Find the tension in the wire from this mass hanging at an angle

AI Thread Summary
The discussion revolves around solving a physics problem involving tension in a wire supporting a 6 kg mass at an angle. Participants emphasize the need for the original poster to show their own attempts at solving the problem, as per forum rules. There is confusion regarding the angles and the setup of the problem, with suggestions to clarify whether the beam is rigid or pivoted. The importance of drawing a free body diagram and resolving forces into components is highlighted as essential for finding the solution. Ultimately, the thread concludes with a reminder that direct answers cannot be provided without the user's effort.
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Homework Statement
Construction is in balance. The weight of the hanging body is 6 kg. The angle created between the wire and the wall is 30 degrees.
Relevant Equations
Find the thread tension in the construction shown in the figure.
help.jpg

I do not understand how this type of exercise is solved because I have not had a solution to such an example. I would ask you to show me the solution of this exercise step by step, so that I understand how to solve it. I appreciate your help. Thank you.
 
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besi said:
Homework Statement:: Construction is in balance. The weight of the hanging body is 6 kg. The angle created between the wire and the wall is 30 degrees.
Relevant Equations:: Find the thread tension in the construction shown in the figure.

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I do not understand how this type of exercise is solved because I have not had a solution to such an example. I would ask you to show me the solution of this exercise step by step, so that I understand how to solve it. I appreciate your help. Thank you.
Hi besi. The forum rules require you to make some sort of attempt first. Take a look:
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Please note, you say “The angle created between the wire and the wall is 30 degrees”. But your diagram indicates this angle is 90º – 20º = 70º. So something is wrong!

The simplest method requires you to know how to resolve a vector into components If you don’t know how to do this already, you could try this video (sorry about the self-promotion!):
Then maybe you can make an attempt at the problem.
 
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Steve4Physics said:
Hi besi. The forum rules require you to make some sort of attempt first. Take a look:
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Please note, you say “The angle created between the wire and the wall is 30 degrees”. But your diagram indicates this angle is 90º – 20º = 70º. So something is wrong!

The simplest method requires you to know how to resolve a vector into components If you don’t know how to do this already, you could try this video (sorry about the self-promotion!):
Then maybe you can make an attempt at the problem.

THANK YOU!
 
besi said:
THANK YOU!
Besi hi! Can u show me how did u solved this example please ?
 
hannah13 said:
Besi hi! Can u show me how did u solved this example please ?
Hi Hannah,
Besi might not be watching this thread any more.
Since the forum rules say anyone seeking assistance needs to show their own effort first, I suggest you do so in a new thread.
 
haruspex said:
Hi Hannah,
Besi might not be watching this thread any more.
Since the forum rules say anyone seeking assistance needs to show their own effort first, I suggest you do so in a new thread.
Hello! My physics professor gave us this kind of exercise but i couldn’t solve it. Please who can help me solve it. Construction is in balance. The weight of the hanging body is 6 kg. The angle is 20°
 

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Hannah, much depends upon how the beam is attached to the wall. Is it embedded in the wall, or is it pivoted at the wall?

Do you know how to draw a free body diagram (FBD)?
 
Dr.D said:
Hannah, much depends upon how the beam is attached to the wall. Is it embedded in the wall, or is it pivoted at the wall?

Do you know how to draw a free body diagram (FBD)?
I think is embedded in the wall. Not sure but it’s exactly like the besi’s type of exercise
 
hannah13 said:
I think is embedded in the wall. Not sure but it’s exactly like the besi’s type of exercise
Yes, it is exactly the same, even in that it is not made clear whether the joint at the wall
  • is rigid, in which case we must assume the string is free to slide over the other end of the beam in order to be able to solve it, or
  • is a pivot, in which case the string must be tied to the end of the beam or it would collapse
In each diagram, it looks rigid, so let's go with that. So the string is free to slide.
Consider the forces acting along the string as it goes around the end of the beam. Since it does not slide, what can you deduce?
 
  • #10
haruspex said:
Yes, it is exactly the same, even in that it is not made clear whether the joint at the wall
  • is rigid, in which case we must assume the string is free to slide over the other end of the beam in order to be able to solve it, or
  • is a pivot, in which case the string must be tied to the end of the beam or it would collapse
In each diagram, it looks rigid, so let's go with that. So the string is free to slide.
Consider the forces acting along the string as it goes around the end of the beam. Since it does not slide, what can you deduce?
I think T might be cos20°+d but I’m not sure. I really need the answer of this exercise. Do u know it? Please
 
  • #11
hannah13 said:
I think T might be cos20°+d but I’m not sure. I really need the answer of this exercise. Do u know it? Please
As I mentioned, forum rules require you to show more effort.
Why do you think it might be cos20°+d, and what is d?
Oh, I should have mentioned two more things.
  1. With the rigid joint assumption you will also need to assume the edge of the beam where the string slides is smooth.
  2. Although I have recommended to go with the rigid joint assumption based on the drawings, in every other problem like this I have ever seen posted on this forum, the string is tied to the end of the beam and the beam is pivoted at the wall. Should I trust these drawings? They are not from the teacher.
 
