Solving Tug of War Physics: Who Wins?

  • Thread starter Thread starter iampaul
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
In a tug of war, the outcome is influenced by the forces exerted by the boys on the ground and the resulting friction, which determines their ability to gain traction. If the ground is frictionless, neither boy can gain an advantage, as they cannot push against the ground effectively. The heavier boy tends to win due to greater force exerted on the ground, leading to a stronger reaction force from the ground. The tension in the rope remains constant, ensuring that the forces acting on each boy are equal, but the one who pushes harder against the ground will prevail. Understanding these dynamics is crucial for solving the physics of tug of war scenarios.
iampaul
Messages
93
Reaction score
0

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!

 
Physics news on Phys.org
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!=)
 
Back
Top