- 22 Nov '05 22:41Suppose you have a ladder of length "L" sitting on a frictionless floor, leaning up against a frictionless wall, held in place by a doorstop-like buttress (heheh, buttress...). The height of the ladder against the wall "h" is given by the following formula:

x^2 + h^2 = L^2

where "x" is the distance from the wall to the foot of the ladder. If you remove the buttress and drag the foot of the ladder away from the wall at a constant velocity "k", we can find the change in height of the top of the ladder with time (vertical velocity) by taking the derivative of the

above formula:

2x*(dx/dt) + 2h*(dh/dt) = 2L*(dL/dt)

which simplifies to:

2xk + 2h*(dh/dt) = 0

and solving for (dh/dt):

(dh/dt) = -xk/h

As the height of the ladder approaches zero, the magnitude of the vertical velocity (the speed) approaches infinity. Is this possible? If not, where is the mistake in the above problem statement (he asked knowingly)? - 22 Nov '05 23:59
*Originally posted by PBE6***Suppose you have a ladder of length "L" sitting on a frictionless floor, leaning up against a frictionless wall, held in place by a doorstop-like buttress (heheh, buttress...). The height of the ladder against the wall "h" is given by the following formula:**

x^2 + h^2 = L^2

where "x" is the distance from the wall to the foot of the ladder. If you r ...[text shortened]... his possible? If not, where is the mistake in the above problem statement (he asked knowingly)?**The height of the ladder against the wall "h" is given by the following formula:**

x^2 + h^2 = L^2

Agreed.

**If you remove the buttress and drag the foot of the ladder away from the wall at a constant velocity "k", we can find the change in height of the top of the ladder with time (vertical velocity) by taking the derivative of the**

above formula:

2x*(dx/dt) + 2h*(dh/dt) = 2L*(dL/dt)

Ok.

**which simplifies to:**

2xk + 2h*(dh/dt) = 0

I'd cancel the 2's.

*xk + h(dh/dt) = 0*

**and solving for (dh/dt):**

(dh/dt) = -xk/h

Agreed...

**As the height of the ladder approaches zero, the magnitude of the vertical velocity (the speed) approaches infinity. Is this possible?**

I don't find any problems with your mathematical analysis. Generally problems like this in which the answer seems intuitively unreasonable involve infinities cancelling one another. In this case, an infinite velocity is approached, but it is only infinite for an infinitesimal amount of time.

There might also be a problem with analyzing motion at relativistic speeds with a purely classical analysis. - 23 Nov '05 00:26

a buttress wouldn't work on a frictionless floor.*Originally posted by PBE6***Suppose you have a ladder of length "L" sitting on a frictionless floor, leaning up against a frictionless wall, held in place by a doorstop-like buttress (heheh, buttress...). The height of the ladder against the wall "h" is given by the following formula:**

x^2 + h^2 = L^2

where "x" is the distance from the wall to the foot of the ladder. If you r ...[text shortened]... his possible? If not, where is the mistake in the above problem statement (he asked knowingly)? - 23 Nov '05 08:12 / 1 edit

Nope! There is definitely a problem with my analysis. Nothing I know of can move at infinite speed, in fact nothing I know of can move faster than the speed of light (approximately 3x10^8 m/s). This includes ladder tops. I'll give you a hint - it's not a problem with the math per se, but the physics of the setup.*Originally posted by AThousandYoung***[b]The height of the ladder against the wall "h" is given by the following formula:**

x^2 + h^2 = L^2

Agreed.

**If you remove the buttress and drag the foot of the ladder away from the wall at a constant velocity "k", we can find the change in height of the top of the ladder with time (vertical velocity) by taking the derivative of the ...[text shortened]... be a problem with analyzing motion at relativistic speeds with a purely classical analysis.**[/b]

(BTW, I shouldn't be a jerk and take credit for this problem 'cause I didn't write it. It was from a math website which I forget the name/address of, but if I find it again I'll post it.) - 23 Nov '05 09:03I think that mathematically there is no problem. Hence, the infinite downward velocity of the end touching the wall is as (im)possible as the constant velocity of the end on the floor. And indeed, if the initial distance was x0, then it is physically impossible to have a constant velocity k at time (l-x0)/k and a zero velocity at the same time (with an infinitesimally small delay). Because that would mean an infinitely large (de)acccelaration, which is possible mathematically, but not physically with a mass involved. Speeds higher than the speed of light are not a problem if no masses are involved.
- 23 Nov '05 19:03

You make a valid point about the initial movement of the foot of the ladder. It's a singularity in the math that needs to be dealt with. Let's fix it by saying the foot of the ladder is smoothly accelerated to a velocity "k", and it's position is given by the following formula:*Originally posted by Mephisto2***I think that mathematically there is no problem. Hence, the infinite downward velocity of the end touching the wall is as (im)possible as the constant velocity of the end on the floor. And indeed, if the initial distance was x0, then it is physically impossible to have a constant velocity k at time (l-x0)/k and a zero velocity at the same time (with an in ...[text shortened]... ass involved. Speeds higher than the speed of light are not a problem if no masses are involved.**

x(t) = x0 + (1/6)*t0*t^3 - (1/12)*t^4

where t0 = (6k)^(1/3) and is the time when we switch to the constant velocity "k". The reason for this complicated formula is that its derivate:

x'(t) = (1/2)*t0*t^2 - (1/3)*t^3

gives values of x'(0) = 0 and x'(t0) = k, which satisfies our requirements for smoothness. The only stipulation is that t0 can't exceed the total time to drag the foot of the ladder from x0 to L, the length of the ladder, otherwise we'd never get to the interesting part of the question. So:

t0 < (L-x0)/k

and:

