Catwalk attached to a wall by a hinged held up by a cord

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In summary, a folding catwalk of uniform density and length 3r is attached to a wall by a hinge. At the other end, it is supported by a cable which makes an angle of θ above the horizontal. If a person of mass m stands a distance r from the wall, the tension, T, in the support cable is equal to 4mg/3sinθ, where g is the gravitational field and the mass of the catwalk is 2m. The net moment at the hinge is 4mg/3=3sinθ. To find T, use the moment equation, N=rF, where N is the moment, r is the distance from the point of rotation, and F is the force acting
  • #1
XxseanxX_22@h
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Homework Statement


A folding catwalk of uniform density and length 3r is attached to a wall by a hinge. At the other end it is supported by a cable which makes an angle of θ above the horizontal. If a person of mass m stands a distance r from the wall what is the tension, T, in the support cable? The gravitational field is a constant g and the mass of the catwalk itself is 2m.

The Attempt at a Solution


i got (3mg/sintheta) but i don't think this is correct
 
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  • #2
Welcome to PF!

Hi XxseanxX_22@h! Welcome to PF! :smile:
XxseanxX_22@h said:
A folding catwalk of uniform density and length 3r is attached to a wall by a hinge. At the other end it is supported by a cable which makes an angle of θ above the horizontal. If a person of mass m stands a distance r from the wall what is the tension, T, in the support cable? The gravitational field is a constant g and the mass of the catwalk itself is 2m.

i got (3mg/sintheta) but i don't think this is correct

hmm … you took vertical components, but you left out the force at the hinge.

Hint: when you've an unknown force like the reaction at the hinge, take moments about some point. :smile:
 
  • #3
thanks i figured it out
 
  • #4
I thought that the net moment at a hinge was 0 as the distance from rotation is 0. Is this wrong?
 
  • #5
Shambles said:
I thought that the net moment at a hinge was 0 as the distance from rotation is 0. Is this wrong?

Hi Shambles!

No, you're absolutely right …

and that's why we take moments about the hinge …

the force there is unknown, and the question doesn't ask for it …

if we took moments about anywhere else, the force at the hinge would come in, and we'd have to use another equation to eliminate it …

by taking moments about the hinge, we reduce the number of equations we need by one! :biggrin:
 
  • #6
I know this post is very old, but I have the identical question on my physics assignment. I found out that the forces in the situation equal 3mg=Ty (Ty being the y-component of the tension in the wire) and for the net moment I have sintheta=4mg/3. Are these right, and if so how can I combine them to find T?
 
  • #7
Hi SeannyBoi71! :smile:

(have a theta: θ :wink:)
SeannyBoi71 said:
… for the net moment I have sintheta=4mg/3

hmm … that's almost right :confused:

show us how you got that, so we can see what went wrong :smile:
 
  • #8
I have moment is N=rF. so moment acting clockwise as 2mg3/2r+mgr, and the opposite 3rsinθ. so simplifying, 4mg=3sinθ, and then 4mg/3=sinθ.
I used 3/2r because the weight of the beam has a uniform density and so the moment acts right in the middle of the beam, did I do that right?
 
  • #9
SeannyBoi71 said:
I have moment is N=rF. so moment acting clockwise as 2mg3/2r+mgr, and the opposite 3rsinθ. so simplifying, 4mg=3sinθ, and then 4mg/3=sinθ.
I used 3/2r because the weight of the beam has a uniform density and so the moment acts right in the middle of the beam, did I do that right?

Yes, I thought you'd done that, and I hoped you'd see what you'd left out once you typed it out …

where's the tension?? :wink:
 
  • #10
I found T= Ty/sinθ, but I don't know how to incorporate that with the rest of the information I found!
 
  • #11
Hi SeannyBoi71! :smile:

(just got up :zzz: …)

SeannyBoi71 said:
I found T= Ty/sinθ, but I don't know how to incorporate that with the rest of the information I found!

ok, when you take https://www.physicsforums.com/library.php?do=view_item&itemid=64" , as you know, every moment has to be a force times a distance, so make sure that that's what your equation has!

in this case, you multiplied your 3r/2 and r by the appropriate force (the weights), but you didn't multiply your 3rsinθ by its force, which is T …

that should give you the equation for T that you're looking for! :wink:

(btw, you'll never need Ty, or any other component, for moments … for moments, you always use the whole force)
 
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  • #12
tiny-tim said:
in this case, you multiplied your 3r/2 and r by the appropriate force (the weights), but you didn't multiply your 3rsinθ by its force, which is T …

that should give you the equation for T that you're looking for! :wink:

(btw, you'll never need Ty, or any other component, for moments … for moments, you always use the whole force)

Ok, but if I'm always using the whole force, why would I even need sinθ? Isn't that one of the component parts of the Tension that is at an angle?? :confused: :bugeye:

As you can see I get very confused with moments and equilibrium and stuff.. it's very hard for me to grasp the concept! But thank you very much for your help so far!
 
