Falling Ladders: Newsletter Challenge (03/04/08)

all 3 will hit the ground at the same time.

regards,

lashsd@yahoo.com

The real question here is: if three ladders fall on a plane and there is no one there to hear them fall, will they make any sound?

Sound is sound only when it is heard, not otherwise. -S. Saran

Maybe the black box flight recorder

P.S. The reason for starting at a thirty-degree angle is an attempt to make the shape of the foot of the ladder irrelevant to the problem. Other than that, the angle is arbitrary.

I'm new to this forum and have not read of the comments yet, my feeling is that the ladder that is on a frictionless plane without any walls or other influence would hit the surface of the plane slightly ahead of the other two ladders, which would both hit the surface at the same time. This is because the frictionless surface would only impose resistance to vertical forces, such as at the feet of the ladder. The ladders will fall in a path via gravity based on their respective centers of gravity. The difference is that the ladders that have a wall or non-slip contact with the surface is that the COG must follow an arc to hit the surface. The other ladder will have the feet slide as the COG follows the most direct path to the surface of the plane; a vertically perpendicular drop to the surface.

None of the three hit the ground. They each hit the surface of the plane at the same time.

-John

Yes, but after they strike the plane they hit the plain, but it depends on what the plane's AGL is above the plain.

Okay, I'm stalling, but if the plane stalls after getting struck by a ladder, then the plane may hit the plain before the ladders.

I would say the one with full slip would hit first. As the other two have a horizontal force applied to them at the base, there is a centrifical force involved. As the ladders rotate through an arc, they will experience an impedment due to the force on the base. Any force acting on a system will affect the acceleration due to gravity (ie the objects are not in free fall).

... As the other two have a horizontal force applied to them at the base, there is a centrifical force involved. ...

We may rephrase it like this: The ladders that experience an horizontal force (reaction from a wall or friction), also experience a smaller torque, as compared to the slipping ladder. Therefore, the 30 degrees rotation needed to turn the ladders horizontal will be slower for the ladders with the impediment at their bases. I need to check the reaction forces from the ground to nail this better...

If this logic is correct, then the ladder that will arrive second, will be the one with the contact against the wall, as it will leave it at a certain point (reminds of the dumbbell challenge) after which it will move under the same conditions as the first ladder.

Another way to see this, maybe better than the above one, is considering that the impeded ladders CGs are forced to move on a circle (at least for part of their "journey"). This means that we will always have a horizontal component in the CG velocity: this in turn "draws" some kinetic energy from the vertical component, consequently causing the CG to reach the ground later.

You seem to be on the right track here Tkot, just one small point:-

The ladders are 30 degrees to the vertical which means they must rotate through 60 degrees to land flatly on the plane.

And yes, this does look like a triple scenario of the barbell challenge which I missed out on.

Well maybe we just did not want to have you feel left out there Maniac.

OK, folks, it's a plain as well, but it needs to be perfectly flat (a plane) - that's probably why the CR4 team left my mistype unchequed. (If you want it to be a flying object, you'd all better follow guessed's advise and add an apostrophe).

;-)

A geometric plane, a (aircraft) plane and a carpenter's plane are spelled the same, but since we didn't know the answer to the puzzle the plain truth is we just had to be silly.

But wouldn't the grass growing on the plain offer any resistance?

I'm going to petition the Challenge editors for a question in metallurgy or process engineering. All these physics questions are giving me a headache. My instinctive and probably incorrect response is that all 3 ladders hit the plain, plane, or ground at the exact same time. I see it similar to if you were to drop a bullet from a height of 1 metre, or fire it dead horizontal from a height of 1 metre both bullets would hit the ground at the same time. OK engineers-how wrong am I?

Depends on the aerodynamics of each bullet, and of course the curvature of the earth, etc.....

as long as we're giving fyz editorial license to have frictionless planes trains and automobiles, my bullets are aerodynamically perfect and the earth is flat

cusnmonkey,

As any good long range shooter knows it doesn't actually matter if the earth is flat or not and the BC of the bullet has no bearing on length of time to drop to earth. The only determining factor is the bullets being released at the same time from captivity, ie. from the end of the barrel and the other device at exactly the same time and under identical conditions, same air density, wind speed, etc... They both strike the earth at the same time. Just one is going a lot faster.

