Statics: Equilibrium

A particle is in equilibrium if it remains in rest or anything at rest, or has a constant velocity if original in motion, basically a particle with zero acceleration and it is represented as ΣF = 0. This follows Newton's second law of motion, indicating that ΣF = ma. Thus, the particle remains at a constant velocity, or remains at rest. 

The next thing I want to discuss is the free body diagram. We need to account for all of the unknown forces that act on the particle by thinking of the particle as isolated, or "free" from it's surroundings. The best way to do this is through a free-body diagram. There are 2 connections in particle equilibrium problems, and I want to discuss both of them.

If a linearly elastic spring of undeformed length l0 is used to support a particle the length of the spring will change in direct proportion to the force F acting on it. A spring constant or stiffness k defines the elasticity of a spring. The magnitude of the spring with stiffness k in a distance s = l - l0 is F = ks. If s is positive (elongation) F must pull, and the opposite for compression. 

The pully, on the other hand, is different. All cables are assumed to have negligible weight. The tension must be constant on a cable to keep the cable in equilibrium. The cable is subjected to a constant tension T as a result. 

So here's the procedure:

We first drew the outlined shape of the particle. Then we show all the forces that act in the particle, which they can be active forces, which induces motion, and reactive forces, which prevents motion. The forces are labelled with their proper magnitudes and directions as a result. Here are some free body diagrams that illustrate the spring, the pulley, and the suspended plate. The last thing I want to do is an example problem.

Here', we want to determine the free body diagram on the force of the sphere.

The sphere is 6kg, we want to determine its FBD. Let's get to it. There are only 3 forces acting on the sphere, its weight and the force of the cord. Now, let's go to the force of the cord.

When the cord is isolated there's only 2 things acting on it, the force of the sphere and the force of the knot.

Now, let's go to the knot, and there are 3 forces, the 2 cordes down and upper right, as well as the force of the sprint. Here's an illustration. 

In the next section, we want to talk about coplanar force systems. If each particle is subjected to coplanar forces, then each force can be resolved into its i and j components as a result.

All forces provide a sum of zero, as indicated by the equation ΣF = 0 or ΣFxi + ΣFyj = 0 in order to achieve this equilibrium. When applying each of these equations to the equilibrium, we must account for the sense of direction of the component by using an algebraic sign corresponding to the arrowhead direction.

We must account for the sense of direction with the algebraic sign. If a force has unknown magnitude, then the sense of force can be assumed. Here's an example, and we can switch the magnitude of the force if negative.

Determine the required length of cord AC in Figure 3-8a so that the 8kg lamp can be suspended below. The undeformed length of the spring AB is l'AB = 0.4m and the spring has stiffness kAB = 300N/m.

So the lamp has a weight of 8(9.81) = 78.5N and we can draw the free body diagram as follows: