How to support a sign on our museum wall?

AI Thread Summary
To support a 12-foot triangular sign weighing 200 pounds on a museum wall, two guy wires anchored 12 feet apart at a 45-degree angle are recommended. Each wire will bear approximately 150 pounds of load, so selecting stainless steel cable with a diameter of at least 4mm (or 3/16") is advisable for safety. It's important to consider the yield strength of the cable and potential adverse weather conditions, which may require stronger materials. Overestimating cable strength is prudent to ensure reliability and safety. Using a thicker cable can prevent failures at anchor points or welds before the cable itself gives way.
agap015
Messages
2
Reaction score
0
Hi all,

I went to an engineering school 40 years ago but never finished, so I kind of know that this is easily solveable.

I have a triangular sign 12' long (attached to wall) that is 2 1/2' thick and sticks out from the wall 6'. If I use 2 guy wires anchored to the wall 12' apart, that meet the sign at a 45* angle at the point, how large should the wires (stainless steel cable) be in diameter?

The sign weighs 200# total.

The sign is to be attached to our local museum announcing an event.



*
* *
*
* *
*
* *
*
* *
*
* *
*
* *
*******************************
* *
* *
*******************************


Thanks

George
 
Physics news on Phys.org
I'm a physicist, but pretty far removed from engineering, so take it for what it's worth. Hopefully, somebody better qualified will come along.

If I understand your setup correctly, each wire is under roughly 150lb of load. (A diagram would help to make sure.) You want this to be within yield strength for the wire, to give yourself a bit of an overhead. Of course, there are steel cables, and then there are steel cables. But you should be able to count on at least 200MPa. That means wire just a touch over 2mm in diameter. So you should go with at least 11 gauge (AWG) for this.

Keeping the load within yield strength should make the structure reliable enough, and if things go wrong, should be able to take nearly twice as much load for a short while. But if you expect adverse weather, such as winds or snows, you'd have to account for that separately.
 
Thanks for the reply.

2MM isn't very large. I will probably use 3/16" cable or about 4MM. The sign is a half scale theater marque. It will showcase our town theater.

Thanks

George
 
I am also an owner of a Museum.

I agree with K^2 that it is simply prudent to "overestimate" on the side of safety in supporting this marque. What's the difference in cost or difficulty of using a "stronger than you think is mimimally necessary cable"?

Bobbywhy
 
4mm really ought to do it. I'd expect anchor points or welds on the marquee to go before a 4mm cable does.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top