Another type of Rotational Equilibrium problem in AP Physics that is commenly seen on exams is the Ladder Syle Problem. Again, this type of problem deals with the concepts of Torques in Equilibrium so you should check out this page before going any further.
Ladders are an example of an object in both static equilibrium and rotational equilibrium. As ladders are rigid bodies they definatley can be affected by torques.
When examining torques with respect to a ladder, it gets slightly more complicated than looking at a beam. There are many forces to consider as the following diagram shows.
This is a force diagram for a ladder propped up against a frictionless wall. In AP Physics all your walls will be frictionless as surreal as it is. This simplifies your force diagram a great deal, which is a good thing for student just learning the topic. Anyways, the forces acting on the ladder have all been draw in red. First we have Fa or the force exerted on the ladder by the wall. This force is paired with the other horizontal force Ff which is the force of friction between the ladder and the ground. Since the ladder is in static equilibrium the net horizontal force (Fa + Ff) must be zero. The same is true for the vertical forces. Fg is the weight of the ladder which is exerted downward while Fn or the Normal Force, exerted by the ground on the ladder must balance each other or the ladder would tumble to the ground.
So for ladder problems it is important to know this force diagram as it yields the torque calculations as well as the following points.
These points will be useful for solving for missing components in ladder problems.
Now let's think torque again. We also know that since the ladder isn't rotating, that the net torque or the sum of all clockwise and counterclockwise torques is zero.
If we pick the bottom of the ladder as our piviot point then Fn and Ff both will not generate torques. Since that eliminates two forces and possible torques right away, this is generally done in ladder problems.
Now examine each of the torques. Fg will create a clockwise torque and Fa will produce a counterclockwise torque.
Next to actually calculate the torque, we must be careful. A common ladder problem mistake is to just use the length of the ladder as the lever arm value in the torque formula. This however is incorrect as the ladder length is NOT the perpendicular distance between the force applied and the pivot point. To understand what I mean, I have illustrated the ACTUAL perpendicular distance from the pivot point for each of the forces in the diagram below. This should give you and idea what the perpendicular distance from the pivot point would be in any ladder problem or at least how to find it.
As you can see from the diagram the lever arm of each force is hard to tell at first glance. I recommend for each problem you do, always draw it out if you are not sure. To get the values of each L (lever arm) you can use right triangle trignometry as the ladder forms a right angle with the ground. If you need further clarification on this part of the problem, you should follow the link at the bottom of the page to some pre-solved sample problems.
To find each torque is now a simple matter of using the formula. In this example, for instance the value for the torques acting in the clockwise direction is Torque CW = Fg * lg and the value of the torques acting in the counterclockwise direction is Torque CCW = Fa * la. We also know that since the Net torque must add to zero since the system is in equilibrium that Torque CW + Torque CCW = 0. Understanding these facts will help you solve for missing bits of information in any ladder problem. They are all basically the same, each with a slighly different twist. Now you are ready to examine the sample problems to see the kind of thing you may be asked to do and try some for yourself. Again, just follow the links.
Problems For You To Try