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.
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
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
We can now obtain the forces of constraint in terms of the function @(t):
A2 = 1 mm + mg
We substitute these into the last equation of motion to obtain the result
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
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