Chris

I see it similar to if you were to drop a bullet from a height of 1 metre, or fire it dead horizontal from a height of 1 metre both bullets would hit the ground at the same time. OK engineers-how wrong am I?

We are talking about different things. The bullets paradigm implies free fall, i.e. in the vertical axis it is only the weight that governs either bullet's motion. On the other hand, in the case of the ladders, the motion is not a free fall.

Imagine two bodies, one free falling and another sliding down a slope. They will both cover equal altitudes eventually reaching equal velocities (if there is no friction) but the trip time will be higher for the sliding body, due to lower acceleration.

As the base of the ladders has been eliminated, I will take a ques and say they fall in the reverse as stated. The one that does not slip being first, as the base will not leave to ground, second the ladder with its base against the wall, as momentum imparted to the centre of gravity will slightly lift the base off the ground, thirdly the the frictionless plain as the ladder will rotate about its centre of gravity lifting the base of the ladder further off the ground delaying contact with the floor.

Its all quite plane?

Regards JD.

As they are in true frictionless environment they may just float off into space because I am unaware of any frictionless environments on earth except supercooled environments in a vacuum and then the questions becomes which ladder breaks first as it hits the plane?

My vote is frictionless one lands first, base contact one second and of course the non-slipping one last as it has the most friction. The more friction the more drag and drag is just another way of saying slower.

Chris

I will thank the God for me not being there while falling.

When the ladder falls through air, it experiences friction in air. As the second ladder falls without friction (no air resistance, as I understand it), it will fall at faster rate.

First ladder will follow the second ladder as it falls on a frictionless area. As it slips fast the moment increases rapidly.

And, of course the third ladder will bthe last to touch the ground.e

The one that falls in an open frictionless area hits first, followed by the one that falls without slipping followed by the one that falls without friction away from the wall it is initially contacting.

While this really needs some maths to be backed up I can't be bothered (and don't think I have the skills to do a proper mathematical analysis) so here is my 'intuitive' reasoning. The one on the frictionless area experiences a rotation around its centre while falling while the other two are experiencing rotation around their ends which is slower. The motion of the one with a fixed end and the one on frictionless floor against a wall are identical up to the point where the one on the frictionless floor leaves the wall at which point its rotation is slower that the one whose end is fixed to the wall. This can also be seen if you trace the path that the end of the ladder takes in all 3 cases with the shortest path being the one on open frictionless area and longest being one initially against wall on frictionless surface.

Bob,

You have a very compelling argument:

For a theoretical infinitely thin line (so all points on the line are equidistant from the center of rotation) rotational inertia is the sum over R, of RδRδM - where R is the distance from the center of rotation, and δR and δM are a 'short increment of radius' and its equivalent mass, as δR tends to zero.

This becomes an integral of RdR, which yields the rotational inertia = (MR²)/2.

The ladder in the open frictionless space, as you say, has its center of rotation at its midpoint: so R = L/2. But it also has two 'arms' rotating about that center. Its total rotational inertia is therefore: 2 * (M(L/2) ²)/ 2 = (ML²)/4.

The other two ladders, having their feet constrained, have their center of rotation at the foot, and so R = L; and they only have one 'arm' rotating about that center. Their total rotational inertia is therefore: (ML²)/2.

The rotational inertia of the first ladder, therefore, is ½ that of the other two – so its rotation will accelerate faster.

Since for all ladders the center has the same distance to fall the time it takes the ladder to rotate from 30° off the vertical to 90° off the vertical is entirely governed by how quickly the ladders' rotational rate can accelerate (because all have their foot level with the plane, and we are assuming that the plane is solid and the foot cannot penetrate). Since the first ladder can rotate faster, sooner, it is the one that will fall to the plane soonest.

1,3,2.

"In what order do they hit the ground?"

Just being a bit pedantic about the phrase "hit the ground". does this mean if the bottom of the ladder does not leave the ground, then a hit is when the centre of gravity strikes the ground, or if the bottom of the ladder does momentary leave the ground is it a strike when it again touches the ground? As you can see if we are referring to the centre of gravity striking the ground, then time will elapse if any end of the ladder strikes first, this will cause the ladder to oscillate from end to end before the centre of gravity rest on the ground.

Regards JD.

Yes I must agree with guest, a simple experiment with a 12" rule, suggest that the bottom does not lift as it rotate about its centre of gravity?

Regards JD.

It seems to me that if they are standing on the plane to begin with that they are allready touching the plane before they fall.

