# People pulling rope, one in a perpendicular direction

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• ChessEnthusiast

#### ChessEnthusiast

Let's say that we have two guys, each weighs 50 kg. They are pulling a piece of against each other as hard as they can. The tension of the rope will therefore be the force of friction acting on each of them (the same magnitude).

Now, let's say that a third person joins the party and grabs the rope right between them and wants to pull it in direction perpendicular to the rope.

1. What information do I need to estimate how much force is required to move the point where the third person holds the rope a certain distance in that direction?
2. How does the tension of the rope after the pull depend on the initial tension, if the point moved the same distance in the perpendicular direction? For example, what difference would it make if the boys were 100 kg each instead?

... each weighs 50 kg. They are pulling a piece of against each other as hard as they can...
Their mass doesn't tell you how strong they are. Do you mean they are at the friction limit, for that bodyweight as normal froce?

1. What information do I need to estimate how much force is required to move the point where the third person holds the rope a certain distance in that direction?
Draw a vector diagram. You need the forces that the two intial guys apply.

1. How does the tension of the rope after the pull depend on the initial tension, if the point moved the same distance in the perpendicular direction? For example, what difference would it make if the boys were 100 kg each instead?
If the force magnitude by the initial two guys doesn't change, the tension doesn't change either.

Their mass doesn't tell you how strong they are. Do you mean they are at the friction limit, for that bodyweight as normal froce?

Draw a vector diagram. You need the forces that the two intial guys apply.

If the force magnitude by the initial two guys doesn't change, the tension doesn't change either.

Yes, this is exactly what I meant. How, then, can I calculate the force required to move the point?

Yes, this is exactly what I meant. How, then, can I calculate the force required to move the point?
Start with the vector diagram. You have guy A over here and guy B over there. You have a distance between them.
Draw a straight dotted line between them where the undeflected rope would be.
Make a mark at the midpoint.
Make a mark where the midpoint of the rope will be once guy M deflects it.
Draw in the two rope segments, one from A to the deflected midpoint and one from B to the deflected midpoint.
Draw in a dotted line between the deflected and undeflected midpoints.

Pick a letter to describe the length of the undeflected rope. Maybe "l" ?
Pick a letter to describe the deflection distance. Maybe "d" ?
Pick a letter to describe the tension applied by guy A and guy B. Maybe "T".

Annotate the drawing with these values. Do not bother to give them numbers. We want to work with symbols.

Consider labelling the unknown angle between the deflected rope and the dotted line where the undeflected rope would be, calling that angle ##\theta##.

Now consider a very small section of rope right where guy M is pulling. What forces act on that small section?

• ChessEnthusiast and russ_watters
Is this diagram fine?

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Is this diagram fine?
No, the friction on the two initial guys is parallel to their rope segments. Just draw the 3 force vectors from all 3 guys. They have to balance (vector sum is zero).

• ChessEnthusiast
No, the friction on the two initial guys is parallel to their rope segments. Just draw the 3 force vectors from all 3 guys. They have to balance (vector sum is zero).

There are only 2 forces acting on A and B - friction and tension. They act in the opposite direction. In addition, neither A nor B does move, therefore, does it mean that Tension = Friction (with respect to magnitude)?

There are only 2 forces acting on A and B - friction and tension. They act in the opposite direction. In addition, neither A nor B does move, therefore, does it mean that Tension = Friction (with respect to magnitude)?
As the third guy starts to pull, the direction of the tension acting on the first two will change and so will the friction force on their feet. Friction is the same magnitude as the tension and in the opposite direction for each guy. You just need three lines on your vector diagram. The force from the third guy will be shared by the y components on the diagram of the forces that the other two guys exert.
Of course, if they started of with limiting friction, the third guy will make them slip until equilibrium is reached once again.

neither A nor B does move
If the rope is in-extensible, they will have to move if the rope is deflected from a straight line, by geometry alone.

If its a rubber rope, extending it will increase the tension. So if they were at limiting friction already, they will slip and thus move as well.

If the rope is in-extensible, they will have to move if the rope is deflected from a straight line, by geometry alone.

