Is My Assumption About Rope Tension Correct?

In summary: In this figure, the 5 N should be the hypotenuse, not one of the other sides. You're using the wrong triangle.Base on the above diagram (arrow edited),5N x sin 60 + 5N x sin 60 - W = 0therefore W= 2(5N x sin 60)So the weight of the object is 8.66 N.
  • #1
stuc
20
0

Homework Statement


1.jpg


Homework Equations


My assumption:-
if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 5N load on it.Is it my assumption is correct?

The Attempt at a Solution


Tension of the other rope = 5N (is it correct)?? base on my assumption [/B]
 
Physics news on Phys.org
  • #2
stuc said:
My assumption:-
if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 5N load on it.Is it my assumption is correct?
I would assume they mean the tension at the instant after cutting the other string.
 
  • Like
Likes stuc
  • #3
Have you studied force vectors? I think your assumption is half correct and your conclusion is wrong. What you are right about is the hanging vertically. All of your statements about force are wrong.
 
  • Like
Likes stuc
  • #4
haruspex said:
I would assume they mean the tension at the instant after cutting the other string.
Hm ... could be. I didn't interpret it that way but I guess that could be right. stuc, my comments are not based on that assumption.
 
  • Like
Likes stuc
  • #5
haruspex said:
I would assume they mean the tension at the instant after cutting the other string.
...the tension at the instant after cutting the other string

how to calculate it? it mean my assumption is wrong

i just check an answer..the answer is 5N...but how to get it
 
  • #6
stuc said:
.the answer is 5N.
Hmm.. that does not fit with either interpretation. But I remain of the view that it's the instantaneous value they're after.
First, can you figure out the weight of the mass?
 
  • Like
Likes stuc
  • #7
You have not answered my question about whether or not you have studied force vectors.
 
  • Like
Likes stuc
  • #8
haruspex said:
Hmm.. that does not fit with either interpretation. But I remain of the view that it's the instantaneous value they're after.
First, can you figure out the weight of the mass?
there is no weight given.:H
 
  • #9
phinds said:
You have not answered my question about whether or not you have studied force vectors.
learn but always confuse when try to solve it
 
  • #10
stuc said:
there is no weight given.:H
What is the definition of a Newton?
 
  • #11
phinds said:
What is the definition of a Newton?
1N = the force needed to accelerate 1kg of mass
 
  • #12
1 ans.jpg


Attached is the calculation..any one can verify my ans (4.33N) because the given ans in 5N.

Update:Wrong Arrow for Fy
 
Last edited:
  • #13
Untitled.png

Attached other option ans
 
  • #14
As I posted, the first step is to determine the weight of the mass. No, it's not given, but there is enough information to find it. To do this, just consider the initial set up with two ropes. Write the equation for balance of forces in the vertical direction.
 
  • Like
Likes stuc
  • #15
stuc said:
Attached is the calculation..any one can verify my ans (4.33N) because the given ans in 5N.

Update:Wrong Arrow for Fy
You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

Chet
 
  • Like
Likes stuc
  • #16
haruspex said:
As I posted, the first step is to determine the weight of the mass. No, it's not given, but there is enough information to find it. To do this, just consider the initial set up with two ropes. Write the equation for balance of forces in the vertical direction.

My step to determine the weight of the mass..is it ok? (weight of the mass = 7.071N)
Snapshot.jpg
 
  • #17
Chestermiller said:
You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

Chet
weight of the mass = 7.071N
Snapshot.jpg
 
  • #18
stuc said:
My step to determine the weight of the mass..is it ok?
View attachment 91455
No, that doesn't work. Pythagoras applies when two vectors at right angles are to be added.
What is the component of each tension in the vertical direction?
 
  • Like
Likes stuc
  • #19
haruspex said:
No, that doesn't work. Pythagoras applies when two vectors at right angles are to be added.
What is the component of each tension in the vertical direction?
:nb)..tq for the info..will be calculate
 
  • #20
stuc said:
:nb)..tq for the info..will be calculate
Hint: you already calculated it in post #12.
 
  • #21
stuc said:
My step to determine the weight of the mass..is it ok? (weight of the mass = 7.071N)
View attachment 91455
In this figure, the 5 N should be the hypotenuse, not one of the other sides. You're using the wrong triangle.
 
  • Like
Likes stuc
  • #22
haruspex said:
Hint: you already calculated it in post #12.
2 ans.jpg


Base on the above diagram (arrow edited),

5N x sin 60 + 5N x sin 60 - W = 0
therefore W= 2(5N x sin 60)
= 8.66 N
 
  • #23
Chestermiller said:
You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

Chet
i get W=8.66N ..but how the answer is 5N:cry:
 
  • #24
stuc said:
i get W=8.66N ..but how the answer is 5N:cry:
Steady on, so far all we have found is the hanging weight. The tension in one string immediately after the other is cut might be different. So leave the weight in the form ##10\sin(60)N## for now.
Now draw a free body diagram for the weight just after one string is cut. What equation can you obtain for the tension in the remaining string.
 
