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

I can't really see the intended difference between ladders 2 and 3. They both have a vertical component of reaction, and also a horizontal one (albeit stated as friction/lack of room to move sideways). For now, I'll just play pick-up sticks. I think 1 is first, then 2 & 3 at the same time (based on path of travel). Since they are ladders, they should follow the Order of the Garter (at least I didn't say the 'b' word )

They are the same until the base of ladder2 parts company with the wall. See post #79.

Let's see if I've learnt my lesson.

Ladder 1 will be the first to hit the floor. (It seems to be agreed that it's centre of Gravity falls straight down, there is no resistance to sideways movement of the base of the ladder and so that ladder base slides gracefully out of the way.)

Ladder 3 would seem to be next. Pivoting about a single point (without sliding) on the plane there is no other resistance to movement. That point of rotation will be the "bottom extreme corner" of the ladder rail rather than the centre line.

Ladder 2 will be last since for all of its rotation, the action of the centre of gravity and that ladder being on a frictionless (horizontal) plane will force the ladder against the wall and thus it experiences the restriction of movement (slowness) that this would create. (Wall was not stated to be of the same "magical" frictionless material as the plane.) Friction in the wall material providing a torque opposing rotation.

Again it seems that I'm running slightly against the mob.

By the way, where can I get some of that frictionless plain stuff? Want it for another project.

"Ladder 2 will be last since for all of its rotation, the action of the centre of gravity and that ladder being on a frictionless (horizontal) plane will force the ladder against the wall and thus it experiences the restriction of movement (slowness) that this would create. (Wall was not stated to be of the same "magical" frictionless material as the plane.) Friction in the wall material providing a torque opposing rotation."

I'm reading the situation for ladder #2 differently, I think. You are seeing it initially leaning against the wall with 30 degrees interior angle, correct? But the question says "The second falls, without friction, away from a wall that the base contacts initially." (my emphasis throughout) I cannot fault your sequence for the condition that you are analyzing, but I don't think it matches the original question. I would think that there is an implicit similarity to the previous "dumbbell-almost-against-a-wall" thread on this forum because that question led to this one, I believe.

Thanks for the feedback, some clarification (I hope)

I also see the ladder top falling away from the wall. What the frictionless plane allows though is the bottom of the ladder will try to slide (like the free standing condition) but be impeded by the wall.

Thus, the extreme bottom corner of the bottom of the ladder will remain in contact with the plane (sliding ever colser as the ladder falls.) and the upper corner of the bottom of the ladder will remain in constant contact with the wall. (and thus experience resistance to movement.)

The phrase "The second falls, without friction" was meant to mean that there was no friction anywhere - i.e. no friction from the wall. The intention was that all cases would be lossless** (at least until the ladders hit the ground, at which point all bets are off). Sorry if this was not clear.

**In practice air losses would not affect the result - but I didn't want to introduce yet another variable.

Fyz

Indeed it did.

Fyz

Here is my angle (Excuse the pun) on the maths of the sliding ladders:-

Let us first look at ladder 1 and ladder 3.

Ladder 1 will experience a rotational torque about it's centre of gravity whereas ladder 3 is experiencing a rotational torque about it's base point. The torque is provided by the mass of the ladder acting vertically downwards through the COG and the opposing normal Reaction provided by the plane. Thus the torque is the same on all ladders for any given angle theta. What is different however, is the rotational Inertia I. As you will see from the calculations below the angular acceleration on ladder 1 is 4 times larger than the angular acceleration on ladder 3 thus ladder 1 must reach a theta of 90 degrees before ladder 3.

Finally ladder 2 will behave like ladder 3 until it reaches that critical angle (As in the barbell challenge) when it's base parts company with the wall. From then on it will behave like ladder 1 because it is no longer rotating around it's base but now rotates around it's COG just like ladder 1. It's angular acceleration should quadruple and again it will rotate faster than ladder 3 and hence hit the ground before ladder 3, but after ladder 1.

Thus my prediction on the sequence for hitting the ground is :-

ladder 1 first, followed by ladder 2 and finally ladder 3

Does the ladder in case 2 leave contact with the wall first or the floor first?

-Sully

The foot of the ladder does not leave the ground while the ladder is falling. This is true for all three cases.

If we performed the same procedure with a light-handled sledgehammer (head at top), case 3 would leave the ground - but I can't honestly say it would then meet the "non-sliding" constraint.

Finally ladder 2 will behave like ladder 3 until it reaches that critical angle (As in the barbell challenge) when it's base parts company with the wall. From then on it will behave like ladder 1 because it is no longer rotating around it's base but now rotates around it's COG just like ladder 1. It's angular acceleration should quadruple and again it will rotate faster than ladder 3 and hence hit the ground before ladder 3, but after ladder 1.

