- 18 May '11 18:23So you are in one of those Helen of Troy kind of siege attacks, ladders run up to the big wall, you get to the top, someone pushes you off with a stick, your ladder starts falling back. Say it gets pushed at 25 cm/second (1/4 m/s) and you are at the top of a 15 meter tall ladder, if you realize it you can start down the ladder with some variable speed, hopefully enough to keep you from getting your head crushed by the ground. So at top speed, say you are wearing gloves and you just let go, you are falling to the ground at your full 9.8 and you grab the ladder at the last moment to break your fall and turn to get out of the way of the ladder which is what you can do. But our guy goes straight down and just hopes he doesn't break his legs. So which one gets to the ground first? The guy falling at 1 G or the ladder going in its 15 meter radius circle?
- 18 May '11 20:58Lets hope the ladder is butted up against the castle wall, otherwise I think its bottom end rises when you grab it.

Assuming it is aginst the castle wall, grabbing it at the last moment has an effect that depends on the mass of the ladder. E.g, if the ladder is as light as a feather your grab pulls the other end down very fast and hardly breaks your fall at-all. - 18 May '11 21:52

Yes but which one hits the ground first? The ladder is middle ages, massive. I would think about 100 Kg, same weight as the person on the ladder.*Originally posted by iamatiger***Lets hope the ladder is butted up against the castle wall, otherwise I think its bottom end rises when you grab it.**

Assuming it is aginst the castle wall, grabbing it at the last moment has an effect that depends on the mass of the ladder. E.g, if the ladder is as light as a feather your grab pulls the other end down very fast and hardly breaks your fall at-all. - 19 May '11 00:56There is not really enough information here, so I will make some assumptions:

The ladder is standing at 90 degrees (it actually would be leaning against the wall at some undescribed angle).

The ground is 90 degrees from the top of the ladder.

The man moving down the ladder would not affect the speed of the ladder. This is, of course, a bad assumption, but since we are not given the mass of the man or the mass of the ladder, we can't make determinations on how his movements will change the speed of the ladder.

The man moves down the ladder at uniform speed.

The falling man does not affect the ladder's speed no matter where he grabs it, or how much he decelerates when he does grab the ladder.

The ladder moves at it's end at a uniform speed of 1/4 m/s. Of course, this would not be true, but due to our not knowing various things, we can't do any better.

The circumference of a circle is 2 x 3.14 x r, or about 90 meters with the 15 meter ladder at the center. At 90 degrees, the ladder will travel 1/4 of it's circumference before it hits the ground, or 1/4 x 90 or about 23 meters.

At 1/4 meter per second, it will take 92 seconds for the ladder to hit the ground, so obviously, the man can travel down the 15 meter ladder much faster than this no matter how he does it.

This puzzle needs more information to do properly. - 19 May '11 14:59im not sure, but this seems more like a physics problem...not quite a puzzle.

And It seems that if you arrived at a solution, you over-simplified the question a good bit.

but here we go

(Situation 1) where the man rides the ladder, staying at the top for the duration.

assumptions:

intitial angle between the ladder and the surface is 90 degrees such that

Ao = pi/2 (rad)

The man is not an extended body, ie it is treated as a point mass. His moment of inertia about the point "C" at the base of the ladder is given by

I_m = m*L^2

where

m= mass of man

L= length of ladder

The monent of inertia of the ladder about the point "C" is estimated by a thin rod and yeilds

I_l = 1/3*m_l*L^2

where

m_l = mass of ladder

and we are going to ignore any other forces that would be acting on the body

using angular verions of the kinematic equation of motion start with

A= Ao + Wo*t +1/2*S* t^2

Wo = intitial angular velocity

S = angular acceleration of the system

Wo = Vo/L

S can be found by summing the moments about "C"

Sum(Mc) = I_total*S

Where the moments are caused by the weight of the man and the weight of the ladder

After some ( what I hope was ) careful algebra The equation of motion for situation 1 is

A = Ao + (Vo/L)*t + (3/4)*g* cos(A)* [(m_l + 2m)/(m_l + 3m)]* t^2

as can be seen its probably pretty readibly solved for t given all the other perameters

note*(g is plugged in as negative value)

For situation 2 where the man is free falling down the ladder we can use the same equation, but coincidentally any term in the analysis attached with the mass of the man is now a time dependent funtion vicariously through length.

that is to say L_m = L(t)

assuming he begins with zero velocity in the -y is standard coordinated

L(t) = L + 1/2*g*t^2

so the situaton 2 equation becomes (warning this is going to look nasty)

A = Ao + (Vo/L)t + (1/2)*{[m_l*g*cosA*L/2 + m*g*cosA*(L + 1/2*g*t^2)]/[m*(L+1/2*g*t^2)^2 + 1/3*m_l*L^2]}*t^2

as can easily be seen this will be a bit tougher to solve for t - 19 May '11 15:40 / 5 edits

I did later say the ladder and the man massed 100 kg. Sorry, I made unstated assumptions about the setup, at first vertical with reference to the ground and the ground indeed 90 degrees from the vertical.*Originally posted by TomCr***There is not really enough information here, so I will make some assumptions:**

The ladder is standing at 90 degrees (it actually would be leaning against the wall at some undescribed angle).

