Hanging a mirror with two off-center nails

[dennish00a]

This is a question from a mathematically-inclined homeowner who knows he /should/ be able to figure it out, but can't.

I need to hang a heavy mirror in a house in Berkeley, CA, which is a seismically-active area. The mirror has a cable on the back of it attached to two sides of the mirror (just like a cable for hanging a picture). The nails need to be placed in studs in case of an earthquake.

The problem is that the mirror needs to be centered on the wall--but the studs aren't centered on the wall. Therefore, I plan to use two off-center nails, placed at different heights, to hang the mirror so that it is level. The final configuration is shown in the attached image.

My question is: given the two distances from each mirror edge to the nail locations (x and y in the diagram), how can I figure out the difference in the height of the two nails that is required to make the mirror level?

Or, if you can't tell me this, can you just tell me how to think about the problem? I don't get how to figure out the amount and direction of tension placed on each cable terminus, given the fact that they are connected to each other by the mirror itself. If I knew this, then I suppose I'd just need to place the same amount of vertical tension on each cable, and I'd be done.

 

[Phrak]

Very cool, and practical problem. The studs are 16 inches apart, by the way. Two or three other items are needed: the distance between the cable ties, and how far off center the miror is to be from being centered on the studs, and probably the cable length.

I think answer is that it depends on a few things-

Do we assume the nails are frictionless?
What is the cable length?
More, the solution could be astable so that any offset from ideal will cause it to tilt from random vibrations. If this is so, astability should be dependent on cable length.

On a practical note, it would be better to find the cable length by measuring how far it is displaced from a straight line when pulled from the center.

Can you do a vector diagram of forces?

 

[dennish00a]

OK, I think I figured this out. I believe that nothing matters except the angle the cable makes with the edge of the mirror. When this angle is the same on both sides, then each side has the same force pulling on a cable with the same configuration, and everything should cancel out, I believe (assuming that half the weight of the mirror is attributable to each cable).

So I divided the cable into 3 segments of lengths x*sec(theta), y*sec(theta), and (h^2 + (internail distance)^2)^0.5, where h is the difference in heights of the two nails (the desired answer). I set h equal to (x-y)tan(theta) and then used QuickMath to solve.

I was unable to solve it symbolically myself! If anybody could show me how to do that, I'd appreciate it. I'd also (still) appreciate an explanation of the physical forces so that I'd know if my intuition is correct.

I also intend to hang the mirror now--I'll let you know how it turns out.

 

[Phrak]

OK, I think I figured this out. I believe that nothing matters except the angle the cable makes with the edge of the mirror. When this angle is the same on both sides, then each side has the same force pulling on a cable with the same configuration, and everything should cancel out, I believe (assuming that half the weight of the mirror is attributable to each cable).

So I divided the cable into 3 segments of lengths x*sec(theta), y*sec(theta), and (h^2 + (internail distance)^2)^0.5, where h is the difference in heights of the two nails (the desired answer). I set h equal to (x-y)tan(theta) and then used QuickMath to solve.

That all sounds good, so long as theta is the angle made by the cable and the top edge of the mirror. sec(theta) = 1/cos(theta)

I was unable to solve it symbolically myself! If anybody could show me how to do that, I'd appreciate it. I'd also (still) appreciate an explanation of the physical forces so that I'd know if my intuition is correct.

I also intend to hang the mirror now--I'll let you know how it turns out.

I don't know what symbolically means, but the vector diagrams of the forces involved is simple enough. The tension (force) in the cable is equal in all parts of the cable.

 

[russ_watters]

You could always wrap some tape around the wire to keep it from slipping past the nails...

 

[Cyrus80772]

Im a construction worker and if I ran into this problem I would say just use some sheetrock screws with the little flipper locks on the end of the threading, then you can put the mirror where ever you want and you won't have to worry about studs, earthquakes, or physics. It will be strong too, I can mount a 70 pound cabinet to 5/8 rock with 4 screws and then hang on it or even do pull ups. lol

 

[Phrak]

Im a construction worker and if I ran into this problem I would say just use some sheetrock screws with the little flipper locks on the end of the threading, then you can put the mirror where ever you want and you won't have to worry about studs, earthquakes, or physics. It will be strong too, I can mount a 70 pound cabinet to 5/8 rock with 4 screws and then hang on it or even do pull ups. lol

I don't think we're allowed to screw through the mirror.