Try swinging a racket or a club. During the period after ladder 2 leaves the wall: some of the horizontal energy of ladder 3 is converted into rotational (and CoG downward) energy. Hence 3 gets there slightly before 2.

Hi Randall,

I did have the same thought as you when I first started to think about the motion of ladder 3, however that would mean that ladder 3's angular acceleration would suddenly increase at that critical angle when the COG's horizontal velocity reaches a maximum and unfortunately for that to happen some extra force would have to appear to make this so. I believe that the horizontal Force of friction, although it changes direction at the "Critical angle" is present at all times to ensure that the COG of ladder 3 is travelling in a circular path. Thus it is providing for the centripetal acceleration needed to keep the COG of the ladder moving in a circular orbit.

Again, try and imagine ladder 3 as being hinged to the ground ( No slip). The torque force which is causing this ladder to rotate about it's base is simply mg.SinΘ . L/2 as I have shown in post #103

This torque is certainly varying with the angle Θ, but it does not suddenly jump at the critical angle when COG horizontal velocity reaches it's maximum.

The only factor that "jumps" or has an instantaneous change at this critical angle is the rotational inertia, I, of ladder 2. Ladder 2's Rotational inertia has an instantaneous change from mL^2_/3 to mL²/12 when it's base looses contact with the wall and as Torque = Rotational Inertia X Angular acceleration, this essentially means that as the Rotational Inertia is instantaneously reduced by a factor of 4, the angular acceleration must instantaneously be increased by a factor of 4. I.E. The Torque must keep changing smoothly as per the sinusoidal function which defines it.

I do hope this makes sense.

High time for another hint, I think:

What is the horizontal velocity of the Centre of Gravity of ladder 1 when it hits the ground?
Ditto ladder 3.
Now what about ladder 2?

What does that imply about the relationship between the vertical velocities of the three ladders as they hit the ground?

Horizontal velocity of ladder 1 and ladder 3 is zero when it hits the ground.

Horizontal velocity of ladder 2 is whatever max velocity it had when it parted company with the wall....

Now you have put me thinking deeply about the significance of this.....Hmmmmm.....

Horizontal velocity of ladder 1... is zero when it hits the ground.

I don't think so. As ladder 1 falls, the pull of gravity on the top will push the bottom to one side.

I conducted a quick experiment to confirm this. I have a six-inch steel ruler that a keep on my desk (it doubles quite well as a back scratcher!) that I held at an angle and released. I also had the tip of my index finger on the other hand marking the point where the base was at the beginning of its fall. Twice the base end stopped about 1 1/2 inches away.

Yes, 3Doug,

The base of ladder 1 will have an initial horizontal velocity and the top of the ladder will have an equal an opposite horizontal velocity but the COG will have zero horizontal velocity. Hence when ladder 1 hits the ground at 90 degrees to the vertical the base and top will also have decelerated to a zero horizontal velocity at this point.

That is what I meant in my reply to Fyz.

Thanks MPM - I think I understood you. I'm pretty certain from your diagrams that you are using the correct conditions. For simplicity of analysis (though it is not strictly necessary to arrive at correct answers) I've defined all three so that there are no losses until the ladders have struck the ground. So everything can be seen in terms of conservation of energy.

Fyz

MPM, I did another experiment where I held the ruler at its C/G and marked a point under the C/G. The ruler lands with the C/G to one side of the mark.

JF, the friction in my experiments is neglible. The ruler is very thin, and the surface of the desk is very smooth.

Well 3Doug,

What can I say, other than it is very difficult to sometimes recreate these idyllic environments in the real world. Perhaps, you may be imparting a slight horizontal velocity as you let go of the ruler - I don't know.

However, I think you will agree, if you think about it, ladder 1's COG must fall vertically downwards because in order for it to move horizontally there would have to be force horizontally applied at some stage during it's fall and as it is falling on a frictionless floor this horizontal force can never exist.

I do, however, agree that it's base and top points will move horizontally but in opposite directions to each other, hence cancelling out any horizontal movement of the ladders centre-point.

Did you run that experiment on a true zero-friction surface?

The friction will bias the test towards the #2 & #3 cases. The horizontal motion should be null in absence of any friction.

I still feel my initial prediction was correct; ladder 1 hits the surface first, with ladders 2 & 3 hitting second at the same time. All 3 ladders are identical and will act identically in identical situations, but we're given 3 situations with different conditions. I also feel that the conditions imposed for ladders 2 & 3 are very similar and have the same influence on the ladder movement. I think we all agree that ladder 1 on the frictionless surface will fall such that its COG will drop straight to the ground while the base of the ladder slides without resistance along the plane. This means all the ladders would do the same thing if they did not have their other influences.

