Physics Friday 120

Here is another classic problem: A ladder (of negligible thickness) has length l and mass m, distributed symmetrically enough so that the center of mass is at the geometric center. The ladder is set on a frictionless floor, and leaned against a frictionless wall. The base of the ladder is a small but negligible distance from the wall, so that the ladder, initially at rest, will have the base slide away from the wall, while the top end slides away from the wall. At what point does the ladder lose contact with the wall? What is the maximum force exerted on the ladder by the wall?

Let us define . While the ladder is in contact with the wall, it forms the hypotenuse of a right triangle. As a corollary of Thales’ theorem (see here), distance between the base of the wall and the center of the ladder is simply R. Thus, until the ladder loses contact, the center of mass moves along a circle of radius R with center at the base of the wall. Let θ be the angle between the ladder and the vertical; and thus also the angle between the ladder and the wall, and the angle between the wall and the above-described radius to the center of mass.

As with the skier and the hill in this problem, contact is lost when the normal force exerted by the wall goes to zero. Now, instead of considering the forces, let us consider the conservation of energy. Initially, the energy is entirely the gravitational potential energy; as the height of the center of mass in that vertical alignment is R, we have as the total energy.
Now, when the ladder is at angle θ, the height of the center of mass is , and so the potential energy has decreased to .
Next, we consider the kinetic energy, which may be divided into the translational energy of the center of mass plus the rotational kinetic energy about the center of mass. As the center of mass is travelling along a circle of radius R, with angle θ from the vertical, we see the magnitude of the center of mass velocity is , and so the translational kinetic energy is . Similarly, since θ is also the angle of the ladder, is also the angular velocity about the center of mass, and the rotational kinetic energy is , where I is the moment of inertia about the axis through the center of mass parallel to both the wall and floor (perpendicular to the plane of the diagram). Combining, and using conservation of energy, we see

Letting us define ξ as the dimensionless quantity , and noting as before that , we see the velocity of the center of mass is
as this is tangent to the circular path of the center of mass, the horizontal component is
. Since the only horizontal force on the ladder is the normal force exerted by the wall, when the normal force goes to zero, the horizontal acceleration goes to zero, and so the horizontal component of the center of mass velocity is at a maximum. Taking the derivative of vx with respect to θ, we see that
Setting this equal to zero, and solving for 0<θ<π/2, we see that
, and thus ≈48.19°, independent of the mass, length, and moment of inertia of the ladder.
[Note that this is the same as the angle at which the skier went airborne.]

Now, as noted before, the normal force from the is the only horizontal force, hence

Using the fact that , we see
, and plugging in , we get
, and so
taking the derivative with respect to θ and setting equal to zero to find the angle where the normal force is maximum, we get
solving this quadratic in , we get
, and the maximum horizontal force occurs when ≈26.73°. Plugging this back into our force equation,


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