# Equilibrium conditions: concept, applications and examples

The **equilibrium conditions** are required so that a body remains at rest or in uniform rectilinear motion. In the first case, the object is said to be in static equilibrium, while in the second it is said to be in dynamic equilibrium.

Assuming that the object in motion is a particle, in this case the dimensions are not taken into account, it is enough that the sum of the forces acting on it is canceled out.

But the vast majority of moving objects are of considerable dimensions; therefore, this condition is not sufficient to guarantee equilibrium, which is, in any case, the absence of acceleration, not movement.

**First and second equilibrium condition**

Let’s see: if the sum of the forces is zero, it’s true that the object won’t move or will move fast, but it can still start to rotate.

Therefore, to avoid rotations, a second condition must be added: that the sum of torques or torques caused by external forces acting on it, at any point, is also canceled out.

In summary, denoting as **F** the net force vector and **τ** or **M** for the net torque vector, we will have:

**first equilibrium condition**

∑ **F** =

Which means that: ∑ F _{x} = 0, ∑ F _{y} = 0 and ∑ F _{z} = 0

**second equilibrium condition**

∑ **τ** =or ∑ **M** =

With torques or moments calculated in relation to any point.

Next, we will assume that the moving object is a rigid body, which does not undergo deformation.

__Forms__

__Forms__

Although motion seems to be the common denominator in the universe, balance is also present in many aspects of nature and in the objects around us.

**isostatic balance**

On a planetary scale, the Earth is in *isostatic equilibrium* , a kind of gravitational equilibrium of the Earth’s crust, whose density is not uniform.

The differences in the densities of the different blocks or areas of the Earth’s crust are offset by the differences in height that characterize the planet’s orography. It works in the same way that different materials are more or less submerged in water according to their density and reach equilibrium.

But since the crustal blocks do not float properly in the water, but in the mantle, which is much more viscous, the balance is not called hydrostatic, but isostatic.

**Core Fusion Operation**

In stars like our Sun, the balance between the force of gravity that compresses them and the hydrostatic pressure that expands them keeps the fusion reactor in the star’s core functioning, keeping it alive. We depend on this balance for the Earth to receive the necessary light and heat.

**Construction**

On a local scale, we want buildings and constructions to remain stable, that is, to obey equilibrium conditions, in particular static equilibrium.

That’s why statics came into being, which is the branch of mechanics devoted to the study of the balance of bodies and everything that’s needed to keep them that way.

**Static Balance Types**

In practice, we find that static equilibrium can be of three types:

**stable balance**

It occurs when the object moves from its position and returns immediately when the force that moved it ceases. The closer an object is to the ground, the more likely it is to be in stable equilibrium.

The ball on the right in figure 2 is a good example, if we take it out of the balanced position at the bottom of the bowl, gravity will make it come back quickly.

**Indifferent or neutral balance**

Occurs when the object, despite being moved, remains in balance. Round objects like the ball, when placed on flat surfaces, are in indifferent balance.

**Unstable balance**

Occurs when, if the object moves from its equilibrium position, it does not return to it. If we move the ball away from the top of the hill to the left, it certainly won’t come back under its own power.

**Example: Particle Static**

Suppose a block of mass *m* on an inclined plane, from which all mass is assumed to be concentrated at its geometric center.

The horizontal component of the weight W _{x} tends to make the block slide downhill; therefore, another opposing force is needed. If we want the block to remain at rest, that force is static friction. But if we allow the block to slide downhill at a constant speed, the force needed will be dynamic friction.

In the absence of friction, the block will slide downhill quickly, in which case there will be no balance.

For the block to be at rest, the forces acting on it: the weight **W** , the normal **N** and the static friction **f **_{s} must be compensated. Thus:

∑ F _{y} = 0 → N – W _{y} = 0

∑ F _{x} = 0 → W _{x} – f _{s} = 0

Static friction balances the horizontal component of weight: W _{x} = f _{s} and therefore:

f _{s} = m. g.sen θ

**Exercise solved**

A 21.5 kg traffic light hangs from a homogeneous AB aluminum bar weighing 12 kg and 7.5 m long, supported by a horizontal rope CD, as shown in the figure. Meet:

a) The voltage of the CD cable

b) The horizontal and vertical components of the rotating force A exerts on the pole.

**Solution**

The diagram of forces applied to the bar is constructed, with the weight **W** , the stress on the chords and the horizontal and vertical components of the pivot reaction, called R _{x} and _{Ry} . So the equilibrium conditions apply.

**first condition**

As a problem in the plane, the first equilibrium condition offers two equations:

ΣF _{x} = 0

ΣF _{y} = 0

From the first:

R _{x} – T = 0

R _{x} = T

And from the second:

R _{y} – 117.6 N – 210.7 N = 0

R _{y} = 328.3 N

The horizontal component of the reaction is of the same magnitude as the voltage T.

**second condition**

Point A in figure 5 is chosen as the center of rotation, so the reaction arm **R** is null, remember that the magnitude of the momentum is given by:

M = F _{┴ } d

Where F _{┴} is the perpendicular component of the force and d is the distance between the axis of rotation and the point of application of the force. We will get an equation:

AM _{A} = 0

(210.7 × sin 53º) AB + (117.6 × sin 53º) (AB / 2) – (T × sin 37º) AD = 0

Distance AD is:

AD = (3.8 m / sin 37º) = 6.3 m

(210.7 × sin 53º N) (7.5 m) + (117.6 × sin 53º N) (3.75 m) – (T × sin 37º N) (6.3 m) = 0

Carrying out the indicated operations:

1262.04 + 352.20 – 3.8T = 0

Resolution for income T:

T = 424.8 N

From the first condition, we had R _{x} = T, so:

R _{x} = 424.8 N

**matters of interest**

First equilibrium condition.

Second equilibrium condition.