A basic rope tension problem

In summary: A doesn't let go"?In summary, the tension in a massless rope must be the same throughout...when you make up these problems, you get into difficulties.
  • #1
FisherDude
23
0

Homework Statement


If a rope were being pulled in opposite directions with two opposing forces, one being 5 N and the other 7 N, what is the tension of the rope?

Homework Equations


The Attempt at a Solution


Since the rope would be accelerating in the direction of the force of 7 N, the reaction of the rope which I think is the tension should be 5 N, though I'm not really sure of this. It can't be 7 N, because otherwise the rope wouldn't move right? (is that called being in equilibrium?)
 
Physics news on Phys.org
  • #2
This question is bit unclear. Does the rope have mass? If so, it must be accelerating with a net force of 2N, and it's tension will be variable. If it has no mass, the problem doesn't make any sense to me. It can't be in equilibrium, either. Please clarify.
 
  • #3
Actually I made up the problem myself in an attempt to understand tension...let's say the rope has no mass, why wouldn't it make any sense? I'm just basing it off the concept of the same problem, except the forces are both 5 N, the rope has no mass, which means, according to my textbook, that the tension is 5 N, unless I'm seriously misreading it. And I meant that it would be in equilibrium if both forces were 7 N...

Sorry, let me rephrase the question as two people playing tug of war on a rope, one person is exerting 5 N on the rope, the other person is exerting 7 N on the rope.
 
Last edited:
  • #4
FisherDude said:
Actually I made up the problem myself in an attempt to understand tension...let's say the rope has no mass, why wouldn't it make any sense? I'm just basing it off the concept of the same problem, except the forces are both 5 N, the rope has no mass, which means, according to my textbook, that the tension is 5 N. And I meant that it would be in equilibrium if both forces were 7 N...
The tension in a massless rope must be the same throughout...when you make up these problems, you get into difficulties. Let's assume you have a 0.5 Kg block (weight = 5N) hanging vertically from the bottom of a rope you are holding. If there is no movement, you must ber pulling up with a force of 5N, and the rope tension is 5N. Now let's say you start to pull up on the block with a force of 7N. The block accelerates upward. The rope tension is 7N throughout. It is not 5N. nor is it 2N. There is no tension change in a rope of negligible mass. I'm not sure though if this is your question.
 
  • #5
Upon further research I discovered that in a tug of war situation the force exerted by both ends are the same...so I guess that's why my question didnt make sense...

Though I don't really see how that's possible.. if a person was stronger than the other person at the other end of the rope, why wouldn't he exert more force when pulling the rope?

Thanks for the response btw!
 
  • #6
FisherDude said:
Upon further research I discovered that in a tug of war situation the force exerted by both ends are the same...so I guess that's why my question didnt make sense...

Though I don't really see how that's possible.. if a person was stronger than the other person at the other end of the rope, why wouldn't he exert more force when pulling the rope?

Thanks for the response btw!
That's a good question. They both pull on the rope with the same force, yet someone wins. That is because another force comes into play: friction between the persons and the ground. The person who is able to push on the ground harder wins. Consider a special case where one person A is standing on an icy frictionless surface, and the other person B at the other end is standing on rough ground. When they each pull on the rope with a force of say 200N each, the person on the ice, A, will move toward the other person, B, who will haul him in hand over hand. They both exert the same force as long as A doesn't let go. A acelerates toward B (there is a net force of 200N acting on him), but B remains stationary (no net force on B since the rope exerts 200N on him, but the ground friction exerts an opposite 200N force on him.
 
  • #7
For your understanding...solve the problem now if the mass of the rope is .5 kg.:p
 
  • #8
I have read you conservation and I still don't understand, why did you write that: "They both pull on the rope with the same force"?
 
  • #9
Lily2502 said:
I have read you conservation and I still don't understand, why did you write that: "They both pull on the rope with the same force"?
Hi, Lily, welcome to PF!
The tension in a rope of negligible mass is always the same at each end of the rope (or in between) , whether the objects or persons are stationary, moving at constant speed, or accelerating. The rope just serves to transmit the force from one end to the other. In a game of tug of war, if I pull on you with a force of 10 N, then you must pull back on me with a force of 10 N. If I increase the tension to 100 N pulling on you, you must exert a force of 100 N on me. In this last example, if no one is moving, the ground must exert a backwards force ( by static friction, or digging your feet into the ground, for example) of 100 N on you, and the ground must exert a backwards force of 100 N on me. Ultimately, assuming you or I don't let go of the rope, as we increase the tension force, the ground friction can no longer support that increase on the person who doesn't provide enough force to the ground, and that person will now start moving in the direction of the ultimate winner, and the tension force in the rope, whatever it is, say 150 N, will still be the same force acting on both of us.

