# Homework Help: Catwalk attached to a wall by a hinged held up by a cord

1. Nov 28, 2008

### XxseanxX_22@h

1. The problem statement, all variables and given/known data
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.

3. The attempt at a solution
i got (3mg/sintheta) but i don't think this is correct

2. Nov 28, 2008

### tiny-tim

Welcome to PF!

Hi XxseanxX_22@h! Welcome to PF!
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.

3. Nov 29, 2008

### XxseanxX_22@h

thanks i figured it out

4. Nov 29, 2008

### Shambles

I thought that the net moment at a hinge was 0 as the distance from rotation is 0. Is this wrong?

5. Nov 30, 2008

### tiny-tim

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!

6. Nov 24, 2010

### SeannyBoi71

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. Nov 25, 2010

### tiny-tim

Hi SeannyBoi71!

(have a theta: θ )
hmm … that's almost right

show us how you got that, so we can see what went wrong

8. Nov 25, 2010

### SeannyBoi71

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. Nov 25, 2010

### tiny-tim

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??

10. Nov 25, 2010

### SeannyBoi71

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

11. Nov 26, 2010

### tiny-tim

Hi SeannyBoi71!

(just got up :zzz: …)

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!

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

Last edited by a moderator: Apr 25, 2017
12. Nov 26, 2010

### SeannyBoi71

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??

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. Nov 26, 2010

### tiny-tim

ah, now I see what you're doing!

let's go back to your …
… 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

… 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 ), 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 : write it all out and see if you get the correct answer, and if you're still worried about anything, come back and ask

Last edited by a moderator: Apr 25, 2017
14. Nov 27, 2010

### SeannyBoi71

Thank you so much!! I finally got the answer. It definitely makes sense now. Thanks again :)