{"id":15441,"date":"2017-10-12T14:59:40","date_gmt":"2017-10-12T18:59:40","guid":{"rendered":"http:\/\/chcollins.com\/100Billion\/?p=15441"},"modified":"2023-03-18T18:30:03","modified_gmt":"2023-03-18T22:30:03","slug":"the-physics-of-hanging-pictures","status":"publish","type":"post","link":"https:\/\/chcollins.com\/100Billion\/2017\/10\/the-physics-of-hanging-pictures\/","title":{"rendered":"The Physics of Hanging Pictures"},"content":{"rendered":"

Asked and Answered 3.3<\/em><\/strong><\/h4>\n

This is the third and final article in my series about hanging picture frames.\u00a0 The first post, Why Frames Tilt Forward<\/em><\/a>, discusses why frames tilt at the top and what you should and should not do about it.\u00a0\u00a0 My next post, The “Hang It with Two Hooks” Calculator<\/a><\/em>, presents an online calculator to help you hang pictures with less forward tilt, using two wall hooks and 45\u00b0 wire angles.\u00a0 This post completes the picture, so to speak.\u00a0 Here I try to illustrate, with intuitive examples, the role of physics in picture hanging.\u00a0 Most of all, I want to help you understand why it is a bad idea to string a wire tightly across a frame to keep it from tilting forward.<\/p>\n

\"2hookicon2\"<\/a>I will also discuss the picture-hanging hardware I like and why.\u00a0 Finally, I will provide another online calculator — this one evaluates your frame’s margin of safety by estimating the tension in the wire and the tendency of your frame to bend.\u00a0 Click the icon at right to go directly to my safety-factor calculator<\/a>.<\/p>\n

Physics-minded readers, relax.\u00a0 This article is for general readership.\u00a0 So I am not going to distinguish between mass and weight, I am ignoring the gravitational constant, and I will use pounds, not newtons, as the unit of force, because that is how the people around here hang pictures.<\/p>\n

The Graphical Physical Tour<\/span><\/h3>\n
\"FIGURE<\/a>

FIGURE 1<\/p><\/div>\n

Everyone else can relax too.\u00a0 I’m going to walk us through some basic physics that my wife and I learned in high school, before<\/em> we started dating.\u00a0 I’m sure she remembers all of this.<\/p>\n

Let’s start with something simple.\u00a0 In Figure 1 (click to zoom) we see a 1o-pound weight hanging from the ceiling via a wire.\u00a0 The weight is at rest, neither rising or falling — this means that the upward force (or tension) in the wire must be exactly equal to the downward force\u00a0w<\/em><\/strong> of the weight.\u00a0 Hanging a 10-pound weight on a single wire produces 10 pounds of tension in the wire.<\/p>\n

\"\"<\/a>

FIGURE 2<\/p><\/div>\n

Onto our next example.\u00a0 In Figure 2 (at right), the weight is the same as before (1o pounds) but it is now hanging by two identical wires instead of one.\u00a0 Once again, the downward force w<\/em><\/strong> of the weight is balanced by the upward pull of the wires.\u00a0 Because there are two wires, each individual wire carries just half the load.\u00a0 So the tension in each wire is now w<\/em><\/strong>\/2<\/strong> (or 5 pounds in this case).<\/p>\n

Okay, time to use your intuition.\u00a0 If we were to weld together the two wires in Figure 2 at the top, would this change the tension?\u00a0 No — the tension in each section of the wire would still be w<\/em><\/strong>\/2<\/strong>.\u00a0 Ponder this until you’re comfortable with the idea.<\/p>\n

Now that (in our minds) the two wires are connected at the top, let us take one more step:\u00a0 slice the weight down the middle, so that each end of the wire supports half the original weight.\u00a0 This action should also have no effect on the tension in the wire.\u00a0 Agree?<\/p>\n

\"FIGURE<\/a>

FIGURE 3<\/p><\/div>\n

We are now prepared to consider this model of a frame suspended from one hook (Figure 3).\u00a0 The total weight w<\/em><\/strong> is the same, but it is divided into two equal weights supported on either end of a single continuous wire.<\/p>\n

The wire passes over pulleys at the top and sides. The top pulley represents the wall hook; the side pulleys are the D-rings attached to the frame.<\/p>\n

The weights create forces that pull downward at the top and inward at the sides.\u00a0 We will take a closer look at this in the diagram below (Figure 4) which focuses on the left side of the setup (the right side is a mirror image).\u00a0 It may be helpful if you click on the figure to view it full size.<\/p>\n

\"\"<\/a>

FIGURE 4<\/p><\/div>\n

Once again, we have a system at rest: the weights are not rising or falling and the pulleys are not moving.\u00a0 This means that the downward forces are balanced by equal and opposite upward forces — and the same is true for the horizontal forces.<\/p>\n

