This started as a calculus problem. The question said that the top of a ladder moves down with some constant speed and asks about the speed of the contact point with the floor. You can see my solution here.
Of course my initial thought was not about derivatives of x and y, but instead I thought: I don’t think the top of a sliding ladder would move down with a constant speed. You know what comes next, right? Yes, a physics problem.
In fact, if you have a ladder on a frictionless floor leaning up against a frictionless wall (which is NOT OSHA approved), the ladder will actually slide down in such a way that it will lose contact with the wall at some point. So, let’s find the angle at which this happens — using physics. It’s going to be fun.
Forces on a Sliding Ladder
Without friction, there are just 3 forces that act on the ladder.
There’s the gravitational force pulling down at the center of mass and then there are the two contact forces — N1 from the wall pushing to the right and N2 from the floor pushing up. Oh, notice that if the ladder is leaning at an angle θ with a length L, then I can find the coordinates of the center of mass as:
If the ladder is a rigid object, then we can model it’s motion by looking at the motion of the center of mass and rotation ABOUT the center of mass. Using Newton’s second law and torque, we get the following three equations.
There’s a problem though. I don’t know the values of either contact force (N1 and N2) since they are forces of constraint. They apply a force to keep the ladder ends from moving through the wall (or floor). So, instead of using Newton’s laws — let’s take another approach.
The Work-Energy Principle
The work energy principle says that the total work done on a system is equal to the change in energy of that…