Transcribed Text
A uniform ladder of length l and mass m has one end on a smooth horizontal floor and the
other end against a smooth vertical wall. The ladder is initially at rest and makes an angle
00 with respect to the horizontal.
y
(x,y)
o
X
Figure 8.5: A ladder sliding down a wall and across a floor.
(a) Make a convenient choice of generalized coordinates and find the Lagrangian.
Solution : I choose as generalized coordinates the Cartesian coordinates (x,y) of the
ladder's center of mass, and the angle 0 it makes with respect to the floor. The
Lagrangian is then
L 1m(i2+y)+1102+mgy
There are two constraints: one enforcing contact along the wall, and the other enfore
ing contact along the floor. These are written
(b) Prove that the ladder leaves the wall when its upper end has fallen to a
The equations of motion are
Thus, we have
mji + mg=,
10=  = Qg.
We now implement the constraints to eliminate I and y in terms of 0. We have
7
We can now obtain the forces of constraint in terms of the function @(t):
A1 =
A2 = 1 mm + mg

We substitute these into the last equation of motion to obtain the result
or
with I This may be integrated once (multiply by
é to convert to a total derivative) to yield
0 = 2w3 sin to
which is of course a statement of energy conservation. This,
We may now obtain () and
Demanding A1 (0) = 0 gives the detachment angle 0 = ea where
Note that 2(d) = mgo/(1 + a) > 0, so the normal force from the floor is always
positive for 0>Bd The time to detachment is
now
+
sin
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