The ground is 90 degrees from the top of the ladder.

The man moving down the ladder would not affect the speed of the ladder. This is, of course, a ba han this no matter how he does it.

This puzzle needs more information to do properly.

You state the man moves down the ladder at a uniform speed, don't you mean a uniform acceleration?

Also, it seems the man would be pushed away from the wall by the ladder on the way down, not much as I think he gets to the ground well before the ladder and the energy it takes to move the man away from the wall would be taken from the kinetic energy of the ladder's fall so that would complicate things a bit! It looks intuitively like the path the man would take, that is to say, the graph of his motion in Z and X would be a parabola.

As T=Sqr root (2S/A) for simple freefall time, 15 meter fall takes very close to 1.75 seconds. 30/9.8= 3.06, square root of that being 1.75 seconds, close enough anyway.

It looks to me like the X movement would be the half the ladders movement in 1.75 seconds and when the ladder is vertical, I think it safe to say for the initial velocity of 0.25 m/s while vertical, the assumption can be made to ignore gravity and just go with that initial push velocity.

So in 1.75 seconds, the top of the ladder would have moved 0.35 meter and as the man falls down he would be traversing sections of the ladder moving progressively less and less where at the bottom the movement in the X direction would be zero so it seems safe to say the ladder would have pushed him away from the wall half of 0.35 or 0.175 meter.

Does these seem accurate numbers? One assumption I made is to ignore friction between the man and the ladder on his downward traverse.

If those numbers are correct and he gets to the ground in less than 2 seconds, it seems for that amount of time he would be slowing down the fall of the ladder so it would have a skewed velocity curve on its downward fall, say 1.75 seconds of velocity A followed by strictly gravitational fall. I don't know how to calculate that one though. - 19 May '11 17:12

plugging in all the variables into the equation and solving for "t" ( using my graphing calculator in both cases.*Originally posted by joe shmo***im not sure, but this seems more like a physics problem...not quite a puzzle.**

And It seems that if you arrived at a solution, you over-simplified the question a good bit.

but here we go

(Situation 1) where the man rides the ladder, staying at the top for the duration.

assumptions:

intitial angle between the ladder and the surface is 90 degrees ...[text shortened]... )^2 + 1/3*m_l*L^2]}*t^2

as can easily be seen this will be a bit tougher to solve for t

situation 1

t= 2.08 seconds

situation 2

t = 1.66 seconds - 19 May '11 18:14 / 1 edit

Did you read my work about the man moving in X direction? Is my assumption safe, when the ladder is vertical, you can ignore gravity so for the initial time, only the 0.25 meter per second movement of the top of the ladder is needed. Also, isn't my free fall solution right, 15 meter drop takes 1.75 seconds, looks like thats what it would round up to.*Originally posted by joe shmo***plugging in all the variables into the equation and solving for "t" ( using my graphing calculator in both cases.**

situation 1

t= 2.08 seconds

situation 2

t = 1.66 seconds - 19 May '11 20:13 / 1 edit

Your time for free fall is ok, but its irrelevant and you can't, or rather shouldn't ignore gravity.*Originally posted by sonhouse***Did you read my work about the man moving in X direction? Is my assumption safe, when the ladder is vertical, you can ignore gravity so for the initial time, only the 0.25 meter per second movement of the top of the ladder is needed. Also, isn't my free fall solution right, 15 meter drop takes 1.75 seconds, looks like thats what it would round up to.** - 20 May '11 12:31 / 1 edit

To be honest, its hard to tell what you are or aren't ignoring. If you could provide more detailed mathematics accompanied by minimum logic, in a step by step format we could work out what you are doing. The secret to understanding Math/Physics lay in thrift, that is to say, cut your response down to the bare essentials. I also feel that it would be benificial if you went through my explaination line by line asking no more than 2 questions per post, it will make it my response to your inquiry more direct.*Originally posted by sonhouse***I'm just ignoring it for a very brief time frame, I think anyway.**

Also, Ignore my analysis of sitiuation 2 as well as the answer of 1.66 seconds, I just realized that it is wrong...