Ladder 2 has its base against a frictionless wall. The wall can only offer resistive force in one horizontal direction, but the ladders base will only want to slide in one direction. The ladders COG path will thus follow an arc because the base can not slide along the floor as with ladder 1 because the wall prevents the horizontal movement with exact and opposite force equal to the ladders force to slide the base. Therefore, the base does not move.

I feel that ladder 3 is similar to ladder 2, except the non-slip surface provides the resistance. The non-slip surface offers resistance to horizontal movement in any direction. Since, we know that ladder 1 wants to fall by allowing the base to slide in one direction, that's all that ladder 3 would do if it weren't for the non-slip surface. So, I feel that the non-slip surface acts the same way as the wall by providing forces equal and opposite to the sliding motion of ladder one, and resulting in ladder 3's COG following the arc of ladder 2. Horizontal forces and velocities from centripetal influences would be the same for ladder 2 & 3, thus causing them to fall at the same rate and path. You can run the numbers to find out what that path would be, but for this question, I feel it's not important. We are asked for the order in which they hit the ground. I think the answer to the question posed is ladder 1 hits the surface first, with ladders 2 & 3 hitting second at the same time.

Please let me appologize for not reading post #79. I'd like to revise my last post with the following.

Based on the info in post 79, think I can see what would happen. I think what your telling us is ladder 1 hitting the surface first, followed by #3, then #2. All 3 ladders are identical and will act identically in identical situations, but we're given 3 situations with different conditions. I also feel that the conditions imposed for ladders 2 & 3 are very similar and have the same influence on the ladder movement. I think we all agree that ladder 1 on the frictionless surface will fall such that its COG will drop straight to the ground while the base of the ladder slides without resistance along the plane. This means all the ladders would do the same thing if they did not have their other influences.

Ladder 2 has its base against a frictionless wall. The wall can only offer resistive force in one horizontal direction, but the ladders base will only want to slide in one direction. The ladders COG path will thus follow an arc because the base can not slide along the floor as with ladder 1 because the wall prevents the horizontal movement with exact and opposite force equal to the ladders force to slide the base. If the centripetal forces cause the base to part company with the wall due to a horizontal force influence away from the wall, the COG would follow an arc of radius L/2 until the horizontal force away from the wall exceeds the force to press the base against the wall. Then the COG would follow a curved path the ground. Since this would be a longer arc than in ladder 3, ladder 2 would be the last ladder to hit the surface.

I feel that ladder 3 is similar to ladder 2, except the non-slip surface provides the resistance. The non-slip surface offers resistance to horizontal movement in any direction. Ladder 3 would then follow the same path as ladder 2 if it weren't for the ability of the non-slip floor to prevent movement away from the initial location of the base, and this causes the COG to fall in an arc with radius L/2.

Sorry for not reading post 79 earlier. My mistake

Hi TL21043,

You say:-

Then the COG would follow a curved path the ground. Since this would be a longer arc than in ladder 3, ladder 2 would be the last ladder to hit the surface.

Yes, you are correct that ladder 2 follows a curved path to ground, and that curved path is outside the circular arc and hence the path is longer, but that does not necessarily mean that it takes a longer time. Remember that ladder 2 will leave the wall with some horizontal velocity V_h and on a frictionless surface this Vh will remain constant unless acted on by a force (Once it leaves the wall, there is no horizontal force now for ladder 2). Remember this Vh will in no way affect how ladder 2 will now fall vertically. If you throw a ball 1 horizontally from a tower with an initially velocity gretaer than zero and drop a similar ball 2 from the same tower, then ignoring drag/losses etc they will both hit the ground at the same time although ball 2 travels a longer path!!!

Now, please don't think that I am advcocating that ladder 2 and ladder 3 hit the ground together, you will need to read my previous posts to discover my thoughts on the landing sequence of the 3 ladders...

Thank you, MPM, for pointing out that oversight I had in the longer arc for ladder 2. I should have remembered that. However, that makes me think since the vertical velocities are constant due to gravity, only a force acting in the vertical direction would influence the ladders velocity. Since none of the situations has any other influence in the vertical direction, wouldn't they all hit at the same time? Horizontal velocities due to rotation, etc., would have no influence on the fall. If that's true, then aren't all the discussions about horizontal velocities, although interesting, irrelavent to the answer?

It so happens that you are right but not so much for the reason of a longer path length.

See my post 146 just before you where I eat humble pie.

You may need to read previous posts to see the complications that rotational motion brings to horizontal and vertical velocities...