(6k)^(1/3) < (L-x0)/k

solving for k we get:

k < [(L-x0)^3/k]^(1/4)

which limits the speed at which we can pull the ladder for this particular smooth acceleration. After time t0, the velocity of the foot of the ladder is given by:

(dx/dt) = k

and we go back to try and solve the problem as before. Mephisto2's comments are actually a giant step forward in solving the problem, although not for the obvious reason. It has to do with the way in which we model the question using math, and whether our models breakdown at any point (like at the very beginning, as Mephisto2 pointed out). - 23 Nov '05 19:19My main comment was not aiming at the start of the movement, but rather at the end. At that point, the movement stops, which means a discontinuity in the horizontal velocity k->0. If that assumption is deleted, then from mathematical point of view, there is no issue. The vertical end will simply switch from vertical movement to horizontal, and follow the other end with a speed k.

Besides that point, if the problem is physical (i.e. mass involved) then the force needed to pull the ladder while maintaining horizontal velocity of k would increase towards infinity to keep the vertical end on the move as the end comes nearer to the floor. And that is as impossible as a velocity>c. - 23 Nov '05 19:36

Actually, I think this is precisely the point. The thing which is holding the ladder against the wall, as well as accelerating it downwards, is gravity. If you can imagine trying this problem in a weightless environment, then as soon as you started dragging the base of the ladder away from the wall, then the top would also move away from the wall.*Originally posted by Mephisto2***My main comment was not aiming at the start of the movement, but rather at the end. At that point, the movement stops, which means a discontinuity in the horizontal velocity k->0. If that assumption is deleted, then from mathematical point of view, there is no issue. The vertical end will simply switch from vertical movement to horizontal, and follow the o ...[text shortened]... end on the move as the end comes nearer to the floor. And that is as impossible as a velocity>c.**

I think this is the critical point to understanding where the mathematical model breaks down. At some point gravity will be insufficient to maintain the top of the ladder against the wall--at that point your equation fails,the ladder is dragged away from the wall, and the ladder is accelerated to the floor purely based upon the force of gravity. - 23 Nov '05 19:54

That is correct! Great job, leisurelysloth!*Originally posted by leisurelysloth***Actually, I think this is precisely the point. The thing which is holding the ladder against the wall, as well as accelerating it downwards, is gravity. If you can imagine trying this problem in a weightless environment, then as soon as you started dragging the base of the ladder away from the wall, then the top would also move away from the wall.**

...[text shortened]... rom the wall, and the ladder is accelerated to the floor purely based upon the force of gravity. - 23 Nov '05 19:58 / 1 edit

not sure if it is gravity, but I agree that this causes some additional effects (important or not depending on the relative numbers). I looked at this this way:*Originally posted by leisurelysloth***Actually, I think this is precisely the point. The thing which is holding the ladder against the wall, as well as accelerating it downwards, is gravity. If you can imagine trying this problem in a weightless environment, then as soon as you started dragging the base of the ladder away from the wall, then the top would also move away from the wall.**

...[text shortened]... rom the wall, and the ladder is accelerated to the floor purely based upon the force of gravity.

- at the floor end, the force pulling along the horizontal axis is decomposed in an upward part (compensated by a force keeping that end on the floor, which could be the gravity or partly the gravity) and a diagonal part along the ladder (because that is the only path to transfer forces to the top end of the ladder). At the top end, the force is again decomposed in a horizontal part (pulling the ladder away from the wall, compensated by structural forced preventing the movement away from the wall) and a vertical downwards part, causing the movement downwards. Now, as the ladder takes on a nearly flat position, the angles become such that the force has to become infinitely large to create the decomposed vertical part. - 23 Nov '05 22:57

So what if both ends were attached to the floor/wall by a wheel connected to some sort of track?*Originally posted by leisurelysloth***Actually, I think this is precisely the point. The thing which is holding the ladder against the wall, as well as accelerating it downwards, is gravity. If you can imagine trying this problem in a weightless environment, then as soon as you started dragging the base of the ladder away from the wall, then the top would also move away from the wall.**

...[text shortened]... rom the wall, and the ladder is accelerated to the floor purely based upon the force of gravity. - 24 Nov '05 14:54

I think the problem is the lateral force required to keep the foot of the ladder going at a constant velocity. When the foot of the ladder gets very close to the floor, this force becomes infinite, so it's not surprising that the speed of the bottom of the ladder is infinite as well.*Originally posted by AThousandYoung***So what if both ends were attached to the floor/wall by a wheel connected to some sort of track?**