Let’s zoom in on the force on the left side.\u00a0 The weight w<\/em><\/strong>\/2<\/strong> exerts a downward force at the pulley, which must be offset by an equal force upward.\u00a0 But the wire does not extend upward — instead it heads away from the pulley on a diagonal.\u00a0 How can a diagonal wire produce an upward force?<\/p>\n

It helps to imagine that the pull of the wire is composed of vertical and horizontal parts that add up, so to speak, to a total force (tension) in the diagonal direction.\u00a0 In the figure, I\u00a0denote the vertical and horizontal parts as Ty<\/span><\/em><\/strong> and Tx<\/span><\/strong><\/em>, respectively.\u00a0 Because there is no net motion in the vertical direction, we know that Ty<\/span><\/em><\/strong> (the upward force) must equal w<\/em><\/strong>\/2<\/strong> (the downward force).
\n<\/em><\/p>\n

How do we find Tx<\/span><\/strong><\/em>, the force in the horizontal direction, and T<\/strong><\/em>, the tension in the wire?\u00a0 Here, we have to use some trigonometry.\u00a0 The wire tension T <\/strong><\/em>equals Ty<\/span><\/em><\/strong> times the cosecant (csc<\/em>) of the wire angle \u03b1<\/strong><\/em>, and the horizontal force Tx<\/span><\/strong><\/em> equals Ty<\/span><\/em><\/strong> times the cotangent (cot<\/em>) of the wire angle \u03b1<\/strong><\/em>.\u00a0 If you did not take trigonometry in school, please accept this on trust.<\/p>\n

Sorry for the math, but I wanted to show how the wire angle has a multiplier<\/em> effect on the tension T<\/strong><\/em>.\u00a0 The smaller the wire angle (that is, the closer to horizontal the wire is strung), the greater the multiplier.<\/p>\n

I have listed the multipliers for various wire angles in Figure 4.\u00a0 The first column of the table is the wire angle, the second column is the tension multiplier, and the third column is the horizontal force multiplier.\u00a0 These multipliers apply to Ty<\/span><\/em><\/strong> (which is w<\/em><\/strong>\/2<\/strong> in this case).<\/p>\n

From the table, we see that if the wire is strung only 5\u00b0 from horizontal, then the tension in the wire will be more than eleven times w<\/em><\/strong>\/2<\/strong>.\u00a0 For our 1o-pound frame, the wire tension would be 57.3 pounds and the inward pull on each side of the frame would be 57.1 pounds!<\/p>\n

But if we were to string the wire 45\u00b0 from horizontal, the wire tension would be 7.1 pounds and the force pulling in on the side would only be 5 pounds.\u00a0 This shows why one should not<\/em> string a wire tightly across a picture frame to reduce its forward tilt.<\/p>\n

I like the idea of using two wall hooks and 45\u00b0 wire angles, as discussed in my other posts, because it reduces both the wire tension and the inward pull on the frame.\u00a0 But this does not mean that horizontal forces go away.\u00a0 In any two-hook installation, there will be a net horizontal force on each hook, pulling them toward the center of the frame.<\/p>\n

\"Figure<\/a>

FIGURE 5<\/p><\/div>\n

The diagram at left (Figure 5) depicts one wall hook in a two-hook setup.\u00a0 The left end of the wire extends diagonally down to the frame, and the right end leads to the other hook.\u00a0 The wire tension T<\/strong><\/em> is the same everywhere along the wire.<\/p>\n

In the figure, the black arrows represent the forces that the wire exerts on the hook.\u00a0 The net force on the hook Tz<\/span><\/strong><\/em> (red arrow) results from adding together the horizontal and vertical components of these forces.<\/p>\n

Again, using some trigonometry, we find that the force Tz<\/span><\/strong><\/em> will always be somewhat higher than Ty<\/span><\/em><\/strong> (the exact formula is shown in Figure 5).\u00a0 The direction of this force relative to vertical is one-half the wire angle.\u00a0 In the case of our preferred 45\u00b0 wire angle, the overall force on each hook would be 0.54 times the frame weight and the force would be directed 22.5\u00b0 inward from vertical.\u00a0\u00a0 The horizontal component of this force would be 0.21 times the frame weight.\u00a0 If one were to use a steeper wire angle — say, 60\u00b0 from horizontal, as\u00a0 some people suggest<\/a> — it would increase the lateral force on each hook by almost 40%.
\n<\/span><\/p>\n

That’s it for the hard-core physics.\u00a0 Let’s talk about what this means for picture framing.<\/p>\n

[ A NOTE FROM THE AUTHOR ]<\/span><\/h4>\n
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