  • #25
haruspex said:
Steady on, so far all we have found is the hanging weight. The tension in one string immediately after the other is cut might be different. So leave the weight in the form ##10\sin(60)N## for now.
Now draw a free body diagram for the weight just after one string is cut. What equation can you obtain for the tension in the remaining string.
3 ans.jpg


the ans is 10N after cut the 5N right rope...huhu near but still cannot get the ans
 
  • #26
stuc said:
View attachment 91592

the ans is 10N after cut the 5N right rope...huhu near but still cannot get the ans
After a rope is cut, the system is no longer stable. There will be an acceleration. The forces in the vertical direction will no longer balance.
You need to consider which way that acceleration will be and apply ##\Sigma F=0## at right angles to it.
 
  • #27
haruspex said:
After a rope is cut, the system is no longer stable. There will be an acceleration. The forces in the vertical direction will no longer balance.
You need to consider which way that acceleration will be and apply ##\Sigma F=0## at right angles to it.
if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 10sin(60)N load on it.Is it my assumption is correct?

it mean all the calculation in step 25 not require especially step 2 &3 ?
 
  • #28
stuc said:
if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 10sin(60)N load on it.Is it my assumption is correct?
Not immediately. Although the question does not make it completely clear, I am very confident that you are supposed to find the tension in the remaining rope the instant after one rope is cut. You can take the remaining rope and the mass as being in exactly the same positions as before, and, for the instant, stationary. But now there is a nonzero acceleration.
Which way will the mass accelerate?
 
  • Like
Likes stuc
  • #29
haruspex said:
Not immediately. Although the question does not make it completely clear, I am very confident that you are supposed to find the tension in the remaining rope the instant after one rope is cut. You can take the remaining rope and the mass as being in exactly the same positions as before, and, for the instant, stationary. But now there is a nonzero acceleration.
Which way will the mass accelerate?

Do you mean the answer is wrong (Given ans: 5N)..i still stuck
 
  • #30
stuc said:
Do you mean the answer is wrong (Given ans: 5N)..i still stuck
Yes, it's wrong. (That would be the right answer if the angle were 45 degrees.)
You have established that the weight is 10sin(60) N. Draw a free body diagram, exactly like the original diagram, but with only one rope.
So the rope is still at 60 degrees to the horizontal.
Let the tension in the rope be T. You know the weight, and hence the mass. The tension and the weight are the only forces acting, agreed?
Which way will the mass accelerate?
What can you therefore say about the forces orthogonal to that direction?
 
  • Like
Likes stuc
  • #31
haruspex said:
Yes, it's wrong. (That would be the right answer if the angle were 45 degrees.)
You have established that the weight is 10sin(60) N. Draw a free body diagram, exactly like the original diagram, but with only one rope.
So the rope is still at 60 degrees to the horizontal.
Let the tension in the rope be T. You know the weight, and hence the mass. The tension and the weight are the only forces acting, agreed?
Which way will the mass accelerate?
What can you therefore say about the forces orthogonal to that direction?

thank you very much for your guide..i will try your idea. i hope you have time to check it
 
  • #32
stuc said:
thank you very much for your guide..i will try your idea. i hope you have time to check it
haruspex said:
Yes, it's wrong. (That would be the right answer if the angle were 45 degrees.)
You have established that the weight is 10sin(60) N. Draw a free body diagram, exactly like the original diagram, but with only one rope.
So the rope is still at 60 degrees to the horizontal.
Let the tension in the rope be T. You know the weight, and hence the mass. The tension and the weight are the only forces acting, agreed?
Which way will the mass accelerate?
What can you therefore say about the forces orthogonal to that direction?
4 ans.jpg


dear haruspex ... is it like this?
 
  • #33
stuc said:
View attachment 92448

dear haruspex ... is it like this?
The free body diagram is right, but you need to think about which way the acceleration must be. In your force analysis, you have summed the vertical components and assumed the total to be zero. In other words, you have supposed the acceleration to be horizontal. Clearly it will not be. Which way will it be?
 

FAQ: Is My Assumption About Rope Tension Correct?

1. What is a rope tension mystery?

A rope tension mystery refers to a situation where the tension in a rope cannot be easily determined or is unexpected. This can occur due to various factors such as the weight of the object being lifted, the angle of the rope, and the type of rope being used.

2. How do you solve a rope tension mystery?

To solve a rope tension mystery, you need to use the principles of physics and engineering. This involves calculating the forces acting on the rope, considering the properties of the rope and the objects it is supporting, and using equations such as Newton's laws of motion to determine the tension in the rope.

3. What are some common factors that can affect rope tension?

The weight of the object being lifted, the angle of the rope, the type and condition of the rope, and any external forces such as wind or friction can all affect the tension in a rope. Other factors may include the length and thickness of the rope, the method of attachment, and the surface the rope is resting on.

4. How can you accurately measure rope tension?

There are several methods for measuring rope tension, including using a tension meter, a dynamometer, or a load cell. These devices can provide a numerical value for the tension in the rope, but it is important to also consider other factors that may affect the tension, such as those mentioned above.

5. What are some real-world applications of solving rope tension mysteries?

Solving rope tension mysteries has practical applications in various industries, such as construction, engineering, and transportation. For example, understanding the tension in cables and ropes used in bridges and cranes is crucial for ensuring their structural integrity and safety. It can also be useful in everyday situations, such as setting up a hammock or securing a load on a truck or trailer.

Back
Top