Suppose you fall from the top of a ladder straight down. You will hit the ground with an amount of force.

Now suppose that you fall over while holding onto the ladder, tipping over in an arc instead of falling straight down. You will hit the ground with another amount of force.

Neglecting the mass of the ladder and air resistance, which impact will have the most force? Falling or tipping?

This has been a bit of a debate between myself, my father, and my grandfather. I believe that they would fall with the same force because, relative to the ground, you start with the same amount of potential energy in both situations. My grandfather and father, however, guess that it would fall with around half the force because the force is dissipated somewhat by the forward motion and the support of the ladder. We would appreciate an answer from a third party to help us find the solution.

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    $\begingroup$ Hi user1583259, and welcome to Physics Stack Exchange! No worries about the tags, I fixed them up for you. $\endgroup$ – David Z Aug 8 '12 at 1:14
  • $\begingroup$ as you see here you've already got two opposite answers. So don't judge only by the words "yes/no", you should follow the calculations and more importantly the assumptions yourself. Are you capable of that? $\endgroup$ – Yrogirg Aug 9 '12 at 7:17
  • $\begingroup$ I think so. Both answers make sense in their own circumstances, but one of them factors in the mass of the ladder, something we were explicitly leaving out. $\endgroup$ – Kenkron Aug 12 '12 at 19:36
  • $\begingroup$ Both of the answers include maths, but only one, John Rennie's, handles it correctly. $\endgroup$ – Thriveth Apr 30 '14 at 21:23
  • $\begingroup$ I have unchecked the answer, out of respect for your confidence, but I am still unconvinced. John Rennie does not account for rotation. I'll have to think about whether that's okay... $\endgroup$ – Kenkron May 2 '14 at 16:23

Both you and your ancestors are wrong. But I bet you would never guess the real answer!

Assuming the base of the ladder doesn't slide you have a rotating system. Just like a freely falling man you convert potential energy to kinetic energy, but for a rotating system the kinetic energy is given by:

$$ T = \frac{1}{2}I\omega^2 $$

where $I$ is the angular momentum and $\omega$ is the angular velocity. Note that $v = r\omega$, where $r$ is the radius (i.e. the length of the ladder). We'll need this shortly.

Let's start by ignoring the mass of the ladder. In that case the moment of inertia of the system is just due to the man and assuming we treat the man as a point mass $I = ml^2$, where $m$ is the mass of the man and $l$ the length of the ladder. Setting the change in potential energy $mgl$ equal to the kinetic energy we get:

$$ mgl = \frac{1}{2}I\omega^2 = \frac{1}{2}ml^2\omega^2 = \frac{1}{2}mv^2 $$

where we get the last step by noting that $l\omega = v$. So:

$$ v^2 = 2gl $$

This is exactly the same result as we get for the man falling straight down, so you hit the ground with the same speed whether you fall straight down or whether you hold onto the ladder.

But now let's include the mass of the ladder, $m_L$. This adds to the potential energy because the centre of gravity of the ladder falls by $0.5l$, so:

$$ V = mgl + \frac{1}{2}m_Lgl $$

Now lets work out the kinetic energy. Since the man and ladder are rotating at the same angular velocity we get:

$$ T = \frac{1}{2}I\omega^2 + \frac{1}{2}I_L\omega^2 $$

For a rod of mass $m$ and length $l$ the moment of inertia is:

$$ I_L = \frac{1}{3}m_Ll^2 $$

So let's set the potential and kinetic energy equal, as as before we'll substitute for $I$ and $I_L$ and set $\omega = v/l$. We get:

$$ mgl + \frac{1}{2}m_Lgl = \frac{1}{2}mv^2 + \frac{1}{6}m_Lv^2$$

and rearranging this gives:

$$ v^2 = 2gl \frac{m + \frac{1}{2}m_L}{m + \frac{1}{3}m_L} $$

and if $m_L \gt 0$ the top of the fraction is greater than the bottom i.e. the velocity is greater than $2gl$. If you hold onto the ladder you actually hit the ground faster than if you let go!