Does this help?
 
  • #10
"Let's assume you have a 0.5 Kg block (weight = 5N) hanging vertically from the bottom of a rope you are holding. If there is no movement, you must ber pulling up with a force of 5N, and the rope tension is 5N. Now let's say you start to pull up on the block with a force of 7N. The block accelerates upward. The rope tension is 7N throughout."

"The rope just serves to transmit the force from one end to the other. "

These two statements don't seem to make sense to me. If the tension in the rope is 7N (the force you use to pull the block upwards), this is transmitted to the block, making it accelerate upwards. This makes sense.

But then if I look at it from the opposite perspective, the block is pulling down on your hand with 5N. Therefore, shouldn't the tension of the rope be 5N from this perspective (in accordance with the second statement)?

I'm getting myself very confused, so hopefully what I said makes sense.
 
  • #11
QuarkyPhysics said:
"Let's assume you have a 0.5 Kg block (weight = 5N) hanging vertically from the bottom of a rope you are holding. If there is no movement, you must ber pulling up with a force of 5N, and the rope tension is 5N. Now let's say you start to pull up on the block with a force of 7N. The block accelerates upward. The rope tension is 7N throughout."

"The rope just serves to transmit the force from one end to the other. "

These two statements don't seem to make sense to me. If the tension in the rope is 7N (the force you use to pull the block upwards), this is transmitted to the block, making it accelerate upwards. This makes sense.

But then if I look at it from the opposite perspective, the block is pulling down on your hand with 5N. Therefore, shouldn't the tension of the rope be 5N from this perspective (in accordance with the second statement)?

I'm getting myself very confused, so hopefully what I said makes sense.
The block would exert a force of 5 N on the rope ONLY if it was not accelerating (that is, only if it were at rest or moving at constant velocity). However, the block IS accelerating, and thus there must be a net force acting on the block. If you draw a free body diagram of the block, there are 2 forces acting on it, it's weight of 5 N acting down, and the tension force acting up. Since the block is accelerating upward, then T must be greater than 5 N. Per Newton 2, Fnet = ma, and Fnet = 7 - 5 = 2 N in this case. And since the person exerts an upward force of 7 N on the block, the block must exert a downward force of 7 N on the man, per Newton 3. Don't confuse Newton 2 with Newton 3.
 

1. What is a basic rope tension problem?

A basic rope tension problem is a physics problem that involves calculating the tension force in a rope that is supporting an object. It is commonly used in engineering and construction to determine the strength and stability of structures.

2. How do you solve a basic rope tension problem?

To solve a basic rope tension problem, you need to identify all the forces acting on the rope and use Newton's second law of motion (F=ma) to calculate the tension force. You will also need to consider the angle and direction of the rope, as well as the weight and mass of the object being supported.

3. What are some real-life applications of basic rope tension problems?

Basic rope tension problems are commonly used in construction and engineering to design and test the strength of structures such as bridges, cranes, and suspension systems. They are also used in rock climbing and mountaineering to determine the safety and stability of ropes and anchors.

4. What factors can affect the tension force in a rope?

The tension force in a rope can be affected by the weight of the object being supported, the mass of the rope itself, the angle and direction of the rope, and any external forces acting on the rope such as wind or friction. The material and thickness of the rope can also play a role in determining the tension force.

5. Are there any common mistakes to avoid when solving a basic rope tension problem?

One common mistake to avoid when solving a basic rope tension problem is forgetting to take into account the weight and mass of the rope itself. Another mistake is not considering the direction and angle of the rope, which can greatly affect the tension force. It is also important to carefully identify and label all the forces acting on the rope to ensure an accurate calculation.

Similar threads

  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
7K
  • Introductory Physics Homework Help
Replies
2
Views
699
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
25
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
1K
Back
Top