A simple way to consider this is to use Einstein's Principle of Equivalence. Imagine conducting this test in an elevator moving upwards through infinite space far away from any large celestial bodies. The upward acceleration of the elevator is 9.8 meters per second per second, essentially the same as 'g' on Earth at sea level.

All three ladders will experience the same 'fall' towards the bottom of the elevator, and all three ladders will experience the same amount of rotation as they fall. In all cases the rotation is imparted by the floor to the ladder.

So all three ladders would hit the floor at the same time -- unfortunately the elevator compartment you chose to use was too small and all three ladders hit the far wall before they could reach the floor. So the test was inconclusive.

Seeing as how the question is looseley worded, one can assume anything, or nothing.

For instance, how far apart are the ladders?

How precise does the measurement have to be?

How far away is the measuring instrument (the observer?) that determines when they hit?

What is the angle to the observer relative to the the ladders? Are the ladders arranged in a radius equidistant from the observer, or are they in a straight line?

If the observer is not equidistant from all ladders, there will be a difference in measurement due to this factor alone.

Are the ladders falling toward or away from the observer, or at an angle?

Are alll ladders made from the same material? (there is a documented case of some materials not following the accelleration rules in free fall)

Do the second and third ladders fall in an open space?This is not specified, so it cannot be assumed.

Please restate the question more precisely, or be willing to accept variations in your answer(s)

A previous guest wrote:

"Are all ladders made from the same material? (there is a documented case of some materials not following the accelleration rules in free fall)."

My reply: You actually have documented evidence of materials that violate the principle of equivalence? That's literally incredible.

--Guest #36

"My reply: You actually have documented evidence of materials that violate the principle of equivalence? That's literally incredible."

Perhaps one, or more, of the ladders are made of feathers. But then, OSHA would go into convulsions, and the said feather ladders would be disallowed from the theoretical challenge problem...

Are you quite sure that 'Elf & Safety won't blow a fuse at the use of ladders without suitable anchors?

Ladder #1 hits first, the only thing slowing it down from free fall is the horizontal movement acrossed the frictionless plane. This movement will be less than the other 2 ladders. Ladder #2 and #3 seem to almost be the same thing, however, the key is "without slipping". The 2nd ladder's angular velocity will cause the top of the ladder to hit the ground before the rest of the ladder. #3 will hit all at once so #2 will beat #3 "by a nose".

3,2,1

But the important thing is that all three ladders, since they have been allowed to fall, must now be destroyed by the safety committee.

I am not really sure how the second case is different from the first case, but gravity is constant. They all fall exactly the same distance and have the exact same shape so the air resistance is the same, assuming sliding doesn't present a different amount of surface area cutting thru the air. They arrive at the same time I think is the answere he wants.

The difference between the three cases is how many degrees of freedom exist. The first case has 3, the horizontal and vertical locations of the base of the ladder and the angle between the ladder and horizontal. Case 2 has 2: the vertical location of the base and the angle. Case 3 has 1 degree of freedom: angle. (That's my take on the problem.)

Being that the potential energy has to be split up into three distinct modes in case 1, there should be less energy in it's downward velocity causing more time to elapse before contact with the ground. Case 2 (against the wall) should have the middle time, and case 3 (fixed base) should put all its potential energy into rotation causing it to contact the ground first. Now this does ignore that fact that the top of ladder 1 might hit the ground first since the vertical translation along with the rotation cause the top to approach the ground, but I'm too lazy to break out Lagrange multiplies tonight.

-Sully

Accudave hit on the same line of thinking as I composed my answer in Notepad. Here's what I concluded:

"Three identical ladders standing squarely on a plane are simultaneously released, starting thirty degrees to vertical."

I take this to mean that the ladders are being held up by hand, then tilted 30 degrees from vertical, and released by their holders at the same moment.

"One falls in an open frictionless area."

I take this to mean that the ladder falls in an area free of walls or any objects that it contact before it hits the ground. The lower end will slide as the upper end falls, and the ground presents no friction (maybe the ground has been slicked by an oil spill. The holder must have stood to one side of the ladder.

"The second falls, without friction, away from a wall that the base contacts initially."

I take this to mean the handler held the ladder against the wall before tilting and releasing it. The base of the ladder stays in contact with the ladder and air drag is ignored, minimal, or non-existant

"The third falls without slipping."