This seems counterintuitive, but it's because left to itself the ladder would rotate faster than the combined system of you and the ladder. In effect the ladder is accelerating you as you and the ladder fall. That's why the final velocity is higher.

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    $\begingroup$ Thank you; I believe that you're right that the mass of the ladder will add to the impact, but the question said "Neglecting the mass of the ladder and air resistance", and I think what Prahar said about rotation is correct. $\endgroup$ – Kenkron Aug 12 '12 at 19:53
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    $\begingroup$ In fact, if you leave out the mass of the ladder, John's (correct) calculations show that both cases will just reduce to the simple free/fall case: $v^2 = 2gl$. So even so, Prahar's answer is wrong. $\endgroup$ – Thriveth Apr 30 '14 at 21:18
  • $\begingroup$ So a ladder with its mass evenly distributed falls faster than one with the mass concentrated at any end: I'm surprised at this. $\endgroup$ – Larry Harson May 5 '14 at 22:35
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    $\begingroup$ @LarryHarson: it's one of the classic problems you give students because their initial guess at the answer is always wrong :-) $\endgroup$ – John Rennie May 6 '14 at 6:51
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    $\begingroup$ Yep, our professor at Classical Mechanics gave us the same problem (not a ladder, some other rod-shaped thingie). The gist of it is that it is the center-of-mass that falls with free-fall velocity, and so if the thing is rotating and the center-of-mass is not at the very end, the end will fall faster thhan this. $\endgroup$ – Thriveth May 11 '14 at 19:58

So the force you would feel is the $F = \frac{\Delta p}{\Delta t}$. Now, when you hit the ground, your momentum basically changes to zero in some sudden time $\Delta t$. Let us assume that the time it takes for the momentum the change to zero ($\Delta t$) is the same for both the cases (tipping over and falling down). Thus $$F = \frac{m \Delta v}{\Delta t} = \frac{m (v-0)}{\Delta t} = \frac{m v }{\Delta t}\;.$$

Now when falling down, all your initial potential energy is converted to KE implies $$ v = \sqrt{2 g h} \implies F = \frac{ m \sqrt{2 g h} }{ \Delta t}\;.$$

When it tips over, you now have both KE and Rotational energy. We then have $$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = m g h = \frac{1}{2} m v^2 + \frac{1}{2} (m h^2) \left(\frac{v}{h} \right)^2 = m v^2 \implies v = \sqrt{gh}\;.$$

We then have $$ F = \frac{m \sqrt{gh}}{\Delta t}\;.$$

Thus $$\frac{ F_\textrm{fall} } {F_\textrm{tip} } = \sqrt{2} \implies F_\textrm{fall } > F_\textrm{tip}\;.$$

So falling down imparts a greater force!

  • $\begingroup$ I think this is the answer I was looking for (I'll double check with dad & g-pa before I confirm it), and though I had to do a bit of work to follow your answer, I think it made sense. It somehow didn't occur to me to factor in rotational energy. $\endgroup$ – Kenkron Aug 12 '12 at 19:36
  • $\begingroup$ For clarification, if you consider the rotational force of the fall as well as the linear force of the fall, and assumed that Δt was the same for rotational force, then would the total impact come out the same? $\endgroup$ – Kenkron Aug 12 '12 at 19:56
  • $\begingroup$ I am pretty sure this answer is not correct. We had an example in classical mechanics that showed that the tip of a rod falling over would fall faster than free-fall velocity. I will try to find it. $\endgroup$ – Thriveth Apr 30 '14 at 20:28
  • $\begingroup$ I have now downvoted this answer, John Rennie shows below correctly and rigorously why Prahars answer is wrong. If you leave out the mass of the ladder, the two cases reduce to the same impact velocity. If you add in the mass of the ladder, tipping over will make you hit the ground faster than free fall, making the impact harder. $\endgroup$ – Thriveth Apr 30 '14 at 21:22
  • $\begingroup$ I think you've double counted the kinetic energy of the person when the ladder tips over. $\endgroup$ – Larry Harson May 5 '14 at 22:32

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