# Tug of war (in a vacuum)

1. Mar 10, 2015

Suppose two people, A and B, are pulling on a rope of mass $M_r$ in space. The force exerted by B is $F_B$ and the force exerted by A is $F_A$ (in magnitude). Suppose B pulls harder than A, we now have the equation of motion:
$$F_B - F_A = Ma_{rope}$$
We also have a constraint; since A, B and the rope are all connected, their accelerations must all be the same. That is $a_A = a_B = a_{rope}$. According to Newton's third law, the force exerted by the rope on B is equal in magnitude and opposite in direction to the force exerted by B on the rope. That is, it is not in the same direction as the acceleration of B, which is confusing because this force (tension) seems to be the only force acting directly on B, thus one would expect the acceleration of B to be in the same direction as the tension acting on B, in accordance with Newton's second law, except it isn't. I am really confused.

2. Mar 10, 2015

### Staff: Mentor

Unless you fix A and B somewhere (like on a planet), they will accelerate towards each other, which gives them different accelerations. They move relative to the rope.
The rope will accelerate, too.

3. Mar 10, 2015

### Staff: Mentor

With the condition that the accelerations are equal the only solution is that the accelerations and forces are 0. That condition does not accurately represent the scenario I suspect that you are really envisioning.

4. Mar 11, 2015

### A.T.

Nope, humans aren’t rigid bodies. The acceleration of the rope and your hand holding it can be different from the acceleration of your bodie's center of mass.

5. Mar 11, 2015

### Staff: Mentor

This is a bit of a mess with impossible assumptions:

mfb said it, but let me try to be more explicit:
This constraint is unacceptable. Because the tension has a single value, it is not possible for the two people to apply different forces to the rope, as per Newton's 3rd law.

Also, in a real game of tug-of-war, the forces arise because the players are pushing against the ground. In space, the only way to generate a force is for them to pull themselves toward each other, accelerating both in opposite directions by moving their arms (which was A.T.'s point).

There is a physics classroom demonstration of this issue (probably youtube videos) where the prof gets the biggest and smallest person in the class, puts them on skateboards, and tells each to push away from the other as hard as they can, with the assumption the bigger person can push harder. But that's impossible and what always happens is they move apart, with accelerations proportional to their different masses.

6. Mar 11, 2015

### A.T.

Not in a massive accelerated rope.

It is possible.

The probably most often misapplied law in physics.

7. Mar 11, 2015

### Staff: Mentor

Oops: rope with mass. I don't think I've ever seen a tug-of-war problem with the mass of the rope included, and I missed it

In either case we still have the undeveloped issue of how they apply the force (if neither are fixed).

The title says "tug-of-war", but the problem really isn't very similar to a tug-of-war.

8. Mar 11, 2015

### A.T.

By pulling with their hands.

So they will accelerate.

9. Mar 11, 2015

So B will accelerate in an opposite direction to the acceleration of the rope, right?

10. Mar 11, 2015

### jbriggs444

Right.

B pulls harder than A, so the rope accelerates toward B. The rope pulls on B so B accelerates toward the rope.

11. Mar 11, 2015

### Staff: Mentor

Possibly. But since the OP's constraints/conclusions contradict each other, we'll have to let him tell us which is true and which is false.