Principles of Equilibrium

Introduction

1. Architecture is a science arising out of many other sciences, and adorned with much and varied learning; by the help of which a judgment is formed of those works which are the result of other arts. Practice and theory are its parents. Practice is the frequent and continued contemplation of the mode of executing any given work, or of the mere operation of the hands, for the conversion of the material in the best and readiest way. Theory is the result of that reasoning which demonstrates and explains that the material wrought has been so converted as to answer the end proposed.
2. Wherefore the mere practical architect is not able to assign sufficient reasons for the forms he adopts; and the theoretic architect also fails, grasping the shadow instead of the substance. He who is theoretic as well as practical, is therefore doubly armed; able not only to prove the propriety of his design, but equally so to carry it into execution.
3. In architecture, as in other arts, two considerations must be constantly kept in view; namely, the intention, and the matter used to express that intention: but the intention is founded on a conviction that the matter wrought will fully suit the purpose; he, therefore, who is not familiar with both branches of the art, has no pretension to the title of the architect. An architect should be ingenious, and apt in the acquisition of knowledge. Deficient in either of these qualities, he cannot be a perfect master. He should be a good writer, a skilful draftsman, versed in geometry and optics, expert at figures, acquainted with history, informed on the principles of natural and moral philosophy, somewhat of a musician, not ignorant of the sciences both of law and physic, nor of the motions, laws, and relations to each other, of the heavenly bodies.

Vitruvius, De Architectura

Outline

Bodies in Equilibrium

Static Equilibrium

Practice with the Concepts
Discussion Points

Equilibrium

Equilibrium occurs when forces on an object are in exact balance so that the body is not accelerating.

Static Equilibirum

We have talked about dynamics (forces) and kinematics (energy) as they apply to moving bodies, that is, where net forces on an object are greater than zero. Now let's look at the situation where the object or system of interest is not moving. Such a system is said to be in static equilibrium. Static comes from the Latin stasis, which means nothing is changing, and equilibrium means equal forces, every force is balanced. Both acceleration and velocity are zero, so in most cases, we are dealing only with situations involving potential energy.

The conditions for static equilibirum are two:

First, the sum of all forces which act through the center of mass and result in translational motion must be zero, and
second, the sum of all torques is zero. (If you need to review torque, check out the University of Guelph tutorial on torques.)

Usually we construct a classical 2 or 3 dimensional co-ordinate system on a situation in order to identify all the forces and break them down into common F_x and F_y (and F_z if we are working in three dimensions) components. Then we use the principle that the sum of all F_x is zero, the sum of all F_y is zero, and the sum of all F_z is zero to determine any unknowns.

Problem solving techniques

Given the basic conditions for static equilibrium, we can now approach a number of problems requiring the analysis of forces on a static object. For all such problems, the common approach to a solution involves these steps:

Identify all forces on the body.
Make a freebody diagram (if you are still having trouble with this, check out this tutorial on freebody diagramming).
Identify all the forces which result in translational motion. These forces act on the same point, the center of mass, and are sometimes called concurrent forces.
Identify all the forces which produce rotationation motion. These torques operation on points other than the center of mass and produce rotation. Because torques are often the result of forces which operate on different points in the object, they are non-concurrent forces.
Choose a convenient coordinate system. Since the sum of the torques is zero, we can consider any point to be on the axis of rotation (where we would normally put the original of the coordinate system). General practice is to put the origin on a point through which one of the forces acts; that way we don't have to consider a torque for that force, because its radial arm component (r in torque = F * r * sin theta) is zero.
Solve for the unknowns.

Practice with the Concepts

There are a number of classic static equilibrium problems, and they fall into three classes. Solved examples of each are in your text, so I won't do a solution here, but instead, look at the common approach for each required.

Teeter-totter problems. These problems involve a beam with both ends free but which has a fixed point of rotation somewhere along its length. For the beam to remain motionless, all forces which do not act through the fixed point must balance--and all such forces will be torques which will cause the beam to rotate.
- Draw a diagram showing all forces--in the example above, there are five. This is often the hardest step, because once you know all the forces, the rest is relatively straightforward.
- Identify which forces cause counterclockwise torques. You may even want to make a table:

F₁	r₁
F₂	r₂
F₃	r₃
F4	r₄
F₅	r₅

Once you've identified all the torques and their directions, set the counterclockwise torques equal to the clockwise torques. Then you can solve any problems for which you have enough information.
In the above diagram, assume that all forces are equal, and that r₁ = r₄. What must r₅ be in order for the beam to remain motionless?

Cantilever problems concern beams with a fixed point at one end and various forces acting on other points. Torques are produced around the fixed point. In addition, forces (such as the weight of the beam, F₃) acting through the center of mass of the beam--which would normally cause translational motion--must be balanced by forces at the fixed point. In this case, all the torques have to cancel, and all the translational forces have to cancel.

Analyze the situation above and write an equation or set of equations which describes the sum of all forces and torques.

Ladder problems are interesting because the "fixed point" can be any convenient place we choose. If we need to calculate normal force at the wall required to keep the ladder from falling down, we would choose the point at which F₃ is exerted to be "fixed", and then determine the effects of torque from F₁ and F₂ about that point, plus the translational force of F₂ and resistant force of F₃.

In this case, why doesn't F₃ exert any torque?

Suspension (chandelier) problems are actually very simple, since the forces act on the common point and there are no torques. In this case, solution is a matter of finding the most convenient set of axis and breaking all forces into their x and y components.
How is F₁ related to F₂ and F₃?

Discussion Points

Can the sum of the torques on an object be zero while the net force on the object is nonzero?
A ladder, leaning against a wall, makes a 60° angle with the ground. When is it more likely to slip: when a person stands on the ladder near the top or near the bottom?

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Physics

Chapter 9: 1-4