# What force is needed to break a rope?

• Lotto
Lotto
Homework Statement
I pull a rope tied to a tree with 10 N. The force needed to break a rope is 15 N. Will I break the rope? What if two people pull the rope from opposite ends with forces 10 N and -10 N. Will the rope break now?
Relevant Equations
Newtons third law of motion
I would say that it will not break in both cases, since tension force is 10 N, no matter if I pull it against a tree or a person.

But I am not sure. How is it?

Lotto said:
since tension force is 10 N, no matter if I pull it against a tree or a person.
You are exactly right.

Lotto said:
Homework Statement: I pull a rope tied to a tree with 10 N. The force needed to break a rope is 15 N. Will I break the rope? What if two people pull the rope from opposite ends with forces 10 N and -10 N. Will the rope break now?
Relevant Equations: Newtons third law of motion

I would say that it will not break in both cases, since tension force is 10 N, no matter if I pull it against a tree or a person.

But I am not sure. How is it?
Sketch a FBD for each case.

In the first, we have a person acting on the rope to, say, the left of 10 N. On the other end we have the tree in a fixed position. Since ideal ropes transmit force, there is a force of 10 N on this tree, acting to the left. Thus, by Newton's 3rd, it is exerting a force of 10 N on rope, acting to the right. The tension in the rope will be will be 10 N, so it will not break.

In the second case, we have almost the same situation. Note that, if the person on the other end didn't move or exert any force, then the tension in the rope is again 10 N. But the person is now exerting a 10 N force to the right, in addition to the force exerted by them by staying still. So now there is an extra 10 N acting on the rope and thus the tension is 20 N. So the rope breaks.

-Dan

PeroK and jbriggs444
topsquark said:
Note that, if the person on the other end didn't move or exert any force,
Then the rope could not be subject to any force from that person. It would be a massless object subject to a non-zero net force. Which is a contradiction to Newton's second law.

Write down the force balance and you will see that to balance forces on the rope, the other person must be exerting a 10 N force, equal and opposite to the force exerted by us. Those two forces are a "second law pair".

The other person exerts the same force on the rope that the tree exerts on the rope. @Lotto has it right. You have it all wrong.

PeroK, Lnewqban and topsquark
jbriggs444 said:
"second law pair".
Shouldn’t it be a third law pair?

topsquark
Frabjous said:
Shouldn’t it be a third law pair?
A third law pair is matching up the force of A directly on B with the force of B directly on A. Yes, the third law states that those must be equal and opposite.

If you have an ideal massless object between A and B (call it R for rope) then you have the force of A on R and the force of B on R. Those are not third law partner forces.

But if you write down Newton's second law: ##\sum F = ma## you get ##F_\text{AR} + F_\text{BR} = m_Ra_R = 0## and you can conclude that ##F_\text{AR} = -F_\text{BR}##. I use the term "second law pair" to refer to the two forces in such a situation.

Admittedly, if you are used to reading "third law partner", the notion of a "second law partner" may be jarring. If so, feel free to ignore the phrase and substitute the second law logic.

Frabjous and topsquark
jbriggs444 said:
Then the rope could not be subject to any force from that person. It would be a massless object subject to a non-zero net force. Which is a contradiction to Newton's second law.

Write down the force balance and you will see that to balance forces on the rope, the other person must be exerting a 10 N force, equal and opposite to the force exerted by us. Those two forces are a "second law pair".

The other person exerts the same force on the rope that the tree exerts on the rope. @Lotto has it right. You have it all wrong.
The issue, then, is that I'm reading the question wrong. I had taken the person to be adding an extra 10 N on top of the 10 N they needed to remain motionless.

Thanks for the catch!

My apologies to all.

-Dan

jbriggs444
Lotto said:
I would say that it will not break in both cases, since tension force is 10 N, no matter if I pull it against a tree or a person.
Right. The person pulls with 10 N of force, the tree pulls back with 10 N of force.
In the other case, a person pulls with 10 N of force and a second person pulls back with 10 N of force.

It makes no difference if it's the tree or the second person that's doing the pulling. You get 10 N of tension in the rope either way.

jbriggs444 and topsquark
jbriggs444 said:
A third law pair is matching up the force of A directly on B with the force of B directly on A. Yes, the third law states that those must be equal and opposite.

If you have an ideal massless object between A and B (call it R for rope) then you have the force of A on R and the force of B on R. Those are not third law partner forces.

But if you write down Newton's second law: ##\sum F = ma## you get ##F_\text{AR} + F_\text{BR} = m_Ra_R = 0## and you can conclude that ##F_\text{AR} = -F_\text{BR}##. I use the term "second law pair" to refer to the two forces in such a situation.

Admittedly, if you are used to reading "third law partner", the notion of a "second law partner" may be jarring. If so, feel free to ignore the phrase and substitute the second law logic.
I thought the third law was that a robot must protect its own existence as long as such protection does not conflict with the first or second law. I don’t see what that has to do with this rope problem.

jbriggs444 and Frabjous

## What factors determine the force needed to break a rope?

The force required to break a rope depends on several factors, including the material of the rope, its diameter, the type of weave or braid, and the conditions under which it is used (e.g., wet or dry, presence of abrasions). The tensile strength of the material is a primary determinant.

## How do you calculate the breaking strength of a rope?

The breaking strength of a rope can be calculated using the tensile strength of the rope's material and its cross-sectional area. Manufacturers often provide a breaking strength rating, which is determined through standardized testing. The formula generally used is: Breaking Strength = Tensile Strength x Cross-sectional Area.

## Does the age of a rope affect its breaking strength?

Yes, the age of a rope can significantly affect its breaking strength. Over time, exposure to environmental factors such as UV radiation, moisture, and physical wear can degrade the rope's fibers, reducing its overall strength and making it more susceptible to breaking under lower forces.

## Can knots in a rope reduce its breaking strength?

Knots can indeed reduce the breaking strength of a rope. When a rope is knotted, the fibers experience uneven stress distribution, which can lead to localized points of high tension. This can cause the rope to break at a lower force than its rated breaking strength. The reduction in strength can vary depending on the type of knot used.

## What safety factors should be considered when determining the force needed to break a rope?

When determining the force needed to break a rope, it is important to consider safety factors to account for uncertainties and variations in real-world conditions. A common safety factor is to use a rope with a breaking strength that is several times greater than the maximum expected load. This helps ensure safety and reliability, especially in critical applications such as climbing, lifting, or rescue operations.

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