You can balance a long stick upright by moving horizontally only. Or you can accelerate its rotation. Of course that is quite obvious when the stick is upright, because the vertical motion is negligible. But what you are doing directly is to influence the rotation, and in principle you can continue to do that whatever angle the stick is at (other than horizontal). If the stick (or in our case ladder) is to remain in contact with the ground, the vertical movement of the Cof G must also change with the rate of rotation. The force that allows this change to the vertical velocity is due to upward pressure from the ground; this is self-adjusting just so long as the rotation allows the ladders to provide any downward pressure.

That's quite a mouthful, but I hope it helps.

Fyz

Aha, The light has dawned...Randall is right...

Hi Randall,

You have my humble admiration. I now believe you are correct and the order of hits is in fact 1, 3 and finally 2 and you actually stated this in post 26.

I think I now see the error of my earlier predictions regarding ladder 2:-

Once ladder 2 starts rotating about it's end point, it continues to rotate about it's end point even after it leaves the wall at an angle of 54.74 degrees (Please tell me your calculations agree with this ). It doesn't suddenly change to rotation about it's centre as I was predicting earlier. (Question for Fyz :-Does the rotation point start to move from the base towards the COG at all before ladder 2 hits the ground? )

Up to this angle of 54.74 degrees, ladder 2 and ladder 3 will be tracking each other exactly.

Now, immediately after this angle, the friction force on ladder 3 will start to decelerate it's COG horizontal velocity and of course this energy must get transferred to the COG vertical velocity ( Thanks for the final velocity hint Fyz and the swinging club analogy Randall ). Hence from 54.74 degrees until 90 degrees, the vertical velocity of ladder 3 will increase more rapidly than the vertical velocity of ladder 2 which has no frictional force to convert that parting horizontal velocity to a vertical component. Hence ladder 3 beats ladder 2 between the 54.74 and 90 degrees when it hits the ground.

Finally I thought this pop-up quote to be most appropriate and worthy of repeat:-

"The greatest thing in this world is not so much where we stand as in what direction we are moving." -- Johann Wolfgang von Goethe (1749-1832), playwright

To finally clarify the happenings of ladder 2, if we revisit the barbell challenge we will see that the barbell which started at an angle of zero degrees to the vertical left the wall at an angle whose cosine equalled 2/3 or 48.1897 degrees. This happens because when you balance out the loss of potential energy to the gain in kinetic and rotational energy, there comes a point when the horizontal velocity reaches a maximum and at this point the falling and rotating object must leave the wall.

Now the barbell challenge had the barbell leaving the wall at y = 2/3.r

If we were to revisit the equations of the barbell challenge and start with an initial angle Θ we would find that y now becomes 2/3 .r.CosΘ. and hence the angle at which the ladder of barbell parts company with the wall is Inverse Cosine(2/3.CosΘ).

I have listed below the contents of an excel spreadsheet that shows how the parting angle varies with the initial release angle. As you will see, when the initial release angle is 30 degrees to the vertical (As in this case), ladder 2 will part company with the wall at an angle of 54.7356 degrees to the vertical.

Furthermore, it is interesting to note that no matter what angle one releases ladder 2 it will always loose contact with the wall before it hits the floor at 90 degrees to the vertical E.G. Releasing at 85 degrees it looses contact at 86.669 degrees

Column 1 = Starting Angle to the vertical in degrees

Column 2 = Cosine of this angle

Column 3 = 2/3 Cosine of this angle

Column 4 = Inverse Cosine of result in column 3 = Departing angle

Column 5 = Angle difference (Departure angle - Initial release angle)