The only thing I can see from this is that the ladder encounters some form of friction and/or air drag.

"In what order do they hit the ground?"

1 - 2 - 3

The first ladder encounters no resistance. The second remains in contact with the wall at it s lower end until the top contacts the ground, then it slides away from the wall. The third encounters resistance of an unspecified type, but it is enough to prevent slipping.

I will have to give Accudave a vote for a Good Answer.

Gravity is the only force applied. Therefore, all of the ladders should hit at the same time.

In the first case, without friction or obstruction, the only thing keeping the ladder's center of gravity from free falling is the rotational inertia of the ladder. So, this one will hit first.

In the second and third cases the base is restricted from slipping horizontally, so the ladders cannot rotate about their CG. In addition to the resistance of rotational inertia, a decreasing portion of the ladder's weight is pressed sideways against the wall or whatever is restricting the base in the third case. So, less force is available to act against the rotational inertia. Furthermore, instead of falling straight down, the CG of the second and third case ladders must trace an arc.

I do not see how the second and third cases are different, until after the ladder hits the ground.

the ladder without friction will hit first.

the slipping ladder will hit nsecond

I submit that ladder #1 hits the ground first. Next, ladders 2 and 3 hit the ground simultaneously.

-John

Yes I agree that number one ladder will hit the ground first, because its centre of gravity will fall in a straight vertical line, but the centre of gravity of ladders two, and three, travels in the same arc, therefore 1 first, 2 & 3 together?

Regards JD.

Hi jdretired,

"therefore 1 first, 2 & 3 together?"

I'm not sure what you meant by posing that as a question. I said that falling order in post # 60. I believe we agree on it. How bout it Physicist?

Hi JD and JohnJohn,

Don't know if Fyz will reply to you without giving too much of the answer away as he set the question, so let me try:-

There is a difference between the path taken by the COG of ladders 2 and 3 and hence probably a difference in timings.

Ladder 3 does not slide at all so it can be considered to be hinged at the base and therefore it's COG traces an arc equal to 1/2 the ladder length.

Ladder 2 however, I believe, will behave like the dumbell and at some point between 30 and 90 degrees it will part company with the wall and slide horizontally across the frictionless floor until it hits the ground.

You said "[ladder 2] will part company with the wall and slide horizontally across the frictionless floor until it hits the ground."

According to the original problem: "The second falls, without friction, away from a wall that the base contacts initially. The third falls without slipping."

You could be right, but I'm with jdretired on this one. Intuitively I think we're got it pinned down (that's what engineers do don't they?).

-John

According to the original problem: "The second falls, without friction, away from a wall that the base contacts initially. The third falls without slipping."

Exactly. To paraphrase what I believe is meant( my interpretation words are in BOLD:-

The second falls (without friction between the ladder's base and the floor/wall) away from a wall that the base of the ladder contacts initially. The third falls without slipping at it's base ."

See diagram in my Post #70.

Hi MPM

So when ladder two falls it generates some centrifugal force, and that has to be provided by g, so the path taken is a resultant? So number three should hit the ground just before 2. interesting.

Regards JD.

Hi john

Yes I agree with your answer, and was meant to give a reason why.

bad habits?

Regards JD.

If you consider the non-slipping ladder, you will see that initially its CofG accelerates away from the base; however, when it hits the ground the horizontal components of velocity are zero. That implies that it will have decelerated again, which would require a horizontal force towards the base; this is not available for case 2 (frictionless and restrained on one side by a wall). So the case2 ladder (like the dumbbells) must at some point part company with the wall. Therefore cases 2 and 3 do not follow the same paths.
BTW, this line of reasoning applies regardless of starting angle and air resistance...

Over to you...

As a follow up to an earlier post, I decided to add some illustrations. To paraphrase an old saying, 1 visual image = 1K chunks of verbiage.

Ladder #1:

Ladder #2:

Ladder #3:

I am sure the first two properly illustrate what the question states. The third one shows a possiblity; the wall may or may not be there at all, or positioned where I have placed it.

Somehow I don't think FYZ had your #3 ladder in mind(?). See post #41. Although he didn't specify your ladder #3 situation specifically (I don't think). I'm not sure what your drawing shows is in the spirit of the problem. Maybe I'm wrong. Actually, I was wrong once before, maybe with the first wife, hmmmmmm

I agree with your analysis, except in the case of ladder 2 I understand that the base is moving AWAY from the wall, or it would be identical to ladder 3.

