29

If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it. This was solved here.

Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it.

I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.)

$$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$

$m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder.

Community
  • 1
cvr
  • 535
  • 6
    Incidentally, even better is to ride the ladder partway by standing on it (but not holding on) until gravity pulls it away from you. – Rex Kerr Aug 19 '16 at 16:00
  • 8
    Tho' both the answer here and at your link ignore the real-world issue of What's Down There. You want to jump off when you're more likely to land on soft ground than rocks or a poison ivy patch. :-) – Carl Witthoft Aug 19 '16 at 16:40
  • 1
    @RexKerr and what do you do if it's falling over backwards? I'm pretty sure it'd be impossible to "stand on it" without holding on. – Doktor J Aug 20 '16 at 05:26
  • 3
    When you say "jump" do you mean "let go" or do you mean "jump" (like, propel yourself away from the ladder with force)? – Jason C Aug 20 '16 at 08:50
  • 1
    From the title I expected this question to be "How long should I stay on a falling ladder before jumping off?" – trichoplax is on Codidact now Aug 20 '16 at 17:38
  • @DoktorJ - You can always push yourself off to the side while in freefall rather than letting the ladder push you down even faster. In theory, you know, when you an infinitesimal point on the ladder. If you have a real ladder: don't fall off; by the time you realize you're in free-fall with a ladder pushing down on you, things are already way too bad. – Rex Kerr Aug 20 '16 at 20:19

4 Answers4

23

You should stand at 2/3 of the height of the ladder.

If you land with the same kinetic energy as without a ladder, then the ladder should land with the same kinetic energy as without you. Equating the kinetic energy of the ladder with its potential energy at the beginning:

$$\frac{1}{2} mgL = \frac{1}{2} I_L \omega^2 = \frac{1}{2} \left(\frac{1}{3} mL^2\right) \omega^2$$ gives: $$\omega = \sqrt{\frac{3g}{L}}$$

where $L$ is the length, $m$ the mass, $I_L$ the moment of inertia and $\omega$ is the angular velocity of the ladder.

For you the same equation holds, but now $\omega$ is known:

$$MgH = \frac{1}{2} I_M \omega^2 = \frac{1}{2} (MH^2) \left(\frac{3g}{L}\right)$$ with $M$ your mass, $I_M$ your moment of inertia and $H$ your height. Solving for $H$ gives:

$$H=\frac{2}{3}L$$

or of course $H=0\;.$

Crimson
  • 2,871
14

We can do this with a minor modification to the calculation described in the earlier question. As before we'll take the ladder length to be $\ell$, but now we'll take your height to be $\alpha\ell$, where $\alpha$ ranges from zero to one. Our reference point is if you let go, in which case your speed when you hit the ground will be:

$$ v^2 = 2g\alpha\ell \tag{1} $$

Now suppose you hold onto the ladder. As before we calculate the total potential energy change of both you and the ladder, which is:

$$ V = mg\alpha\ell + \frac{1}{2}m_Lg\ell \tag{2} $$

And this must be equal to the increase in angular kinetic energy $\tfrac{1}{2}I\omega^2$. The combined moment of inertia of you and ladder is:

$$ I = m(\alpha\ell)^2 + \frac{1}{3}m_L\ell^2 $$

And setting the kinetic energy equal to the potential energy gives:

$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\omega^2 $$

And since $v=r\omega$ your velocity is $v=\alpha\ell\omega$ giving:

$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\frac{v^2}{\alpha^2\ell^2} $$

Which rearranges to:

$$ v^2 = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$

And finally substitute for $v$ from equation (1) to get:

$$ 2g\alpha\ell = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$

And this rearranges to:

$$ \alpha = \frac{2}{3} $$

So if you are more than $\tfrac{2}{3}$ of the way up the ladder you should let go, while if you are lower than $\tfrac{2}{3}$ of the way up the ladder you should hang on.