 

[Cyrus80772]

I don't think we're allowed to screw through the mirror.

That’s correct. You probably would not want to screw through the mirror into the wall. Typically you would not want to try and nail a mirror into the wall ether, actually this is something you would want to avoid doing, especially when the mirror has a cable on the back of it.

 

[stewartcs]

That’s correct. You probably would not want to screw through the mirror into the wall. Typically you would not want to try and nail a mirror into the wall ether, actually this is something you would want to avoid doing, especially when the mirror has a cable on the back of it.

How is the cable fixed to the mirror? Have you considered just using hooks position at the correct elevations on the back of the mirror instead of a wire?

CS

 

[Phrak]

That’s correct. You probably would not want to screw through the mirror into the wall. Typically you would not want to try and nail a mirror into the wall ether, actually this is something you would want to avoid doing, especially when the mirror has a cable on the back of it.

What are flipper locks?

 

[Cyrus80772]

How is the cable fixed to the mirror? Have you considered just using hooks position at the correct elevations on the back of the mirror instead of a wire?

CS

“On the back of it attached to two sides of the mirror (just like a cable for hanging a picture).”

I haven’t considered adding hooks but I would go with the wire instead of the hooks because you would have to find really flat and small hooks or you will be pulling the mirror out of plumb.

If you did have adjustable elevation on the hooks behind the mirror then that still would not solve the problem of the studs preventing you from securing it on the walls center line using nails nailed into the studs to hold the wire.

Not quit sure what your saying here but if your going to install the mirror level, then regardless of the elevation of the nails, there is nothing you can do to center the mirror on the wall using nails in the studs to secure the wire, if the studs don’t allow for that by their positioning to begin with.

There is no such thing as “flipper locks” for screws LOL. I was talking about a type of a sheetrock anchor, they “flip” back go through the pilot hole and “lock” the sheetrock screw to the wall. Was just trying to explain it so a physicist would understand.

 

[Phrak]

Not quit sure what your saying here but if your going to install the mirror level, then regardless of the elevation of the nails, there is nothing you can do to center the mirror on the wall using nails in the studs to secure the wire, if the studs don’t allow for that by their positioning to begin with.

Actually there is, "LOL", but I couldn't explain it to a rock hanger.

There is no such thing as “flipper locks” for screws LOL. I was talking about a type of a sheetrock anchor, they “flip” back go through the pilot hole and “lock” the sheetrock screw to the wall. Was just trying to explain it so a physicist would understand.[/QUOTE]

 

[Subductionzon]

You don't even need toggle or molly boltshttp://www.naturalhandyman.com/iip/inffastener/infanchor/infanchor.html" there are simple picture hangers (its too late and I cannot remember the right term) that will hold up to 50 pounds each when hammered into dry wall. These are simple devices that are commonly called "picture hangers". They consist of a piece of flat metal which is bent at the bottom into a picture hook and at the top it is bent so it will hold a nail (proper size is supplied) so it enters the dry wall at 45 degrees. I have hung hundreds of mirrors with these without a single failure. In fact the one failure that I know of was due to the wire on the back of the picture involved pulling the screws out that attached it to its frame. I would always use two of these with each one rated at least the weight of whatever I was hanging. The advantage of these is that they make a very small hole in the dry wall. And even though you can place them very accurately I would use some wires on the mirror, just 3 or 4 inches long on either side. Then if I was off a bit instead of making another hole in the wall I would just adjust one wire or the other a little bit.

 

[stewartcs]

“On the back of it attached to two sides of the mirror (just like a cable for hanging a picture).”

I haven’t considered adding hooks but I would go with the wire instead of the hooks because you would have to find really flat and small hooks or you will be pulling the mirror out of plumb.

If you did have adjustable elevation on the hooks behind the mirror then that still would not solve the problem of the studs preventing you from securing it on the walls center line using nails nailed into the studs to hold the wire.

Not quit sure what your saying here but if your going to install the mirror level, then regardless of the elevation of the nails, there is nothing you can do to center the mirror on the wall using nails in the studs to secure the wire, if the studs don’t allow for that by their positioning to begin with.