Regardless in cases 2 and 3 the COG's acceleration has some horizontal component that I believe to be the same in both cases.

In case 1 though, the COG will drop straight down vertically.

Intuitively in case 1 the forces acting against the fall are lower.

I say 1 hits first, then 2&3 together.

Three identical ladders standing squarely on a plane are simultaneously released, starting thirty degrees to vertical. One falls in an open frictionless area. The second falls, without friction, away from a wall that the base contacts initially. The third falls without slipping. In what order do they hit the ground?

Hi jfcayron,

A teacher of mine once said the first 3 things you should do in any exam are:-

1. Read the question

2. Read the question

3. Read the question

Now, having read the question again, how can you possibly interpret that ladder 2 has it's base moving away from the wall?

Remember the ladders are at 30 degrees to the vertical when released, therfore whether you interpret "base" to be base of ladder or wall, it both amounts to the fact that ladder 2 falls so that it's top is moving away from the wall.

And no, it does not mean that it is the same as ladder number 3. Look carefully at my diagram again in post #70

You are right, I misread the question.

It does not change the predicted outcome: 1 then 2&3 together.

I don't think the bottom of the ladders in cases 1 and 2 stay in contact with the plane. In fact the top of the ladder may overtake the bottom and contact the plane while the bottom is in the air.

Good question, but does it matter which part of the ladder strikes the ground first, I have sympathy for what you are proposing, but I believe that any part of the ladder striking the ground is classed as a hit. I strongly suspect that the base of ladder 2 leaves the ground, but does it land flat allowing ladder three to hit first, or does the top of the ladder 2 accelerate about its COG so the top of ladder 2, and the flat of ladder 3, hit at the same time?

Regards JD.

For ladder 2's base to leave the ground, it would gave to describe an arc through the wall.

Hi Kris

Get a 12" rule, place your finger at the bottom and let it fall. as it reaches the desk it will spring forward, most of it after the rule lands, in my mind the centripetal and centrifugal forces come into balance at approx 22½° before hitting the ground, it is then a matter of the whether the base slides or lifts, does the centrifugal force being generated by g acting in line with the ladder give an accelerated resultant, like when a ladder is falling down against the wall the bottom accelerates away from the wall faster than the top falls, so is this a case in reverse? I don't know, but put it up as a possibility? If it slides ladder 3 hits first, else how knows.

Regards JD

Hi JD,

I just tried a ruler (against a 'wall') on my desk, and was pretty sure the thing only moved it's bottom end after hitting horizontal, but I'm not 100% sure of what I could see. It's like a variation of the dumbbell question. The ruler experiment definitely ends with it moving (in the direction of lean), but some brave soul will have to do the math to convince me of what's going on. An unconstrained ruler slides back against the angle of lean. My only conclusion so far is that it's worth playing with a ruler to see how it behaves.

Cheers, Kris

You are right that the bottom of the ladder does not leave the ground until the ladder bounces - and even then only if either the ladder or the ground is rough (so that a point between the base and the CofG is the first to hit the ground).

I'll leave typing the sums for another day, though.

I think I have the answer regarding experiments: yesterday's CR4 output included a link to "Phun: a 2D physics sandbox". I cannot see why the experiment can't be run in a virtual world where we can set degrees of freedom / constraint to our liking, set coefficients of friction to 0 (zero) or 1.00 or anywhere between, etc. Nor can I see where a 2D world differs from looking at the ladder precisely edge-on, a projection plane view. So: anybody downloaded it yet? Learned how to use it? Have time to use it?

"So: anybody downloaded it yet? Learned how to use it? Have time to use it?"

Done the 1st, items 2 & 3 will have to fit in when I can manage it, looks like fun though.

Brilliant took about 1 minute to create this solo ping pong game:-

It's a real gas ! I was having a hard time positioning my 'ladder' with no friction, but didn't spend a lot of time on it. I'm fairly sure this question can be emulated on the thing, but that depends on how it was programmed to behave. It's worth a quick play with. Live dangerous and try it.

Hey, Really easy to do the bouncing ball simulation. One minute to do it half an hour to create this composite pic. Don't know why all the small balls are different colours.

Cool ! I didn't get the significance of the colour change either. I shall definately play with this prog some more when I have time. A whole bunch of previous challenge questions can be simulated on it.