# Hang heavy picture with multiple hooks

Suppose I have a 5kg framed picture that I want to hang on hooks, each of which can hold 2kg. Is there some arrangement of hooks that will make this safe? How many hooks would I need? I thought that I could arrange the hooks as something like a catenary to distribute the load, but I'm having trouble calculating the force on each hook.

What are the techniques I need to use to solve this problem?

• This looks like a fun homework question for a first year statics class. What have you been taught and how have you attempted to solve it thus far?
– ChP
Mar 5, 2018 at 7:47
• Did you have a look at en.wikipedia.org/wiki/Free_body_diagram? Mar 5, 2018 at 8:29
• First thought is 3 with the load equally distributed... Mar 5, 2018 at 9:09
• @SolarMike yeah but that's a comment like the Monty Python "how to play a flute: blow here and wiggle your fingers there" Mar 5, 2018 at 16:24
• @CarlWitthoft Unfortunately, if a university's engineering department makes it, it will fail at exactly 2 kg always :P
– ChP
Mar 5, 2018 at 19:40

I couldn't resist, so: First, you should note:

• The tension in the string will be the same along its entire length (assuming frictionless conditions)
• To use the hooks optimally, you should arrange them that they carry equal load

Taking this into consideration, the angle that the string bends at each hook should be the same (the wall hooks should be spaced equally on the arc of a circle between the picture hooks, not a catenary as noted).

The picture below illustrates:

$$ø = \frac{2θ}{3}$$

Or more generally: $$ø = \frac{2θ}{n}$$ where n = the number of hooks

Below is the FBD at one of the frame hooks:

• $F_t$ is the tension in the string.
• $F_p$ is the vertical force on the frame hook.

From this, the tension in the string can be calculated as:

$$F_t = \frac{F_p}{sin(θ)}$$

The FBD at one of the hooks is as follows:

With $F_h$ as the resultant force on the hook.

$F_h$ can be calculated as:

$$F_h = 2F_tsin(0.5ø)$$

This can be generalised as:

$$F_h = \frac{2F_p}{sin(θ)}*sin(\frac{θ}{n})$$

From this you can see that as $θ$ approaches $0$, $F_h$ approaches $2F_p/n$, but $F_t$ approaches infinity. Thus you should consider the tensile strength of the string and the forces that the frame and frame hooks can handle as well.

You can hang your picture with 3 hooks with $θ = 62.11$ degrees, giving you $F_t = 28.285N$ and $F_h = 19.999N$. This will be "safe" assuming that the gravitational constant is 10, wihch it isn't, but that's just nit picking.

• +1 for a good answer. -1 for doing his/her homework. Mar 5, 2018 at 21:11
• Thank you, this is awesome. I feel inspired to make an online tool that implements this. And in case there was any doubt: this was not a homework question; I really have a picture to hang.
– z0r
Mar 6, 2018 at 2:47
• @Transistor +1, Yeah, I probably shouldn't have, but it's a well worded question with explanatory diagrams and one less user scared away with unhelpful comments such as "you should do your own homework" - I know, I'm one of those commentors. And we have just enabled the OP to further enrich the world by using his own skill set to create a tool to help the rest of the world.
– ChP
Mar 6, 2018 at 6:14
• I see quite a few questions on here which I strongly suspect are homework questions but disguised as "prototype design problems". Those are very basic first year problems that gets more attention because it's explicitly labled as something "non-homework".
– ChP
Mar 6, 2018 at 6:18