There is no such thing as “flipper locks” for screws LOL. I was talking about a type of a sheetrock anchor, they “flip” back go through the pilot hole and “lock” the sheetrock screw to the wall. Was just trying to explain it so a physicist would understand.

Ah, didn't read your first post carefully enough!

Are you completely against sheet-rock anchors? I've used them for many a heavy item. Each one (depending on the brand you buy) can hold 50 lbs. So two of them would hold 100 lbs. How much does your mirror way? The ones I use are called the Gripper (I think) and are available at Lowes or Home Depot.

You might, depending on the spacing, be able to use one stud and one anchor too.

Unfortunately with a cable, depending on the offset, you will put more of a load on one of the screws due to friction between the cable and the screw. So even though you may get it level you might overload one of them...just something to consider.

CS

 

[marcusl]

Had you enquired at, say, doityourself forums instead of physics forums, you might have heard about alternatives to nails and cables. Versions of the hanger cleats here

http://govart.com/hardware_cleat.html"

for instance, are available from many vendors (including amazon) in all different lengths. You can attach a long one all the way across your mirror, and anchor the mates to multiple studs. It's then trivial to rigidly hang your mirror level to the ground. I've used other types as well--the downside to these approaches is that you don't get to use vector force diagrams...

 

[Naty1]

Cut the single cable into two pieces. Take a guess at the nail heights and bang them in. Then tie the cables individually to each nail...adjust length as required to get the mirror level.

 

[pallidin]

Why not just put double stick tape-pads on the left and right inside bottom corners, hang the mirror using a carpenters level, press the bottom to the wall and be done with it?
Or maybe I'm not understanding the question.

 

[jdwood]

OK, I think I figured this out. I believe that nothing matters except the angle the cable makes with the edge of the mirror. When this angle is the same on both sides, then each side has the same force pulling on a cable with the same configuration, and everything should cancel out, I believe (assuming that half the weight of the mirror is attributable to each cable).

So I divided the cable into 3 segments of lengths x*sec(theta), y*sec(theta), and (h^2 + (internail distance)^2)^0.5, where h is the difference in heights of the two nails (the desired answer). I set h equal to (x-y)tan(theta) and then used QuickMath to solve.

I was unable to solve it symbolically myself! If anybody could show me how to do that, I'd appreciate it. I'd also (still) appreciate an explanation of the physical forces so that I'd know if my intuition is correct.

I also intend to hang the mirror now--I'll let you know how it turns out.

I have been doing this for years and have just recently worked out javascript to solve the problem. Dennish00a, you have it right. The way to solve the equation is to use the numerical analysis technique of iteration.

See a discussion and calculation forms at:

http://members.surfbest.net/jdwood@surfbest.net/asymt.htm

 

[Danger]

I don't think we're allowed to screw through the mirror.

I take it that you never saw the porn version of 'Alice in Wonderland' with Kristine deBelle... :uhh:

 

[jimbeard]

I have been doing this for years and have just recently worked out javascript to solve the problem. Dennish00a, you have it right. The way to solve the equation is to use the numerical analysis technique of iteration.

See a discussion and calculation forms at:

http://members.surfbest.net/jdwood@surfbest.net/asymt.htm

Is this calculator cool or what?! I went on line to deal with what turns out to be Case III and found much satisfaction.

 

[jimbeard]

Why not just put double stick tape-pads on the left and right inside bottom corners, hang the mirror using a carpenters level, press the bottom to the wall and be done with it?
Or maybe I'm not understanding the question.

The problems are:
1) It's a heavy mirror, so
2) It needs to be hung from the studs, and
3) The desired position of the center of the mirror is not halfway between two studs.

 

[arigger]

I get the question... There are several real-world solutions that would work, [even in California] but I think the OP wants a math exercise... I say that hanging it by two nails and a continuous string won't work so well in reality. I'm with an earlier poster who suggested cutting the string and making two legs of the appropriate length. That's a simpler math problem and will be more stable in the end.

 

[diazona]

I take it that you never saw the porn version of 'Alice in Wonderland' with Kristine deBelle... :uhh:

That just elicited a ROFL... congratulations, now I hate myself :rofl: