T.R | Title | User | Personal Name | Date | Lines |
---|
442.1 | How do you getit started? | METOO::YARBROUGH | | Wed Feb 12 1986 10:13 | 6 |
| Hmm. The problem is not very well specified. As stated, the bottom
of the ladder experiences an infinite acceleration in order to attain
an instantaneous velocity of V. This will tend to separate the top
of the ladder from the bottom. Also vaporize the ladder. In these
conditions it is not clear that the top will come down at all.
|
442.2 | nits | CLT::GILBERT | Juggler of Noterdom | Wed Feb 12 1986 17:39 | 8 |
| Okay:
A ladder of length L is nearly upright, leaning against a wall.
The bottom of the ladder is being pulled away from the wall at
velocity V.
As before:
What is the velocity of the *top* of the ladder when it's
at height L/2? What is it's velocity as it hits the ground?
|
442.3 | If Einstein were a noter | LATOUR::APPELLOF | Carl J. Appellof | Thu Feb 13 1986 07:44 | 3 |
| How close is the velocity V to the speed of light? Must we include
relativistic effects here?
|
442.4 | It's still not well specified | METOO::YARBROUGH | | Thu Feb 13 1986 10:11 | 6 |
| Aside from relativistic effects, this is really a nasty problem.
Long objects which fall over tend to bend; this is why smokestacks
usually break when pushed over. You also need to know how large
V is w/r/t the length of the ladder in order to determine if the
ladder will remain in contact with the wall as it descends. So there
are still a lot of undefined assumptions in the problem statement.
|
442.5 | | CLT::GILBERT | Juggler of Noterdom | Thu Feb 13 1986 11:54 | 1 |
| Well, I guess nobody's willing to do any real work on this.
|
442.6 | A Start | BEING::POSTPISCHIL | Always mount a scratch monkey. | Thu Feb 13 1986 17:23 | 25 |
| I've looked at the problem. Some partial results follow.
When the ladder is at an angle O from the ground, the bottom of the
ladder is traveling at a velocity of V cos O in a direction parallel to
the ladder and V sin O in a direction normal to the ladder. Assuming
the ladder to be a rigid body, the top of the ladder must also be
traveling at a velocity of V cos O in the direction parallel to the
ladder. However, its normal component is unknown.
The above includes the assumption that the bottom of the ladder remains
on the ground. I have not yet figured out under what circumstances
this is true.
If we knew how much energy the ladder had, it would be simple to
determine the last component of the ladder's velocity. However, energy
may be added or removed by three things, the force which causes the
bottom of the ladder to move at velocity V, the force on the ladder
from the ground, and the force on the ladder from the wall. One might
assume these last two forces to do no work on the ladder, since the
ladder does not move because of them, but that is not necessarily true
-- even though the end of the ladder touching the wall/ground does not
move, the rest of the ladder is affected by these forces.
-- edp
|
442.7 | Rotational Energy | LYRA::THALLER | Kurt (Tex) Thaller | Thu Feb 13 1986 19:22 | 3 |
| Be very careful if you try to solve this problem using energy.
You must be certain to include rotational energy the ladder acquires
when it is falling.
|
442.8 | Smokestacks break predictably | LATOUR::AMARTIN | Alan H. Martin | Fri Feb 14 1986 02:12 | 12 |
| Re .4:
Prove:
A smokestack always breaks apart at the point 1/3rd of the way
up from the bottom when it is dynamited, and starts tipping over.
Sounds under-specified? Someone noticed this from inspecting photos
and films of falling smokestacks, and it was proved in some place like
the Amateur Scientist or Mathematical Games section of Scientific American.
(Does this ring a bell with someone?).
/AHM
|
442.9 | | CLT::GILBERT | Juggler of Noterdom | Wed Feb 19 1986 01:35 | 27 |
| Let the mass of the ladder be M, assume this is uniformly distributed
along its length, and let the acceleration of gravity be G. The ladder
is, of course, assumed to be rigid.
The potential enery available as the ladder stands is:
L MGl MGL
Integral --- dl = ---
0 L 2
(i.e., recall that force=mass*acceleration, and energy=force*distance,
and we're integrating over the length of the ladder).
Now if the top of the ladder were constrained to remain against the
wall as the bottom is moved, it's height at time t would be:
sqrt (L� - t�V�)
and it's velocity would be:
- t / sqrt (L� - t�V�)
(negative because it's moving downward; this is the derivative of the
height with respect to time).
Well, it's a start.
|
442.10 | Hint | BEING::POSTPISCHIL | Always mount a scratch monkey. | Thu Feb 27 1986 17:20 | 6 |
| How about considering the rotational position, velocity, and
acceleration of the ladder around the bottom of the ladder? With
that focus, there would be only one force to consider: gravity.
-- edp
|