If its a rubber rope, extending it will increase the tension. So if they were at limiting friction already, they will slip and thus move as well.
One possible simplifying assumption is that the guys evenly and equally pay out as much extra rope as is needed while maintaining the fixed tension. As long as the deflection is small, it does not matter much what simplifying assumption is used.

I suggest we wait for the full information and avoid stumbling down blind alleys. I am sorry for asking a problem without providing all necessary information - I keep forgetting that Physics is science and every detail matters.

The rope is inextensible. Also, it follows straight from the triangle inequality that these guys will have to move - but what direction will it be? Along the rope, or along the AB segment?

I have one more question about this problem. Let's say that we have the force the third guy applies to the rope, as well as the initial force A and B apply. Now, how do I calculate the tension of the rope after the third force has been applied?

Can I assume that the component along the AB segment of the new tension is identical to the initial tension?

The rope is inextensible. Also, it follows straight from the triangle inequality that these guys will have to move - but what direction will it be? Along the rope, or along the AB segment?
It's for them to decide.

Let's say that we have the force the third guy applies to the rope, as well as the initial force A and B apply. Now, how do I calculate the tension of the rope after the third force has been applied?
The force magnitude A and B apply is the tension

Can I assume that the component along the AB segment of the new tension is identical to the initial tension?
Why would that be?

Also, it follows straight from the triangle inequality that these guys will have to move - but what direction will it be? Along the rope, or along the AB segment?
For a slowly applied deflection force, the natural assumption is that they are drawn in the current direction of the partially deflected rope while their feet skid on the floor. This would result in a pursuit curve.

@A.T.
Since tension is the magnitude of the force A and B apply, let's consider such scenario:
A and B pulls with 50 N and have the same mass.
Then, the tension is 50 N and both of them are stationary. The forces acting on each of them are: tension and friction.

The third guy applies his force, say, 50 N and they start skiding.

The only forces acting on A and B are, again, tension of friction.
Since friction has not changed - it's still 50 N, and they move - tension must have increased.

Is my assumption incorrect?

The third guy applies his force, say, 50 N and they start skiding.
The only forces acting on A and B are, again, tension of friction.
Since friction has not changed - it's still 50 N, and they move - tension must have increased.
So one is suddenly applying a 50N force in the middle of the taut rope. If A and B are not massless and have infinitely tight grips on the ideal rope then the tension in the rope must be momentarily infinite. A and B must momentarily experience infinite acceleration. After the transients have settled out, an equilibrium will plausibly be attained with everyone at rest and the tension in the rope back at 50N.

Without careful calculation, it is not immediately obvious whether there are conditions such that A and B would coast past their expected equilibrium positions, resulting in a final tension less than 50N.

Is this diagram fine?

It's not very helpful, even if the above-mentioned errors are corrected.

A better approach might be to draw a diagram of the forces exerted on the point on the rope where the third guy touches it. Thus you have only three forces, call them ##\vec{T}_1##, ##\vec{T}_2##, and ##\vec{T}_3##. The three forces must sum to zero. Remember that just because Guys 1 and 2 have the same mass, it does not mean that ##\vec{T}_1## is necessarily equal in magnitude to ##\vec{T}_2##. As you say, those magnitudes are equal to the friction forces. Guy 1 might be wearing cleats and Guy 2 roller skates. If Guy 3 were to use a pulley with negligible friction then ##\vec{T}_1## and ##\vec{T}_2## would have equal magnitudes.

What information do I need to estimate how much force is required to move the point where the third person holds the rope a certain distance in that direction?

Draw your diagram, as described above, but note that the arrows representing ##\vec{T}_1## and ##\vec{T}_2## will now point in opposite directions. Solve for the magnitude of ##\vec{T}_3##.

How does the tension of the rope after the pull depend on the initial tension, if the point moved the same distance in the perpendicular direction? For example, what difference would it make if the boys were 100 kg each instead?

This pair of questions puzzles me. I don't understand how the scenario described in the second question is an example of the first.

I'm also puzzled by the phrase "same distance". Same distance as what?