# Rotational equilibrium: formulas and equations, examples, exercises

We explain rotational equilibrium: formulas and equations, examples, exercises. An extended body is said to be in **rotational equilibrium** when the sum of the torques acting on it is zero.Â This does not mean that the object is necessarily at rest, but rather that there is no net tendency to change its state of motion for another. rotational equilibrium

An object that moves with constant velocity does so along a straight line and we can consider it in rotational equilibrium.Â Now, objects rotate because there are forces acting on them in such a way that a rotation results.Â The ability of a force to produce rotation, called torqueÂ *,Â *Â depends not only on the intensity of the force, but also where it is applied.

We recognize this immediately when a closed door is to be opened: force is never applied near the hinges, but away from them, that is why the handle is placed as far as possible, on the opposite side of the door. rotational equilibrium

The door’s axis of rotation passes through the hinges. Insisting on pushing it very close to the hinges, it takes a great effort to get the door to move even a little. rotational equilibrium

In the literature , torque is found under different names: torque, torsion, moment of a force, and torque. They are all synonymous.

So, we need to know the torques acting on an object to establish the rotational equilibrium condition.

**Rotational equilibrium condition**

The rotational equilibrium condition is:

*The sum of all the moments or torques that act on a body, calculated with respect to any axis, must be zero.*

The object in question must be stretched, since particles, by definition, only have translational equilibrium.

There may be forces applied to the body and still exist rotational equilibrium, as long as the forces do not make it rotate.

There can also be movement, even accelerated, but always along a straight line, since not all forces cause the appearance of torques. These appear when the forces do not all act along the same line of action. rotational equilibrium

**Torque or moment of a force**

Torque is denoted by the Greek letterÂ **Ï„**Â , inÂ **bold**Â because it is a vector and thus we distinguish it from its magnitude or module, which is a scalar.Â It depends on the applied forceÂ **F**Â , on the vectorÂ **r** that is directed from the axis of rotation O to the point of application of the force and finally, on the angle between these two vectors. rotational equilibrium

The vector product establishes the appropriate relationship between these quantities:

**Ï„ = r**Â xÂ **F**

And the torque module, denoted without bold, is:

Ï„ = râ‹…Fâ‹…sen Î¸

Where Î¸ is the angle betweenÂ **r**Â andÂ **F**Â .Â The units of torque are simply Nâ‹…m in the International System.

In the figure there is a spanner with which it is intended to turn a nut counterclockwise (counter clockwise).Â For this test two forcesÂ **FÂ **_{A}Â andÂ **FÂ **_{B}Â .

**FÂ **_{A}Â is closer to O and has aÂ shorterÂ vectorÂ **rÂ **_{A}Â or lever arm, therefore it does not produce as much torque as the forceÂ **FÂ **_{B}Â , which has the same magnitude, but has aÂ largerÂ vectorÂ **rÂ **_{B.}

Note that if you want to turn the nut clockwise, you must apply the forces in the opposite direction as shown in the figure.

**Direction and sense of torque**

As the torque results from the cross product between the force and position vectors, and these are in the plane of the wrench, the torque has to be a vector perpendicular to said plane, that is, directed towards the reader or towards the inside of the page.

By convention, torque is positive if it rotates counterclockwise, and negative if it rotates clockwise. rotational equilibrium

The direction and sense of the resulting torque is easily determined by the right hand rule shown below:

The index finger points according to the position vectorÂ **r**Â , the middle finger points according to the forceÂ **F**Â and the thumb points the direction and the sense of the torqueÂ **Ï„** .Â In this example, the torque is directed along the x-axis, based on the coordinate axes drawing. rotational equilibrium

**Formulas and equations rotational equilibrium**

If the torquesÂ **Ï„Â **_{1}Â ,Â **Ï„Â **_{2,Â }**Ï„Â **_{3â€¦Â }**Ï„Â **_{i}Â act on a bodyÂ , the net or resultant torqueÂ **Ï„Â **_{n}Â is the vector sum of all of them:

**Ï„Â **_{nÂ }**= Ï„Â **_{1}Â +Â **Ï„Â **_{2}Â +Â **Ï„Â **_{3}Â +Â _{â€¦Â }**Ï„Â **_{i}_{Â }

With summation notation it remains:

**Ï„Â **_{n}Â = âˆ‘Â **Ï„Â **_{i}

The equilibrium condition is expressed mathematically as follows:

**Ï„Â **_{n}Â =Â **0**

O well:

âˆ‘Â **Ï„Â **_{i}Â =Â **0**

Where the torqueÂ **Ï„,**Â with respect to a certain axis O, is calculated by:

**Ï„ = r**Â xÂ **F**

And whose magnitude is:

Ï„ = râ‹…Fâ‹…sen Î¸

**Examples rotational equilibrium**

-In humans and animals,Â weightÂ is a force that can cause torque and spin and fall.

People generally maintain a posture such that when walking, they are kept in rotational balance, unless they practice sports activities, such as gymnastics, skating or sports in general.

Two children who managed to stay in horizontalÂ *swing*Â orÂ *seesaw*Â are balanced rotation.

-When the balance pans are balanced, the system is in rotational equilibrium.

-The notices and traffic lights that hang on streets and avenues are also in rotational balance. If the cables that hold them break, this balance is lost and the sign hangs or falls. rotational equilibrium

-The suspension bridges like the Golden Gate in San Francisco and the bridge in figure 1.

**Exercise resolved rotational equilibrium**

The support bar shown in the figure is very light.Â The force exerted by the support isÂ **F**Â and the forceÂ **A**Â is applied to the extreme rightÂ .

It is requested to calculate the magnitudes of these forces considering that the system is in translational and rotational equilibrium.

As the system does not move, the summation of forces is canceled. They are all vertical and can be worked with the magnitudes. The positive direction is upwards and the negative direction is downwards, therefore: rotational equilibrium

F – 80 – A = 0

Now the rotational equilibrium condition is applied, for which we must choose an arbitrary axis of rotation.Â In this case, the right end is chosen, so that the vectorÂ **rÂ **_{A}Â is null, in this way the torque exerted byÂ **A**Â does not appearÂ , but only those ofÂ **F**Â and the force on the left.

The torque produced byÂ **F** is, according to the right hand rule and the coordinate system shown: rotational equilibrium

**Ï„Â **_{F}Â =Â **rÂ **_{F}Â xÂ **F**Â = 0.9 F (-Â **k**Â ) Nm

It is directed towards the inside of the screen and has a negative sign.Â While the torque produced by the 80 N force is:

**Ï„ =**Â 80 x 1.20 (Â **k**Â ) Nâ‹…m =Â **Â **96 (Â **k**Â ) Nâ‹…m

This torque is directed away from the screen and is assigned a positive sign.Â How there is rotational equilibrium:

96 – 0.9â‹…F = 0

The magnitude ofÂ **F**Â is:

F = (96 / 0.9) N = 106.7 N

And since the system is in translational equilibrium, the summation of the forces cancels out.Â This allows us to solve for the magnitude ofÂ **A**Â :

F – A – 80 N = 0

Therefore: rotational equilibrium

A = 106.7 – 80 N = 26.7 N.