Tug of War: Force, Equilibrium, and the Third Principle

[Luigi Fortunati]

A pull the rope to the left with force -F, and B pull it to the right with force F:
A<-------->B

The intensity of the force exerted by A on B (by means of the rope) is equal and contrary to the force exerted by B on A (third principle).

If B increases its force from F to F+X, can A maintain its strength -F?

 

[PeroK]

If B increases its force from F to F+X, can A maintain its strength -F?

Why not?

What other forces are involved?

 

[Luigi Fortunati]

Why not?

What other forces are involved?

If B exerts the force F+X on the rope, the rope must also exert the same force -(F+X) on B.

And if A exercises the force -F on the rope, the rope must also exert the force F (equal and opposite) on A.

But the rope can not exercise on A *different* force than the one on B!

 

[PeroK]

If B exerts the force F+X on the rope, the rope must also exert the same force -(F+X) on B.

And if A exercises the force -F on the rope, the rope must also exert the force F (equal and opposite) on A.

But the rope can not exercise on A *different* force than the one on B!

There is the same tension in the rope for A and B. That is true. But, there are other forces involved. Think about the difference between:

a) a tug of war on Earth

b) a tug of war in space, where there is just A, B and the rope.

This is a good question, in fact.

 

[Nidum]

This is a non steady state problem .

Each team is trying to pull the other one towards it and so win the match .

So as one side pulls harder the other side pulls harder until such time as one side reaches the limit of it's pulling strength and falls over or starts slipping along the ground .

Note that the actual interaction mechanics is quite complicated when analysed in detail .

 

[Luigi Fortunati]

There is the same tension in the rope for A and B. That is true. But, there are other forces involved. Think about the difference between:

a) a tug of war on Earth

b) a tug of war in space, where there is just A, B and the rope.

This is a good question, in fact.

What changes in the forces between A, B and the rope, whether they are on Earth or in space?

 

[PeroK]

What changes in the forces between A, B and the rope, whether they are on Earth or in space?

Can you imagine doing this in space?

 

[Nugatory]

What changes in the forces between A, B and the rope, whether they are on Earth or in space?

When you do the experiment on earth, there are additional forces involved: forces between A and the earth, and forces between B and the earth.

 

[Luigi Fortunati]

Can you imagine doing this in space?

For example, with a space tug of war: astronaut A (with his hand out of a window) pulls one end of the rope to the left while the other astronaut B (on the other window) pulls the other end to the right.

 

[PeroK]

For example, with a space tug of war: astronaut A (with his hand out of a window) pulls one end of the rope to the left while the other astronaut B (on the other window) pulls the other end to the right.

Okay, forget space. They are pushing on the ground with their legs. Their arms are straight. If B pulls with more force than A, then A moves - a little thing called acceleration occurs.

In that sense, pulling your opponent is no different from pulling a block on a frictional surface. Once you pull harder than the maximum frictional force, the block moves.

 

[Luigi Fortunati]

Okay, forget space. They are pushing on the ground with their legs. Their arms are straight. If B pulls with more force than A, then A moves - a little thing called acceleration occurs.

In that sense, pulling your opponent is no different from pulling a block on a frictional surface. Once you pull harder than the maximum frictional force, the block moves.

And when the block moves, its strength on the string is less than that of the puller on the other side?

And in the arm wrestling, when the hand A prevails on the hand B, the force of A on B is greater than that of B on A?

 

[PeroK]

And when the block moves, its strength on the string is less than that of the puller on the other side?

In your thinking you need to include the concept of acceleration. Imagine the block is on a frictionless surface. It has no way to generate any force. All of the tension in the string results in acceleration. The person pulling the string will feel the reaction force, but this is not the block pulling back. It is the block accelerating. Technically, you do not show resultant acceleration as a "force" in a diagram. If you assume that the object being accelerated is generating a force equal to the force that is accelerating it, then this would cancel the accelerating force and motion would be impossible.

If you pull harder in a tug of war two things can happen:

1) Your opponent will start to move/accelerate.

2) Your opponent will lose their grip on the rope. In this context, that is a different sort of force. It's an internal force that keeps a system together. In the same way, if you pull hard enough you might pull your opponent's arms off!

Another example to think about is being pulled uphill on a ski-tow. You are doing no work and technically provide none of the force to move uphill. But, you do need to be able to hold onto the bars. If you were tied to the ski-tow, you wouldn't even need to do that. All you would need is that your arms can support that tension without being pulled out of your body. None of this represents a force that you can generate. It's a tension that your body can support.

