# Second equilibrium condition: explanation, examples, exercises

The **second equilibrium condition** establishes that the sum of torques or moments produced by all forces acting on a body, regardless of the point at which they are calculated, must be canceled so that the body is in static or dynamic equilibrium.

Denoting the torque or moment of force by the Greek letter **τ** , it is mathematically expressed like this:

∑ **τ** =

The letter in bold indicates the vectorial nature of the moment, which must be canceled in relation to any point chosen as the center of rotation. In this way, canceling the net torque ensures that the object does not start to rotate or tip over.

However, if the object was already rotating before and the net torque suddenly disappears, the rotation will continue, but with a constant angular velocity.

The second equilibrium condition is used in conjunction with the first condition, which says that the sum of forces on a body must be zero, so that it does not move or, if it does, it must be with uniform rectilinear movement:

∑ **F** =

Both conditions apply to extended bodies, those whose dimensions are measurable. When an object is supposed to be a particle, it makes no sense to talk about rotations, and the first condition is sufficient to guarantee equilibrium.

__Examples__

__Examples__

The second equilibrium condition is evident in numerous situations:

**Climbing the ladder**

When supporting a ladder on the floor and wall, we need enough friction, especially on the floor, to ensure that the ladder does not slip. If we try to climb a ladder supported by an oily, wet or slippery floor, it is not difficult to predict that we will fall.

To use the ladder confidently, it must be in static balance when climbing and on the required step.

**Moving a closet**

When you want to move a tall piece of furniture, such as a wardrobe or any item whose height is greater than its width, it is convenient to press a low point to avoid tipping over, so the piece is more likely to slip than rotate. and lie down.

In such circumstances, the furniture is not necessarily in balance, as it can move quickly, but at least it doesn’t tip over.

**Balconies**

Balconies that protrude from buildings should be constructed, ensuring that even if there are a lot of people at the top, they do not tip over and collapse.

**Dielectrics in External Electric Fields**

When placing a dielectric material in an external electric field, the molecules move and rotate to an equilibrium position, creating an electric field within the material.

This effect increases the capacitance of a capacitor when a material such as glass, rubber, paper or oil is introduced between its armatures.

**Plates and lamps**

It is common for many locals to hang signs on the building wall so that they are visible to passersby.

The poster is held by a bar and a cable, both attached to the wall by brackets. The various forces at work must ensure that the cartel does not fall, for which the two equilibrium conditions come into play.

A reflector can also be placed this way in a park, as in the following figure:

**How to calculate the net torque or net moment of a force?**

The torque or moment of a force, denoted by **τ** or **M** in some texts, is always calculated in relation to some point through which the axis of rotation passes.

It is defined as the cross product between the position vector **r** , which is directed from that axis to the point of application of the force and the force **F** :

**τ** = **r ** × **F**

As a vector, it is necessary to express the torque, giving its magnitude, direction and direction. The magnitude is given by:

*τ = rF.sen θ*

**Right hand rule for vector product**

When the problem is in the plane, the torque direction is perpendicular to the paper or canvas and the direction is determined by the right hand rule, in which the index finger points to **r** , the middle finger to **F** and the thumb points in or out of the paper.

When torque points away from the paper, rotation is counterclockwise and a positive sign is assigned by convention. If instead the torque is directed into the blade, the rotation will be clockwise and negative.

To find the net torque, a suitable point is chosen for the calculation, which can be the one at which the greatest amount of force acts. In this case, the moment of these forces is null, as it has a position vector **r** of magnitude 0.

You can choose any point that provides enough information to clear the unknown that prompts you to solve the problem. We’ll see it in more detail below.

__Exercise solved__

__Exercise solved__

The reflector in the figure below has 20 kg of mass and is supported by a thin horizontal bar, of negligible mass and length L, which is hinged to a pole. The cable, also lightweight, which helps to support the reflector, forms an angle θ = 30º with the bar. Calculate:

a) The tension in the cable

b) The magnitude of the force F that the pole exerts on the bar through the hinge.

**Solution**

Let’s apply the first equilibrium condition ∑ **F** = to the forces shown in the diagram:

**F** + **T** + **W = 0**

Note that the magnitude and direction of **F** have not yet been determined, but we assume that it has two components: F _{x} and F _{y} . In this way, we obtain two equations:

F _{x} -T. cos θ = 0

F _{y} – W + T⋅ sin θ = 0

Now let’s apply the second equilibrium condition, choosing point A, and we don’t know the magnitude of **F** or **T** . When choosing this point, the vector **r **_{A} is null; therefore, the moment of **F** is null and the magnitude of **F will** not appear in the equation:

-W⋅L + T⋅sen θ⋅L = 0

Therefore:

T.sen θ.L = WL

T = W / sin θ = (20 kg x 9.8 m / s ^{2} ) / sin 30º = 392 N

Knowing the magnitude of T, we can solve the F _{x} component :

F _{x} = T⋅ cos θ = 392 cos 30º N = 339. 5 N

And then component F _{and} :

F _{y} = W – T⋅ sin θ = (20 kg x 9.8 m / s ^{2} ) – 392⋅sen at 30º = 0

So we can express **F** like this:

**F =** 339.5 N **x**

It is therefore a horizontal force. This is because we consider the bar to have negligible weight.

If the point C was chosen to calculate the resulting moment, the vectors **r **_{T} and **r **_{W} are null, therefore:

M = F and _{⋅} L = 0

We conclude that F _{y} = 0. Thus:

– W + T⋅ sin θ = 0

T = W / sin θ

Which is the same result obtained initially by choosing point A as the location where the axis of rotation passes.