# Duke Physics Challenge: Let Go or Hang On?

This isn't really a homework question, I'm just doing it because I'm curious and the result isn't immediately obvious. I'm hoping someone else can verify the solution.
From http://www.phy.duke.edu/~hsg/physics-challenges/challenges.html

## Homework Statement

A painter is high up on a ladder, painting a house, when unfortunately the ladder starts to fall over from the vertical. Determine which is the less harmful action for the painter: to let go of the ladder right away and fall to the ground, or to hang on to the ladder all the way to the ground.

## Homework Equations

$$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta$$
$$K = \frac{1}{2}I \omega^2$$
$$I_{rod} = \frac{1}{3}ML^2$$
$$v = r\omega$$
and so on

## The Attempt at a Solution

Let the painter have a mass m, the ladder have mass M and length L. $\theta$ is the angle the ladder makes with the wall as it falls.
I'm going to say the criterion for judging the "safety" of the fall is the man's velocity when he hits the ground--he wants a small velocity please!
1: The man stays on the ladder.
I imagined that the ladder was a uniform rod rotating about its fixed end (which is on the ground). Then, I found the work done by the force mg acting on the man at a distance L from the axis of rotation and Mg acting at a distance L/2 (center of mass of the ladder) from the axis of rotation.
$$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta = \int_{0}^{\pi/2} mgL\sin{\theta}\,d\theta + \int_{0}^{\pi/2}Mg(L/2)\sin{\theta}\,d\theta = mgL + MgL/2$$
This work goes into changing the rotational kinetic energy of the rod-painter system, whose moment of inertia is $$I_{rod} + I_{man} = \frac{1}{3}ML^2 + mL^2$$, so
$$W = K = \frac{1}{2}I\omega^2$$
$$mgL + MgL/2 = \frac{1}{2}(\frac{1}{3}ML^2 + mL^2)\omega^2$$
$$2mgL + MgL = (\frac{1}{3}ML^2 + mL^2)\omega^2$$
$$g(2m + M) = L(\frac{1}{3}M + m)\omega^2$$
$$\omega = \sqrt{\frac{g(2m+M)}{L(M/3+m)}}$$
Now we find the velocity of the man v from the relationship $v = r\omega$
$$v = L\sqrt{\frac{g(2m+M)}{L(M/3+m)}}$$
And finally
$$v_{ladder} = \sqrt{\frac{gL(2m+M)}{(M/3+m)}}$$

2: The man jumps off.
This one's easy...
$$\frac{1}{2}mv^2 = mgh$$
$$v = \sqrt{2gh}$$
So
$$v_{freefall} = \sqrt{2gL}$$

Now we can compare these results by the ratio
$$\frac{v_{ladder}}{v_{freefall}} = \sqrt{\frac{gL(2m + M)/(M/3+m)}{2gL}} = \sqrt{\frac{2m+M}{2M/3 + 2m}}$$

So if this ratio is greater than unity, $v_{ladder}$ is greater than $v_{freefall}$ and the man should jump off instead of stay on. And actually, it turns out that the ratio will ALWAYS be slightly greater than unity (one of the masses has to be negative to make it otherwise). I thought this result was surprising, since I expected the ladder to help slow him down somehow (don't ask me why though ;P). Does this look right?

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Dick
Homework Helper
Your analysis is correct. I think your intuition that the ladder will slow him down comes from the fact that he will hit the ground later if he hangs on, but as you've shown, he will wind up hitting it faster. Look at what happens if the mass of the ladder is tiny, then he hits at v~sqrt(2*g*L), the same as if he had hung on. If the ladder has most of the mass, he hits at v~sqrt(3*g*L). His best strategy may be to climb down the ladder as fast as he can.

Dick
Homework Helper
Unfortunately that's not one of the options offered in the Duke Physics Challenge. It IS a poor reference for ladder safety issues. And probably many other things as well.

You've missed two things here.

Both are frictional.

One: There is a dependence of air friction on velocity, see the Bernoulli equation for further details. Note that this decreases the rate of acceleration for both cases slightly, but *more* so for the initially slow hold-onto-your-ladder case as you are accelerating for longer.

Second: We have hinge friction. Depending on the soil, how far the ladder has stuck into the ground, and so on, if you hold onto your ladder this could slow you down a little.

Though as joleeco said, neither of these are really what you want to do if you actually fall from a ladder like that.

Some years back as I was driving down the road, I chanced to see a man painting from a ladder. I'd estimate the ladder was approximately 20' long and he was quite near the top meaning his head was approximately 20' off the ground and his feet about 14'. The ladder was at a proper lean angle, maybe 30 degrees or so.

As I was watching, he lost his footing and fell.

He managed to do a 'fireman' slide down the ladder; feet on the outside of the rails, hands sliding down the rails...beautiful!

When his feet hit the ground, you could see his shin snap...

You fall 20', straight down or in an arc and you're gonna hurt...bad...

I fell last summer from a standing position so my head moved about 6' before hitting the ground. I was instantly knocked unconscious and suffered a minor concussion that took about 18 months to (mostly) recover from.

I'd say breaking your leg is better than busting your head so jumping rather then arcing to the ground is likely to do less serious damage.

That doesn't answer the physics question but may address the strategy to use when falling

Thanks, dog', but i'm trying to answer the physics question.

To be more specific, I mean that the relevant equations should include a KE reduction based on friction, eg K = 1/2*I*w^2 - mkv^2, with m mass, and where k is a constant based on relative density of air, ladder/worker falling object, and frictional form factor of the ladder/worker object.

Essentially, we're adding in Bernoulli's equation for realistic physics (NOT realistic what-do-i-do-'s). As much as i basically agree with the real-world considerations, we're not seeking engineering solutions.

Sorry, I'm a chemist...pretending to be a computer programmer...not much real help. I 'Stumbled' into this thread. Best of luck solving the problem.

Dogu (not Dog)

The answer[to simplified problem where painter is point mass and ladder is rod] is... stand on ladder, so that ladder can only put upward force on painter.

The ladder's falling acceleration is going to vary with time. Initially, it is obvious that ladder's acceleration is less than free fall acceleration, so the ladder is going to put upward directed force on the painter. Near end of fall, however, because ladder works as sort of lever, top of ladder could be falling with acceleration greater than free fall acceleration. If the painter is standing on ladder, however, painter is going to simply get detached from ladder once ladder's acceleration exceeds free fall acceleration.

edit: by the way, original problem is ill specified.
There is at least 3 ways how ladder could fall. Ladder's end on ground may begin to slip. Ladder may fall to the side, and there the end on ground may slip or not slip (the case when wall miraculously disappear is same as fall to side).
All 3 cases have different fall velocities.
Even more, there may be or not be friction; in physics you cant assume there's no friction if you're not given any specific value; furthermore how the hell painter can get up ladder in first place if there is no friction?
Physics isn't mathematics, you have to solve with some constant q for friction coefficient, not assume its zero just because you were not told exact value. (Just like you don't assume gravity is zero even though problem statement doesn't tell you explicitly that it isn't)

Given just how badly the problem is specified, it looks like "stand on ladder" is the only answer that could be given.

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