Community
  • 1
John Rennie
  • 355,118
  • Thanks for both replies. I hadn't really "expected" the masses of ladder and man to be left out of the solution. – cvr Aug 19 '16 at 15:04
  • 1
    @ycc_swe: Actually I was a bit surprised to find that everything cancelled out in the last equation. It looked as though it was going to be complicated then all the terms cancelled! – John Rennie Aug 19 '16 at 15:06
  • 1
    "You should let go" - not necessarily, if you are at the top of the ladder you have some angular momentum as the ladder tips, making it easier to transfer the impact energy into an acrobatic roll. Granted, you'd better have your wits about you. – David Wilkins Aug 19 '16 at 16:21
  • @DavidWilkins - If your size is negligible compared to the size of the ladder (and, let's face it, who doesn't go around climbing ladders that allow you to be approximated as a dimensionless point, when one finds oneself in a physics problem), then your velocity when the ladder hits will be exactly normal to the ground. No acrobatic rolling for you! If you let go partway down, then yes, you may be able to reduce the forces in the lateral direction by rolling (thus extending your deceleration time). – Rex Kerr Aug 20 '16 at 02:04
  • 3
    I can only say that having jumped off a roof with forward momentum, which translated to angular momentum as gravity influenced my trajectory, and having rolled out of the fall instead of falling flat, that it does make a difference. Yes I scared a few bystanders, but physically it made sense. And worked at the time – David Wilkins Aug 20 '16 at 02:11
  • @DavidWilkins - I've done it myself. The biomechanics are complicated and have a lot to do with using your legs as much as they can, and then absorbing the rest with shoulder/back which are tougher than scrunched-up legs--but you can only roll onto tougher parts properly if you have enough forward momentum to get enough angular momentum from your feet to make it. And you start with forward momentum, which you won't have if you hang on to the ladder the whole way--it'll pull you down with it. Anyway, it's really interesting that it works at all, and it's really hard to demonstrate why. – Rex Kerr Aug 20 '16 at 02:28
  • I did parachuting in the 70's when impact with the ground was still something to count with (more vertical speed in the parachutes then). It was definitely advised not to land with horizontal speed straight forward or straight backward. You roll more safely, slightly sideways. I can't remember the instruction on the following, but I would also say that going absolutely straight down should be avoided, since a roll is what you want to achieve. / A few lessons in judo also comes to mind. They slam the arm into the mat just before they hit it and also roll a lot. – cvr Aug 20 '16 at 06:18
  • So making the ladder significantly taller than needed for the job is a safety feature! – JDługosz Aug 20 '16 at 07:32
  • If the ladder falls backwards I think "hang on" and "let go" end up being the same thing, unless you somehow slip through one of the holes between the rungs. – Jason C Aug 20 '16 at 08:55
  • @DavidWilkins Well, sort of necessarily, if you think about how you stand on a ladder. If it falls sideways you're not rolling anywhere unless you have the grace to slip into a cartwheel and not snag an ankle on the ladder. If it falls backwards I can't see how you'd start a backwards roll, with your back facing the ground and a ladder a fraction of an inch above you. If it falls forwards well... watch out for the debris from the wall that just collapsed in front of you (lol). I can't think of any way to roll in any direction off a falling ladder unless you forcefully jump away from it. – Jason C Aug 20 '16 at 08:59
0

If one has the presence of mind, I vote for sliding down the ladder at any cost - gloves and shoes acting as a control brake. Shortening the diameter (height of your feet from ground) will surely reduce any impact. The risk is loss of control in the vertical decent. Gloves might be a recommended safety feature of any ladder climbing to keep confidence while executing a rapid slide. I would think you could get 6 feet lower before getting into the lateral arc. I admit that my own experience is limited to sliding off a wet cedar shingle roof and kicking the ladder away as I slid into it. It was only 10 feet onto damp lawn so I was saved by that and walked away. Yeah, I've had the training but I landed on my butt. No roll was possible.

randybump
  • 11
-2

You people are crazy with your useless calculations.

Doesnt take science to realise that if you hold on the ladder you will have a higher chances of falling on your back or ass and hit your head against the ground now I wouldnt know peoples preferences but yesterday I was at the top of the ladder which is about 3 meters I was trying to screw a piece of mdf sheet when I lost balance and gravity was pulling me backwards so I simply jumped off It and landed on my legs, now my left leg heel is killing me when I step on it but I rather have it this way than ended up in hospital with a concussion or broken back. Ps; the floor was concrete.

So yeah no matter your calculations it's better to jump and fall on your legs than hit any other part of the body, trust me.

Danny
  • 1