Theta cos(Theta) 2/3 cos(Th) Inv Cos(2/3Cos(Th)) Angle Difference
0 1.0000 0.6667 48.1897 48.1897
1 0.9998 0.6666 48.1975 47.1975
2 0.9994 0.6663 48.2209 46.2209
3 0.9986 0.6658 48.2599 45.2599
4 0.9976 0.6650 48.3144 44.3144
5 0.9962 0.6641 48.3844 43.3844
6 0.9945 0.6630 48.4698 42.4698
7 0.9925 0.6617 48.5705 41.5705
8 0.9903 0.6602 48.6865 40.6865
9 0.9877 0.6585 48.8176 39.8176
10 0.9848 0.6565 48.9636 38.9636
11 0.9816 0.6544 49.1245 38.1245
12 0.9781 0.6521 49.3000 37.3000
13 0.9744 0.6496 49.4901 36.4901
14 0.9703 0.6469 49.6944 35.6944
15 0.9659 0.6440 49.9130 34.9130
16 0.9613 0.6408 50.1454 34.1454
17 0.9563 0.6375 50.3916 33.3916
18 0.9511 0.6340 50.6514 32.6514
19 0.9455 0.6303 50.9244 31.9244
20 0.9397 0.6265 51.2104 31.2104
21 0.9336 0.6224 51.5093 30.5093
22 0.9272 0.6181 51.8208 29.8208
23 0.9205 0.6137 52.1447 29.1447
24 0.9135 0.6090 52.4806 28.4806
25 0.9063 0.6042 52.8283 27.8283
26 0.8988 0.5992 53.1877 27.1877
27 0.8910 0.5940 53.5583 26.5583
28 0.8829 0.5886 53.9400 25.9400
29 0.8746 0.5831 54.3325 25.3325
30 0.8660 0.5774 54.7356 24.7356
31 0.8572 0.5714 55.1490 24.1490
32 0.8480 0.5654 55.5723 23.5723
33 0.8387 0.5591 56.0055 23.0055
34 0.8290 0.5527 56.4481 22.4481
35 0.8192 0.5461 56.9000 21.9000
36 0.8090 0.5393 57.3610 21.3610
37 0.7986 0.5324 57.8306 20.8306
38 0.7880 0.5253 58.3088 20.3088
39 0.7771 0.5181 58.7953 19.7953
40 0.7660 0.5107 59.2898 19.2898
41 0.7547 0.5031 59.7921 18.7921
42 0.7431 0.4954 60.3019 18.3019
43 0.7314 0.4876 60.8191 17.8191
44 0.7193 0.4796 61.3433 17.3433
45 0.7071 0.4714 61.8745 16.8745
46 0.6947 0.4631 62.4123 16.4123
47 0.6820 0.4547 62.9566 15.9566
48 0.6691 0.4461 63.5071 15.5071
49 0.6561 0.4374 64.0636 15.0636
50 0.6428 0.4285 64.6260 14.6260
51 0.6293 0.4195 65.1940 14.1940
52 0.6157 0.4104 65.7675 13.7675
53 0.6018 0.4012 66.3462 13.3462
54 0.5878 0.3919 66.9299 12.9299
55 0.5736 0.3824 67.5186 12.5186
56 0.5592 0.3728 68.1119 12.1119
57 0.5446 0.3631 68.7097 11.7097
58 0.5299 0.3533 69.3120 11.3120
59 0.5150 0.3434 69.9184 10.9184
60 0.5000 0.3333 70.5288 10.5288
61 0.4848 0.3232 71.1431 10.1431
62 0.4695 0.3130 71.7610 9.7610
63 0.4540 0.3027 72.3825 9.3825
64 0.4384 0.2922 73.0074 9.0074
65 0.4226 0.2817 73.6356 8.6356
66 0.4067 0.2712 74.2668 8.2668
67 0.3907 0.2605 74.9010 7.9010
68 0.3746 0.2497 75.5380 7.5380
69 0.3584 0.2389 76.1777 7.1777
70 0.3420 0.2280 76.8199 6.8199
71 0.3256 0.2170 77.4644 6.4644
72 0.3090 0.2060 78.1113 6.1113
73 0.2924 0.1949 78.7603 5.7603
74 0.2756 0.1838 79.4113 5.4113
75 0.2588 0.1725 80.0641 5.0641
76 0.2419 0.1613 80.7187 4.7187
77 0.2250 0.1500 81.3750 4.3750
78 0.2079 0.1386 82.0327 4.0327
79 0.1908 0.1272 82.6918 3.6918
80 0.1736 0.1158 83.3522 3.3522
81 0.1564 0.1043 84.0138 3.0138
82 0.1392 0.0928 84.6763 2.6763
83 0.1219 0.0812 85.3398 2.3398
84 0.1045 0.0697 86.0041 2.0041
85 0.0872 0.0581 86.6690 1.6690
86 0.0698 0.0465 87.3345 1.3345
87 0.0523 0.0349 88.0005 1.0005
88 0.0349 0.0233 88.6668 0.6668
89 0.0175 0.0116 89.3334 0.3334
90 0.0000 0.0000 90.0000 0.0000

I don't think that ladder #2 and the dumbell can be compared. The base of the ladder is in contact with the wall at the beginning of its fall, and has no reason to move from that position until the momentum of the top end pulls it away. The rotation of the ladder around its CG will push the bottom into the wall unitil the motion of the top away from the wall pulls the entire ladder that direction.

With the duimbell, the top stays in contact with the wall until the momentum of the bottom pulls it away.

Hi 3Doug,

OK, let me try and convince you that we can compare ladder2 and the dumbell:-

Randall drew a nice graph in the dumbell challenge which clearly shows that the COG of the dumbell traces an arc of a circle while the contact with the wall is present. See below:

Now surely you will agree that if that same dumbell fell so that it's top moved away from the wall instead of the base( Just like ladder 2) then the COG will trace precisely the same arc of circle as shown above. Hence the maths and physics exhibit a symmetry that causes the dumbell to leave the wall at the same angle regardless of whether the bottom or the top moves away from the wall.

The Top of the ladders 2 and 3 rotate about their bases, the resultant direction of the COG is a product of three forces, centripetal, centrifugal and gravity, the resultant direction progressing from towards the wall, graduating to a direction away from the wall, the rotation of the ladder about it s COG is the reaction with the plane, this is brought about by the force of gravity which is evenly spread over the full length of the ladder, this being true for both ladder 2 and 3. it then becomes a matter of whether ladder 2's rotation about its base varies from that of ladder 3, when ladder 2 leaves the wall?

In the case of ladder 2 some of the energy becomes kinetic moving the ladder away from the wall, it then becomes a question of whether that same force in ladder 2 that became kinetic, acts to retard ladder 3's acceleration, when ladder 3 is restricted from moving away from the wall a new force is introduced into the resultant? Where as ladder 2's resultant is down and away from the wall, ladder 3's new resultant is up (restricted force) and away from the wall, and as it cannot move away from the wall the upward resultant retards its fall. thereby allowing ladder 2 and 3 to land at the same time.

Answer 1 first 2&3 together.

Regards JD.

The answer to this challenge will be posted on April 2nd, right here on CR4. Can't wait til then? Then check out these weekly CR4 Challenge Questions:

Is there something happening that I've missed ?

Ladder 1 hits the ground first, falling straight down.

Ladder 2 hits the ground second, falling on an arc.

Ladder 3 never hits the ground. The wall keeps it from falling past its 30 degree angle and there is no slip. This is how a ladder is used. Remember, common sense.

What wall?

Neither of the other two is set up to be usable, so why do you assume this one should be?

could be the wall stated in condition 2.

but then again, how else would you use a free standing ladder?

Since the purpose of the challenge is to determine their rates of fall, the question posed tells you that the wall is in contact with the base of the ladder. Since all 3 ladders start at a 30 degree angle, it would be impossible to have the ladder leaning against the wall while the wall was in contact with the base. Also, the question tells you all three are falling.

Agreed - and the safe angle for using a ladder is appreciably smaller than 30^o.

The question is not the rates of fall. The question is "In what order do they hit the ground?"

Nothing says ladder 3 actually has to hit the ground, it could be supported in a position which is its intended use.

If hits the ground means when do the ladders contact the ground, well that could be at the initial condition (t=0).

As to the falling part, it could start at a 30 degree angle @ t=0 and end at say a 32 degree angle @ t=1.

In a follow up reply Physicist stated

"This type of question is to be answered according to the usual "common sense" rule - i.e.: If there is an interpretation that allows a reasonable answer without additional information, that is the correct one."

What is the intended use of a ladder? How are ladders normally used? Sometimes the simplest answer can be the correct answer.

Hmm. The question says "the third one falls without slipping".

And hopefully common sense would also preclude using a ladder at an angle so far from the vertical. (See this safety guide).

It is beyond me why every engineering challenge has to be converted to a children's riddle, and so often in ways that neither the wording nor practicality would support.

CR4 Admin: Replaced broken link (http://www.hse.gov.uk/pubns/indg402.pdf) with link to updated brochure

http://www.hse.gov.uk/pubns/indg455.pdf

I made a similar assumption first but if you read the question it says for #2 "away from a wall that the base contacts initially", and mentions nothing else for #3.
One should assume it is also leaning away from the wall.

Furthermore the question states "The third falls [...]" which would be incorrect with your assumption (it does not fall).

The question leaves to interpretation, but I think the above makes sense.

The ladder in the frictionless environment has a direct path to the ground for its CG therefore striking the ground first.

The other two ladders' CG's fall through a complete arch (same distance) only because the one ladder "falls without slipping"...if that ladder's base slipped then it would hit prior to the ladder with the wall at the base because its CG would arc to a 'point of no return' and then the base would slip and its CG would then travel straight down toward the plane (or much closer to 'straight down toward the plane'). The only way these two ladders dont strike the plane at the same time is if there is some kind of friction applied because the base is against a wall...any thoughts on that aspect?

Errr, one, then another, and then the last?

First of all, I see the initial statement as contradictory - how can a ladder be standing squarely on a plane if it is at 30 degrees to vertical? Regardless, I will first state my assumptions:

1. The ladders do not have non-slip feet but rather have four points of contact (two per 'leg').

2. The first and second ladders slip (only the third ladder is identified as not slipping which would lead one to presume that slippage IS a factor with ladders 1 and 2.)

3. The ladders are non-adjustable - i.o.w. no second ladder on top of the first which can be raised and lowered.

4. All three ladders are affected equally by the pull of gravity.

5. The friction of air, the friction of the wall surface and the friction of the floor surface are all equal in magnitude.

Now for the answer:

Ladder 1 will hit the ground first - no friction of air opposing the pull of gravity, and no friction opposing the slippage.

Ladder 2 will hit the ground second - the ladder will experience only the force of friction of the floor and the wall against it. However the wall will impose resistance only for 15 degrees (vertical) on two surfaces of minimal area (the ladder feet), and the floor will impose resistance only for a short distance (horizontal) on two surfaces of minimal area (the ladder feet.)

Ladder 3 will hit the ground last - the pull of gravity is opposed only by the force of friction of the air against the ladder. However, the force of friction is on the entire surface area of the ladder.

As a side note, if you picture a 30, 60, 90 triangle, the two short legs belong to ladder 2 and the hypotenuse belongs to ladder 3. Therefore, clearly ladder 3 has a greater force against it than ladder 2 since the hypotenuse of a triangle is greater than the two legs combined.

If (colloquially) I ask someone to place a real ladder squarely (say) against a wall, they will ensure that both legs contact the ground. That may be a regional interpretation, but it is what was meant here.

1 & 3 - for economy of presentation, the questions are posed so that the simplest possible interpretation is the one intended. Thus, you rightly assume single section ladders, and just as for real ladders on a flat surface there is a line along which the contact points of both legs runs. Not that this last is essential when the ladders start from 30-degrees.

2. Both the first and second ladders are expressly stated not only to fall, but without friction.

4. Yes.

5. explicitly zero (1 and 2) or enough to prevent any slipping occuring (3).

6. If you include air resistance, the properties of the air should be the same for all three ladders. Or you can assume it to be zero (no air resistance specified). So long as the atmosphere has the same properties for all three ladders, the level of resistance makes no difference to the order in which the ladders hit the ground.

7. I can't see where 15-degrees comes from - the ladders all start at 30-degrees to the vertical.

In case it helps, Randall has created a dynamic illustration (using phun) that corresponds to my intentions.

The 'official' answer is very disappointing. It's what my Classical Mechanics professor would have called 'hand waving'. Where are the equations that prove these statements? And frankly, I'm not convinced that the logic is correct in distinguishing the actions of ladders 2 and 3.

The statement is made: The KE is the sum of three components: vertical, lateral, and inertial. Excuse me, inertial? What the heck is inertial kinetic energy? Do you mean Potential Energy? In actuality, the situation here is that gravitational potential energy is being converted into sideways KE, downward KE and rotational KE.

Another statement is: However, once the angle becomes flat enough, the horizontal component of the CofG's velocity will need to reduce if the base is to remain stationary; for the non-slipping ladder, that is indeed what happens, but there is nothing to restrain movement of the second ladder away from the wall, so it departs with the CofG maintaining constant lateral velocity. Thus, once the lateral velocity of the second and third ladders reaches their peaks, the vertical KE of the non-slipping ladder can outstrip that of the second ladder, so the third ladder goes ahead of the second.

The second ladder has one initial constraint (the base of the wall). Once the second ladder it released it free-falls away from the wall. The third ladder has a constant constraint (some type of pivot point that prevents the bottom from moving). For the third ladder to strike the ground before the second ladder does, it must have a greater downward acceleration (achieve a faster velocity downward) than the second ladder. I don't see anywhere in your argument that the pivot point will cause the third ladder to accelerate faster than the second ladder. The fact that the second ladder continues its lateral movement does not necessarily mean that it will fall slower, since (as we all know in the classic fired-bullet vs dropped-bullet experiment) lateral velocity does not affect downward acceleration during free-fall. Given two objects that start in similar initial conditions, if one is in free-fall the other must be accelerated downward if it is to fall faster than the one in free-fall.

Also, the path of the COG of the ladders does not predict which one will land first. An obvious comparion here is a small ball bearing that rolls down a straight inclined plane vs the same ball bearing rolling down a hypercycloid curve. The path along the hypercycloid curve is much longer, but the ball bearing gets to the bottom much faster than the ball bearing rolling down the straight inclined plane. (You may remember this from sophomore year Physics).

Guest March 18, 4:30 PM

Well, I am a little disappointed in the answer too, although I do believe the final answer vis-a-vis the order is indeed correct, I would have liked to have seen the full mathematical analysis for all three ladders which should be able to show the times, t1, t2 and t3 , in terms of acceleration due to gravity, g , and ladder length L (It would be independent of the mass of the ladder).