  • #12
haruspex said:
As I mentioned, forum rules require you to show more effort.
Why do you think it might be cos20°+d, and what is d?
Oh, I should have mentioned two more things.
  1. With the rigid joint assumption you will also need to assume the edge of the beam where the string slides is smooth.
  2. Although I have recommended to go with the rigid joint assumption based on the drawings, in every other problem like this I have ever seen posted on this forum, the string is tied to the end of the beam and the beam is pivoted at the wall. Should I trust these drawings? They are not from the teacher.
d is the 4 m. The drawings are from the teacher but the problem is that I can’t solve it 😅. Also I can I show more effort?
 
  • #13
hannah13 said:
d is the 4 m. The drawings are from the teacher
Ok, we'll stick with the rigid joint then.
As I wrote, take the end of the beam the string slides over as smooth. So you can think of it as a pulley.
Do you know how to take moments about an axis?
 
  • #14
haruspex said:
Ok, we'll stick with the rigid joint then.
As I wrote, take the end of the beam the string slides over as smooth. So you can think of it as a pulley.
Do you know how to take moments about an axis?
Yes I know how to take moments about an axis but not on this type of exercise. I don’t know , I’m really confused because calculate the forces
 
  • #15
hannah13 said:
Yes I know how to take moments about an axis but not on this type of exercise. I don’t know , I’m really confused because calculate the forces
I can’t calculate the forces *
 
  • #16
hannah13 said:
Yes I know how to take moments about an axis but not on this type of exercise. I don’t know , I’m really confused because calculate the forces
As I wrote, you can think of the string as running over a pulley at the end of the beam.
There are two tensions in the string, T pulling anticlockwise and the weight pulling clockwise. The 'pulley' doesn’t spin. What does that tell you?
 
  • #17
haruspex said:
As I wrote, you can think of the string as running over a pulley at the end of the beam.
There are two tensions in the string, T pulling anticlockwise and the weight pulling clockwise. The 'pulley' doesn’t spin. What does that tell you?
This means that the construction is in balance. T + P = 0
P = mg
that's all i can do
 
  • #18
hannah13 said:
This means that the construction is in balance. T + P = 0
P = mg
that's all i can do
You are very close.
The sum of the moments is zero, but they act in opposite directions. So what equation relates the tensions?
 
  • #19
haruspex said:
You are very close.
The sum of the moments is zero, but they act in opposite directions. So what equation relates the tensions?
I think I would be T= 60N + 4*sin30°
I don’t know if it’s right , so please can u tell me the answer if u know it . 😢
 
  • #20
hannah13 said:
I think I would be T= 60N + 4*sin30°
That makes no sense. 4m sin 30 would be a distance. You cannot add a distance to a force.
As I posted, you were very close before. Please don't go off in some other direction.
The two torques balance, but your equation in post #17 would make one of the tensions negative.
The tensions are both positive. They balance because they act in opposite directions around the 'pulley'. So what should your equation be?
 
  • #21
haruspex said:
That makes no sense. 4m sin 30 would be a distance. You cannot add a distance to a force.
As I posted, you were very close before. Please don't go off in some other direction.
The two torques balance, but your equation in post #17 would make one of the tensions negative.
The tensions are both positive. They balance because they act in opposite directions around the 'pulley'. So what should your equation be?
Is T= -P or something ?
 
  • #22
hannah13 said:
Is T= -P or something ?
No, that's what you wrote before, as T+P=0.
That equation would make one positive and the other negative.
The two tensions are pulling in opposite directions around the end of the beam, and balance, so each is, as a tension, positive. What tiny change do you need to make?
 
  • #23
haruspex said:
No, that's what you wrote before, as T+P=0.
That equation would make one positive and the other negative.
The two tensions are pulling in opposite directions around the end of the beam, and balance, so each is, as a tension, positive. What tiny change do you need to make?
I don’t know. I tried over and over but i still can’t do it. Can u help me pls . Seems like u know the answer. I would be happy if u tell me what’s the answer.
 
  • #24
hannah13 said:
I don’t know. I tried over and over but i still can’t do it. Can u help me pls . Seems like u know the answer. I would be happy if u tell me what’s the answer.
This homework thread is closed.

@hannah13 -- First please STOP using text speak in your replies -- that is against the PF rules. Second, we do NOT give out answers here. That is also in the PF rules. You can re-read the PF rules that you agreed to when you joined the PF by clicking on INFO at the top of the page.

If you would like help with this problem or other schoolwork-type problems, please start a new thread in the Homework Help forums and post your best effort to work on the problem. We can only provide tutorial help after you show lots of effort on your part. Thank you.
 
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