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Topic: Infinite fast ladder (Read 5236 times) |
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rloginunix
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Re: Infinite fast ladder
« Reply #25 on: Nov 27th, 2014, 11:08am » |
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To develop a better feel for the role of the force FV let us reword the problem slightly. Let us assume that everything else remains the same but there is no FV this time. In other words the ladder, after it starts its motion, is left to its own devices and its both edges move "by themselves" with no external forces acting on either of them.
In that case, prior to separation of the top edge of the ladder from the wall, the right triangle holds:
x + y= L (3.1)
where x is B's coordinate and y is A's coordinate. Differentiate (3.1) over time once:
xx' + yy' = 0
y' = -x'*(x/y) = -(x'/y)* (L- y) (3.2)
where we took x from (3.1). In this case the energy in the IRF wall-floor is conserved:
x'+ y'+ 2gy = 2gL
where I canceled out the m's and multiplied both sides of the equation by 2. Put y'from (3.2) into the above equation and carry the 2gy over the equal sign:
x'+ (x'/y)(L- y) = 2g(L - y)
(x'/y)*L= 2g(L - y)
x'= (2g/L)(Ly- y ) (3.3)
Differentiate (3.3) over time once to obtain B's acceleration:
2x'x'' = (2g/L)(2Lyy' - 3yy') = y'(2g/L)(2Ly - 3y)
Now use y' from (3.2):
x'x'' = -x'(g/L)(2L - 3y)(L- y)
x'' = -(g/L)(2L - 3y)(L- y) (3.4)
We see now that B's acceleration is positive - B speeds up - when (2L - 3y)  0:
y  (2/3)L
B's acceleration is 0 when:
y = (2/3)L
incidentally that's when A breaks contact with the wall in this particular case. Lastly, B's acceleration is negative - B slows down - when:
y (2/3)L
And now we can better understand the role of the constant velocity force FV in the original problem. Prior to A's escape this force varies in such a way that it "eats" B's growing acceleration and slows it down to a constant speed. That is why I pointed FV "against" V.
At the moment of A's escape FV becomes zero.
After A's escape FV's vector flips 180 degrees, points "with" V and FV varies in such a way that it adds just enough force to compensate B's loss of speed but in a careful dosage that ensures that B moves at constant speed.
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rloginunix
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Re: Infinite fast ladder
« Reply #26 on: Nov 27th, 2014, 8:48pm » |
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on Nov 26th, 2014, 10:46pm, towr wrote:| Did we need all that to arrive at that conclusion? |
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I needed it to arrive at the formula via a path that uses the idea of equivalence mentioned by Icarus. Another way to get the same result would be to move into B's IRF and put down the equation for A:
(m/2)u2/L = (m/2)V2/(LSin2( ))= (m/2)gSin()
Either way the same finite velocity results.
on Nov 26th, 2014, 10:46pm, towr wrote:| So, then if I attach one edge to a vertical rail, and the other to a horizontal rail, so that it must always form a right triangle as the ladder moves. Then it would have infinite velocity? |
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Good question. The answer is no. That is the conclusion I have reached for the concrete problem statement as given by BNC. I mean if the socks were permanently attached to Einstein's feet he (I guess) would have came up with a different solution. Otherwise he (I guess) chose the simplest solution that he could physically carry out.
I did the same. The simplest solution in this particular case - when neither edge of the ladder is attached permanently to a surface - is to detach the top edge. The formula for a finite speed followed.
If you want to change the problem statement - and solve a different problem - that is perfectly fine. While we are on the subject let us summarize the few reasonable permutations.
I). No edge attachments, external force, V  0 (BNC formulation):
uf = (V+ 3(gLV)2/3) (I.1)
f = arctan((gLV-2)1/3(3)) (I.2)
II). Top edge is attached to a surface, no external forces: FV = 0:
uf = (2gL) (II.1)
f = 90 (II.2)
III). No edge attachments, no external forces: FV = 0 (my example above):
uf = ((52/27)gL) (III.1)
f = arccos(26-1/2) (III.2)
IV). Top edge is attached to a surface, external force, V 0.
Need to think about that one.
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towr
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Re: Infinite fast ladder
« Reply #27 on: Nov 27th, 2014, 11:08pm » |
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on Nov 27th, 2014, 8:48pm, rloginunix wrote:| If you want to change the problem statement - and solve a different problem - that is perfectly fine. |
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I don't think I changed the problem statement any more than you did. The calculation that is given in the problem statement describes the situation where the top edge stays on the wall, and that's where the problem comes from. And while a first impulse might be to say that the calculation is wrong because the top edge detaches, it's not a greater change to say that attachment to the wall is enforced and the calculation is correct (in theory, if not a plausible universe).
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| « Last Edit: Nov 27th, 2014, 11:11pm by towr » |
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rloginunix
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Re: Infinite fast ladder
« Reply #28 on: Nov 28th, 2014, 9:38am » |
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I think I see what is going on. A disconnect in the problem statement interpretation (how many times did that happen?).
BNC's problem statement does not explicitly state that the ladder's top edge is attached to a wall so (when I first read it) I interpreted it as such. I also did no read the hidden text. Now I did - aero_guy mentions the top edge attachment but BNC makes no comment.
May be the problem is stated elsewhere and I missed it since this is the only place I looked?
Anyway, please confirm (or deny or correct) a proper intended problem statement:
"A ladder of length L rests on a frictionless floor and leans against a wall forming an initial angle 0 with the floor. The ladder is held on place by a mysterious external force (MEF) which is of no consequence. The floor and the wall are orthogonal. The ladder's top edge is attached to a wall via a frictionless rail and can not separate from the wall during any instance of its movement.
At some point MEF disappears, the ladder starts falling down in such a way that its bottom edge moves with a constant velocity V along the floor.
Find the velocity of the ladder's top edge when it comes in contact with the floor."
Also, in the light of all of this do you want me to delete my previous posts?
Thanks.
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towr
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Re: Infinite fast ladder
« Reply #29 on: Nov 28th, 2014, 1:14pm » |
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Hmm, reading back earlier comments, BNC does state in reply #7 that "The ladder is not connected to the wall".
So, err, objections/contentions withdrawn.
(And no, never delete anything, ever)
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rloginunix
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Re: Infinite fast ladder
« Reply #30 on: Dec 6th, 2014, 2:05pm » |
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It is good to have towr as a tough customer to please. The truth comes out in the process.
A few relevant remarks about the cases mentioned earlier.
Case II), top edge is attached to a surface, no external forces.
Putting (3.3) into (3.2) we get:
y'(y) = -(1/L)*(2g)*(y3 - Ly2 - L2y + L3) (II.3)
When the ladder hits the floor y = 0:
y'(y = 0) =-(1/L)*(2g)*(0 - L*0 - L*0 + L3) = -(2g)*L-1*L3/2 = -(2gL) (II.4)
We get the (II.1) formula barring the negative sign which tell us that the velocity of the ladder's top edge points against the chosen positive direction of the Y-axis.
Let us investigate how the bottom edge moves. From (3.2) we get:
x' = -y'(y/x) = -y'(L2/x2 - 1) (II.5)
From the equation of conservation of energy (just below (3.2) in Reply#25) we find y'2:
y'2 = 2g(L - y) - x'2 (II.6)
Putting it back into (II.5) and solving for x' (after some algebraic manipulation) we get:
x'(x) = (2g(L - (L2 - x2))(1 - x2/L2) (II.7)
When the ladder hits the floor x = L and:
x'(x = L) = (2g(L - (L2 - L2))(1 - L2/L2) = 0 (II.8)
And if we take B's acceleration from (3.4) when y = 0 we get:
x'' = -(g/L2)(2L - 3*0)(L2 - 0) = -2g (II.9)
In other words when the ladder hits the floor its bottom edge is yanked towards the wall.
Below I have graphed the velocity of the bottom edge of the ladder taking L = 1, g = 1:
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rloginunix
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Re: Infinite fast ladder
« Reply #31 on: Dec 6th, 2014, 2:17pm » |
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Case IV), top edge is attached to a surface, external force. I think rmsgrey handled it in Reply#20. I will only add that in academic parlance it sounds like a poorly worded problem statement.
Case III), no edge attachments, no external forces. The ladder's top edge detaches from the wall when y = (2/3)L. At that vertical coordinate B's velocity from (3.3) is:
x'Be = ((8/27)gL) (III.3)
The horizontal coordinate of the ladder's CM at that moment is XCM = xB/2 and its horizontal velocity is, consequently, X'CMe = x'Be/2 or:
X'CMe = ((2/27)gL) (III.4)
Once the top edge of the ladder detaches from the wall its CM has a constant horizontal velocity* since there are only two forces that have non-zero X-projections - the internal tension forces T - but they cancel each other out when projected on the X-axis because they are equal in absolute value but point toward each other. So when the ladder hits the floor the initial potential energy of the top (mgL) is transformed into the kinetic energy of both edges:
2gL = uf2 + vB2 (III.5)
However, B's velocity at that moment must be equal to the horizontal component of CM's velocity (rigid body argument plus B has only one velocity component):
uf2 = 2gL - vB2
uf2 = 2gL - (2/27)gL
uf = ((52/27)gL) (III.6)
*
Note here an interesting situation with CM's velocities. In the original (BNC's) formulation the ladder's CM moved with a constant horizontal velocity prior to escape and with a variable x-velocity afterward. In the absence of FV this situation is reversed - prior to escape VCMX varies but becomes constant after the escape.
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SWF
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Re: Infinite fast ladder
« Reply #32 on: Dec 20th, 2014, 5:12pm » |
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If we go back to the idea of the ends sliding along the lines x=0 and y=0, this reminds me of one of these gizmos:
https://www.youtube.com/watch?v=hNPO2MosQNQ
If you treat it as point masses of mass, m, at each end connected by a rigid massless link, the force required to move the end along the x axis at constant velocity V is found to be:
L2V3tm/(L2-V2t2)2
where t is the time, with t=0 being for being at a vertical position. This blows up when Vt=L, or it is horizontal, which shouldn't be surprising.
When it reaches a horizontal position it is like a mechanism with infinite mechanical advantage.
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