pushes boy1(Fa), rope pulls boy1(A')
Boy2: Ground pushes boy2(Fb), rope pulls boy2(B')
Rope:Boy1 and boy2 pull on rope(A and B)
A=-B if the rope is in equilibrium or if the rope is massless
I don't know what's next!
Should i treat the rope massless?
 
The system moves to the side of boy 1 if fa>A'=A>B=B'>Fb?
Fa>fb
 
iampaul said:
Boy1: Ground pushes boy1(Fa), rope pulls boy1(A')
Boy2: Ground pushes boy2(Fb), rope pulls boy2(B')
Rope:Boy1 and boy2 pull on rope(A and B)
A=-B if the rope is in equilibrium or if the rope is massless
I don't know what's next!
Should i treat the rope massless?

The tension in the rope ensures that all forces involving the rope have the same size.

Now to the boys: The size with which the rope pulls them is equal, so it is the one who is pushed more strongly by the ground that wins.

And of course Newtons Third Law tells us that the ground pushes most strongly on the boy that pushes most strongly on the ground.
So he who pushes the ground the hardest wins.
 
Last edited:
thanks a lot!
 
Last edited:
  • #10
"The tension in the rope ensures that all forces involving the rope have the same size."

Wouldn't that happen only if the rope is in equilibrium or if the rope is massless?
Please reply! Last question =)
 
  • #11
iampaul said:
"The tension in the rope ensures that all forces involving the rope have the same size."

Wouldn't that happen only if the rope is in equilibrium or if the rope is massless?
Please reply! Last question =)

The boys would simply not be able to accelerate at the rate required to vary the tension in the rope - especially since the rope has a much lower mass than each of the boys. In fact in a tug of war, motion is generally very slow

If the rope was as heavy as the boys [50+ kg] and one of them took off faster than an olympic sprinter you might notice something.

We can demonstrate by changing the situation to a mass-less rope with a 2kg mass in the middle [so two ropes actually] and the boys having mass 49 kg. [this is to mimic a rope of mass 2kg]

suppose Boy 1 wins, by pushing on the ground with a force 20N more than Boy 2. The net force on the system is thus 20N so the 100 kg system will accelerate at 0.2 m/s^2.

Individually, the 2 kg mass has an unbalanced force of 0.4N acting [it is accelerating at 0.2 m/s^2 also remember]

That would mean the tension in the rope from Boy 1 to the mass is 0.4N more than the tension in the rope from the mass to boy 2.

Now to accelerate at 0.2 m/s is an extremely high rate of acceleration - not the sort of acceleration one associates with a tug of war, so the effects are probably much less than this, indicating that the tension will be pretty well constant even with a real rope.
 
  • #12
Thanks! Now i get it!=)
 

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