I also argue that the following statement in the answer, although not affecting the result, is incorrect:-

"Because the relation between the height of the CofG and the angle of the ladders is the same for all three ladders, they all display the same relation between vertical and inertial KEs"

I would argue that because ladder 1 is rotating around it's centre of gravity and ladders 2 and 3 are rotating about their base points, ladder 1 has only a quarter of the "Rotational Inertia" which ladders 2 and 3 have, hence even in the simulation given by Randall for the PHUN software program, we see that ladder 1 hits the ground way ahead of ladder 3.

Fyz, I would welcome your comments...

With regard to ladder three, you are within your rights to consider it as a purely rotating about the base - which would give the MofI you propose. But I don't find this helpful when comparing the three ladders. What makes the ladder2/ladder3 part of the problem tractable is that
1) you can in all cases consider the motion in terms of two components of linear motion of the centre of gravity and the angular motion about the centre of gravity
2) the relationship between the angular velocity and the vertical velocity is the same for all three ladders at any given height (ditto the equivalent components of kinetic energy).

N.B. The difference in presentation is equivalent to the theorem that the moment of inertia of an object about a remote point is the sum of the moment of inertia of a point object of equal mass located at the centre of gravity and the moment of inertia of the object about its centre of gravity.

If you still need equations, they are not difficult - but what I thought was interesting about this problem (as an ancillary to the dumbbells) was that everything (other than the horizontal velocity itself) was in the concept - the equations are simply a formalisation.

It's not just arm waving, but it does suffer from poor editing (my fault - I used a word processor for a substitution and didn't check sufficiently).

So we agree that the kinetic energy of a rigid body moving in two dimensions can be split into three orthogonal parts - those due to linear motion of the centre of gravity in the horizontal and vertical directions respectively, and that due to rotational motion. Because we are considering situations without loss, the sum of these three components depends only on the height of the centre of gravity (and the properties of the ladders, of course - but we can neglect these for this part of the argument, because the ladders have identical properties).

Now, as shown in the dumbbell challenge, for any given angle the relationship between vertical velocity of the CofG and the angular velocity is fixed. That means that the component of KE due to vertical movement reduces as the proportion of KE due to horizontal movement increases. So it is sufficient to show that the horizontal component of KE is in the order {1}<={3}<={2} at all times, which I believe you already accepted.

I believe that your professor would have regarded the addition of the equations as an exercise for the student. If not, let me know where the problems lie.

BTW, I certainly wouldn't rely on the path length alone.

Fyz

I don't see anywhere in your argument that the pivot point will cause the third ladder to accelerate faster than the second ladder.

Where do you think the horizontal energy is going?

I liked this challenge because it could be solved conclusively by "hand waving".

Although I believe that you an I are in complete agreement as to the substance, I think we have a linguistic issue here.

To my mind, hand-waving means non-rigorous and therefore inconclusive. Solutions involving equations can be completely non-rigorous, even when the manipulation of the equations is immaculate (they can indeed be wrong, as we have sometimes seen in various attempts at solution on CR4). The contrast here is that we have a situation where the logic requires to set up the equations leads to a rigorous solution - and without any need for the equations to be made explicit.

I personally believe that the ability to handle this type of problem lies at the heart of good engineering and physics. Unfortunately, because such the solutions to such problems are expressed in language that can be variable, they are hard to examine (and even harder to demonstrate that the examinations have been properly marked); as a result, these problems are woefully under-represented in many of the places they should gain the highest regard - the educational institutions. My hope is that guest's ex-professor would appreciate the solution (once correctly reworded); sad as it may be, I also fear he would find him/her-self unable to mark a solution expressed in those terms.

Fyz

Before I accept this solution I must be shown that the ladder first contacts the ground in a horizontal position. Since the ladders are rotating as they fall, we must consider the possibility that the top of the ladder overtakes the rest of the ladder and contacts the ground first. If so, the height of the fall is reduced by the distance of the center of gravity from the ground when the top hits.

A stronger constraint is that the base does not leave the ground.

=> all we need to show is that the force applied at the base is upwards whenever the ladders are contacting the ground.

We don't need to do this for the first case, as rotation can only make these ladders hit the ground earlier.
The second ladder will not leave the ground if the third one doesn't.
So we only need to show that the third ladder does not leave the ground. This will depend on its mass distribution. A simple analysis shows that if we release a ladder with uniform mass distribution from the vertical position the vertical force on the base will just reach zero and then increase again - it will not change sign. We are starting 30-degrees from the vertical, so we would need a very non-uniform distribution of the ladders were to lift - and all ladders I have seen are close (enough) to uniform.