Sliding Ladder

A ladder of mass m and length 2l stands against a frictionless wall with its feet on a frictionless floor. If it is let go with initial angle α₀, what will be the angle when the ladder loses contact with the wall (see Figure 1.1)?

Figure 1.1

SOLUTION

Let us watch the ladder until it leaves the wall. Forces N₁ and N₂ are normal reactions of the wall and floor, respectively; W = mg is the weight of the ladder; x_c and y_c are the coordinates of the center of mass (see Figure.1.2). First, find the Lagrangian of the system. The kinetic energy is the moment of inertia relative to the center of mass

Figure.1.2

(1)

From Lagrange’s equations

(2)

In addition, from energy conservation

(3)

We will assume that the ladder loses contact with the wall before it does so with the floor. (This has to be checked). Up until the ladder slides away from the wall, there are constraints of the form

(4)

(5)

Since N₁ is the only force acting in the x direction, N₁ = mẍ_c When the ladder loses contact with the wall, N₁ = 0. Differentiating (4) twice gives

(6)

From (6)  and substituting it into (2), we have for the angle the ladder leaves the wall

(7)

From (3) and (7), we obtain

(8)

We have assumed that the ladder loses contact with the wall first. Let us check this assumption. It implies that N₂ > 0 at all times before the ladder leaves the wall

(9)

From (5) and (6), we have

(10)

Therefore

(11)

At the time the ladder leaves the wall

On the other hand, N₂ is monotonically decreasing while is decreasing (see (3)). So, our assumption was right and indeed α = sin^-1(2/3 sin α₀).