Why?

because, the base kinematic equation I used is only valid for**constant**angular acceleration, I will have to find a new approach to solve that scenario, but the rest should be ok. - 21 May '11 15:07

It would be nice to have a blackboard or something to write things out and help with visualizations. It seemed pretty simple to me, the ladder is upright at T0, a thrust is given to move the ladder at 0.25 meters per second and I assume the gravitational effect would be minimum when vertical, no angular acceleration at all so for the first 10 or 20 degrees of angular movement, I am ignoring gravity for that angle of movement so I visualize the guy freefalling but moved in the X direction by contact with the ladder till he hits the ground in 1.74 seconds, 15 meters of freefall, frictionless contact with the ladder. Since the ladder is 15 meters hight, that represents the radius of curvature, approx. 100 meters in circumferance. If the ladder has only moved between 350 mm and 500 mm, that represents an angle of around 1 to 2 degrees of angular movement max, since one meter of movement would = about 3.6 degrees (100 meters circumferance=~360 degrees). Considering the small angle at that point it seems safe for me to say you can ignore gravitational angular acceleration of the ladder for the timeframe of the freefall.*Originally posted by joe shmo***To be honest, its hard to tell what you are or aren't ignoring. If you could provide more detailed mathematics accompanied by minimum logic, in a step by step format we could work out what you are doing. The secret to understanding Math/Physics lay in thrift, that is to say, cut your response down to the bare essentials. I also feel that it would be beni ...[text shortened]... leration, I will have to find a new approach to solve that scenario, but the rest should be ok.**

Does that logic hold up? If there are flaws in this quickie analysis, please show me where I went wrong. - 21 May '11 17:51 / 1 editI guess another assumption would be the ladder is on a frictionless bearing at the bottom and the starting position is perfectly vertical, 90 degrees from the horizontal ground and falls with no air friction or mechanical friction falling down to horizontal.

I think I mentioned the ladder and the man massing 100 Kg each. - 21 May '11 22:17

You know, I just realized neither case has constant angular accerleration...which means both of my results are*Originally posted by sonhouse***It would be nice to have a blackboard or something to write things out and help with visualizations. It seemed pretty simple to me, the ladder is upright at T0, a thrust is given to move the ladder at 0.25 meters per second and I assume the gravitational effect would be minimum when vertical, no angular acceleration at all so for the first 10 or 20 degrees ...[text shortened]... t logic hold up? If there are flaws in this quickie analysis, please show me where I went wrong.****incorrect**.

Anyhow, you have a contradiction.

If you are to assume that there is no angular acceleration, then you are neglecting both the mass of the man, and the mass of the ladder. But you stated the mass of of both.

And If you neglect these. it takes the massless man 94 seconds and some change to get to the ground, and surprise, he gently touches the ground at .25 m/s.

Thats a pretty graceful landing speed for a 45-50ft fall!!!

So you see, you solved the problem for a man on the ladder in space, not on the surface of a planet. - 22 May '11 02:45

I didn't say I totally ignored angular acceleration, just for the first couple of degrees of movement of the ladder. I stated there would not be much in the way of angular acceleration in that time frame, I think the velocity does not follow a freefall since it is clear there is not much acceleration in the first few degrees. Later, 10 degrees, 20 degrees, it's a different story.*Originally posted by joe shmo***You know, I just realized neither case has constant angular accerleration...which means both of my results are [b]incorrect**.

Anyhow, you have a contradiction.

If you are to assume that there is no angular acceleration, then you are neglecting both the mass of the man, and the mass of the ladder. But you stated the mass of of both.

And If you ...[text shortened]... u see, you solved the problem for a man on the ladder in space, not on the surface of a planet.[/b]

For instance, for the time the ladder is fully vertical, 90 degrees to the horizontal ground, there is zero angular acceleration by definition since it is not moving and without the gentle push of the 0.25 m/s shove, would remain vertical unless there was another force like wind or earthquake to set it in motion to the ground.

All I was doing was simplifying the freefall situation. I even added a fudge factor, saying in effect, even with angular acceleration which clearly would be very small within a few degrees of vertical, I said 1 or 2 degrees of movement when the 0.25 m/s shove was added then adding a fudge factor of a reasonable estimate of angular acceleration, it still would be only about 2 degrees which is less than a meter of movement anyway, so even factoring in a hopefully reasonable estimate of angular acceleration at that point it would not have gone more than 2 degrees in 1.75 seconds so would be approximately vertical when the man hits the ground.

When he gets to the ground, maybe he could even stop the motion and put the ladder back in place to continue the attack on my mythological middle age castle.