It is stated that an extended body is in **rotational balance** when the sum of the torques acting on it is zero. 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 to another.

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

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

Through the hinges passes the axis of rotation of the door. By insisting on pushing it very close to the hinges, it takes a lot of effort to get the door to move even a little.

Torque is found in the literature under different names: torque, torsion, moment of force, and torque. They are all synonyms.

So, we need to know the torques that act on an object to establish the condition of equilibrium of rotation.

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**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 extended, since particles, by definition, only have translational equilibrium.

There can be forces applied to the body and still exist in rotational equilibrium, as long as the forces do not cause it to 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.

**Torque or moment of a force**

Torque is denoted by the Greek letter **τ**in** bold font **because it is a vector and thus we distinguish it from its magnitude or module, which is a scalar. Depends on applied force **F**vector** r **that is directed from the axis of rotation O to the point of application of the force and finally, of the angle between these two vectors.

The vector product establishes the appropriate relationship between these magnitudes:

**τ = r **x** F**

And the torque module, denoted without bold, is:

τ = r⋅F⋅sin θ

where θ is the angle between **r** and** F**. The units of torque are simply N⋅m in the SI.

In the figure there is a wrench with which it is intended to turn a nut in an anticlockwise direction (counterclockwise). For this, two forces are tested **F**oh and **F**b.

**F**A is closer to O and has a vector **r**A or shorter lever arm, therefore does not produce as much torque as force **F**B, which has the same magnitude, but has a vector **r**B bigger.

Note that if you want to make the nut turn clockwise, you have to apply the forces in the opposite direction to how they appear in the figure.

**Direction and sense of torque**

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

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

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 vector **r**the middle finger according to strength **F** and the thumb indicates the direction and sense of the torque **τ**. In this example, the torque is directed along the x axis, according to the drawing of the coordinate axes.

**Formulas and equations**

If the torques act on a body **τ**1, **τ**2, **τ**3 … **τ**i, the net or resultant torque **τ**n is the vector sum of all of them:

**τ**no **= τ**1+ **τ**2 + **τ**3 + … **τ**Yo

With summation notation it remains:

**τ**n = ∑ **τ**Yo

The equilibrium condition is expressed mathematically as follows:

**τ**n = **0**

O well:

∑ **τ**i = **0**

where the torque** τ, **with respect to a certain axis O, is calculated by:

**τ = r **x** F**

and whose magnitude is:

τ = r⋅F⋅sin θ

**examples**

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

People generally maintain a posture such that when walking, it keeps them in rotational balance, unless sports activities are practiced, such as gymnastics, skating or sports in general.

-Two children who managed to stay horizontal in the *rocker* either *up and down* They are in equilibrium of rotation.

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

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

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

**solved exercise**

The bar supported on a support shown in the figure is very light. The force exerted by the support is **F** and on the far right the force is applied **TO**.

It is requested to calculate the magnitudes of these forces considering that the system is in balance of translation and rotation.

**Solution**

Since the system does not translate, the sum of forces vanishes. They are all vertical and can be worked with magnitudes. The positive direction is up and the negative direction is down, therefore:

F – 80 – A = 0

Now the rotational equilibrium condition is applied, for which an arbitrary axis of rotation must be chosen. In this case, it is chosen at the extreme right, so that the vector **r**A is zero, thus the torque exerted by **TO**but only those of **F** and the strength of the left.

The torque produced by **F** is, according to the right-hand rule and the coordinate system shown:

**τ**F = **r**f x **F** = 0.9F (–**what**) Nm

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

**τ = **80 x 1.20 (**what**) N⋅m = ** **96 (**what**) N⋅m

This torque is directed out of the screen and is assigned a positive sign. Since there is rotational equilibrium:

96 – 0.9⋅F = 0

The magnitude of **F** is:

F = (96/0.9) N = 106.7 N

And since the system is in translational equilibrium, the sum of the forces cancels out. This allows us to solve for the magnitude of **TO**:

F – A – 80 N = 0

Therefore:

A = 106.7 – 80 N = 26.7 N.

**References**

Rex, A. 2011. Fundamentals of Physics. pearson.

Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1.7ma. Ed. Cengage Learning.

Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1. Pearson.

Tipler, P. (2006) Physics for Science and Technology. 5th Ed. Volume 1. Editorial Reverté.

Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. McGraw Hill.