 

[Luigi Fortunati]

In your thinking you need to include the concept of acceleration. Imagine the block is on a frictionless surface. It has no way to generate any force. All of the tension in the string results in acceleration. The person pulling the string will feel the reaction force, but this is not the block pulling back. It is the block accelerating.

No, it is the block that *refuses* to accelerate and exerts a * real * reaction force!

Technically, you do not show resultant acceleration as a "force" in a diagram. If you assume that the object being accelerated is generating a force equal to the force that is accelerating it, then this would cancel the accelerating force and motion would be impossible.

No, the block is not accelerating *compared* to those who "pull"!

The person who pulls and the block are *at rest* with respect to each other, precisely because their mutual forces are identically the same and opposite.

It is relative to the ground that the block accelerates (and vice versa).

 

[Herman Trivilino]

A pull the rope to the left with force -F, and B pull it to the right with force F:
A<-------->B

The intensity of the force exerted by A on B (by means of the rope) is equal and contrary to the force exerted by B on A (third principle).

That is not true! Newton's Third Law involves a pair of equal-but-opposite forces, but those two forces never act on the same object!

If B increases its force from F to F+X, can A maintain its strength -F?

If B exerts a force that is larger than the force that A exerts there will be a net force on the rope, and B will win!

The usual way to propose this riddle is to to state that A pulls on B, and B pulls on A. These two forces are indeed equal-but-opposite in accordance with Newton's Third Law. (If the rope you mentioned above is idealized as having negligible mass, then we have an equivalent scenario and the comments made by the other posters apply). Now, to figure out who will win you must look at the strength of the friction force exerted on A by the ground, and on B by the ground. The team with the larger of these will win. For example, give B cleats and put A on roller skates, B will surely win even though the strength of the force A exerts on B equals the strength of the force that B exerts on A.

 

[Luigi Fortunati]

That is not true! Newton's Third Law involves a pair of equal-but-opposite forces, but those two forces never act on the same object!
If B exerts a force that is larger than the force that A exerts there will be a net force on the rope, and B will win!

The usual way to propose this riddle is to to state that A pulls on B, and B pulls on A. These two forces are indeed equal-but-opposite in accordance with Newton's Third Law. (If the rope you mentioned above is idealized as having negligible mass, then we have an equivalent scenario and the comments made by the other posters apply). Now, to figure out who will win you must look at the strength of the friction force exerted on A by the ground, and on B by the ground. The team with the larger of these will win. For example, give B cleats and put A on roller skates, B will surely win even though the strength of the force A exerts on B equals the strength of the force that B exerts on A.

Ok, you're right. In the next discussion I will eliminate the rope.

 

[PeroK]

We need some calculations. We have team A on the right, mass m_A, applying a force (to the ground) of F_A and team B on the left.

1) If $F_A = F_B$ then we have a static equilibrium and the tension in the rope is $T = F_A = F_B$.

2) If $F_A > F_B$, then the acceleration of the system is $a = \frac{F_A - F_B}{m_A + m_B}$ and the tension in the rope is $T = \frac{m_AF_B + m_BF_A}{m_A + m_B} >$F_B##$3) In the special case where$F_B = 0$, we have$T = \frac{m_BF_A}{m_A + m_B}##. In this case, A still feels the rope pulling them, even though team B is exerting no force on the system.

4) If team B let's go of the rope, then the tension in the rope disappears.

If you take the rope out of the system and have just two people pulling each other, then nothing changes, except the tension is simply in the arms of the two players.

 

[Luigi Fortunati]

We need some calculations. We have team A on the right, mass m_A, applying a force (to the ground) of F_A and team B on the left.

1) If $F_A = F_B$ then we have a static equilibrium and the tension in the rope is $T = F_A = F_B$.

2) If $F_A > F_B$, then the acceleration of the system is $a = \frac{F_A - F_B}{m_A + m_B}$ and the tension in the rope is $T = \frac{m_AF_B + m_BF_A}{m_A + m_B} >$F_A####

In this case (2), how much is the force of team B on the rope (and vice versa)?

 

[PeroK]

In this case (2), how much is the force of team B on the rope (and vice versa)?

I'll let you work that out. After all, physics is as much about calculations as about theory.

Note that I had an error in the post which I've now fixed.