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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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