  • #13
SeannyBoi71 said:
Ok, but if I'm always using the whole force, why would I even need sinθ? Isn't that one of the component parts of the Tension that is at an angle?? :confused: :bugeye:

ah, now I see what you're doing!

let's go back to your …
SeannyBoi71 said:
I found out that the forces in the situation equal 3mg=Ty (Ty being the y-component of the tension in the wire) and for the net moment I have sintheta=4mg/3. Are these right, and if so how can I combine them to find T?
SeannyBoi71 said:
I found T= Ty/sinθ, but I don't know how to incorporate that with the rest of the information I found!

… you thought you could find T by finding Ty first.

Unfortunately, that doesn't help in this case because
i] you don't have any equation for Tx, to finish the job!
ii] your 3mg = Ty was wrong anyway, because you left out the y-component of the https://www.physicsforums.com/library.php?do=view_item&itemid=73" at the hinge :redface:

Instead, your only useful equation will be the moments equation (about the hinge), which you did use in your next post …
I have moment is N=rF. so moment acting clockwise as 2mg3/2r+mgr, and the opposite 3rsinθ. so simplifying, 4mg=3sinθ, and then 4mg/3=sinθ.

… and for some reason you got the correct distance 3rsinθ, but you didn't mutliply it by the force, T: that would have given you the equation 4mg=3Tsinθ.

The sinθ doesn't come from "Ty = Tsinθ", it comes from the distance 3rsinθ (which you yourself got :wink:), so you must multiply it by the whole of T.

(You can split T into Tx and Ty, and then use a distance of 3r instead of 3rsinθ, but that isn't the way you actually did it, and although it works I don't recommend it since I think it's a bit confusing and might lead to mistakes.)

ok, sorry if that's a bit confusing :redface:: write it all out and see if you get the correct answer, and if you're still worried about anything, come back and ask :smile:
 
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  • #14
Thank you so much! I finally got the answer. It definitely makes sense now. Thanks again :)
 

Related to Catwalk attached to a wall by a hinged held up by a cord

1. How does a catwalk attached to a wall by a hinged held up by a cord work?

A catwalk attached to a wall by a hinged held up by a cord works by utilizing the hinge to allow for movement and flexibility in the catwalk. The cord acts as a support, holding the catwalk up and preventing it from falling.

2. What materials are typically used to construct a catwalk attached to a wall by a hinged held up by a cord?

The materials used to construct a catwalk attached to a wall by a hinged held up by a cord can vary, but they typically include metal or wood for the catwalk itself, and strong, durable cords for support. The wall and hinge may also be made of metal or wood, depending on the specific design and purpose of the catwalk.

3. Are there any safety precautions to consider when using a catwalk attached to a wall by a hinged held up by a cord?

Yes, there are several safety precautions to consider when using a catwalk attached to a wall by a hinged held up by a cord. It is important to regularly inspect the catwalk, cords, and hinge for any signs of damage or wear. Additionally, the weight limit of the catwalk should be strictly adhered to, and proper safety gear, such as harnesses and hard hats, should be worn when using the catwalk.

4. How is a catwalk attached to a wall by a hinged held up by a cord installed?

The installation process for a catwalk attached to a wall by a hinged held up by a cord will vary depending on the specific design and location. However, in general, the hinge will be securely attached to the wall, and the cords will be attached to the hinge and the catwalk. Proper measurements and structural support are crucial during the installation process to ensure the catwalk is safe and stable.

5. What are some common uses for a catwalk attached to a wall by a hinged held up by a cord?

A catwalk attached to a wall by a hinged held up by a cord can be used in a variety of settings, such as construction sites, theatrical productions, or industrial facilities. It provides a safe and stable platform for workers or performers to access elevated areas. It can also be used as a temporary solution for situations where a permanent walkway is not feasible.

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