The particle equilibrium It is a state in which a particle is found when the external forces that act on them are mutually canceled. This assumes that it maintains a constant state, in such a way that it can occur in two different ways depending on the specific situation.
The first is to be in static equilibrium, in which the particle is immobile; and the second is the dynamic equilibrium, where the summation of forces is canceled, but nevertheless the particle has uniform rectilinear motion.
The particle model is a very useful approximation to study the motion of a body. It consists in assuming that all the mass of the body is concentrated in a single point, regardless of the size of the object. In this way you can represent a planet, a car, an electron or a billiard ball.
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The point that represents the object is where the forces that affect it act. All these forces can be replaced by a single one that does the same effect, which is called resultant force or net force and is denoted as FR or FN.
According to Newton's second law, when there is an unbalanced resultant force, the body experiences an acceleration proportional to the force:
FR = m.a
Where to is the acceleration that the object acquires thanks to the action of force and m is the mass of the object. What happens if the body is not accelerated? Precisely what was indicated at the beginning: the body is at rest or moves with uniform rectilinear motion, which lacks acceleration.
For a particle in equilibrium it is valid to ensure that:
FR = 0
Since adding vectors does not necessarily imply adding the modules, the vectors must be decomposed. Thus, it is valid to express:
Fx = m.ax = 0; FY = m.aY = 0; Fz = m.az = 0
In order to visualize the forces that act on the particle, it is convenient to make a free-body diagram, in which all the forces that act on the object are represented by arrows..
The above equations are vector in nature. When decomposing the forces, they are distinguished by signs. In this way it is possible for the sum of its components to be zero.
The following are important guidelines to make the drawing useful:
- Choose a reference system in which the greatest amount of forces are located on the coordinate axes.
- Weight is always drawn vertically down.
- In the case of two or more surfaces in contact, there are normal forces, which are always drawn by pushing the body and perpendicular to the surface that exerts it..
- For a particle in equilibrium there can be frictions parallel to the contact surface and opposing the possible movement, if the particle is considered at rest, or definitely in opposition, if the particle moves with MRU (uniform rectilinear movement).
- If there is a rope, the tension is always drawn along it and pulling the body.
Figure 2 shows a particle on which two forces act. In the figure on the left, the particle receives the action of two forces F1 and Ftwo that have the same magnitude and act in the same direction and in opposite directions.
The particle is in equilibrium, but nevertheless with the information supplied it is not possible to know if the equilibrium is static or dynamic. More information is needed about the inertial frame of reference from which the object is observed.
The figure in the center shows the same particle, which this time is not in equilibrium, since the magnitude of the force Ftwo is greater than that of F1. Therefore there is an unbalanced force and the object has an acceleration in the same direction as Ftwo.
Finally, in the figure on the right, we see a body that is not in equilibrium either. Despite the fact that F1 and Ftwo have equal magnitude, the force Ftwo is not in the same direction as 1. The vertical component of Ftwo is not counteracted by any other and the particle experiences an acceleration in that direction.
Can a particle subjected to three forces be in equilibrium? Yes, as long as when placing the end and end of each one, the resulting figure is a triangle. In this case the vector sum is zero.
A force that frequently intervenes in the equilibrium of the particle is static friction. It is due to the interaction of the object represented by the particle with the surface of another. For example, a book in static equilibrium on an inclined table is modeled as a particle and has a free-body diagram like the following:
The force that prevents the book from sliding across the surface of the inclined plane and remaining at rest is static friction. It depends on the nature of the surfaces in contact, which microscopically present roughness with peaks that interlock with each other, making movement difficult..
The maximum value of static friction is proportional to the normal force, the force exerted by the surface on the supported object, but perpendicular to said surface. In the example in the book it is indicated in blue. Mathematically it is expressed like this:
Fs more∝ N
The constant of proportionality is the static friction coefficient μs, which is determined experimentally, is dimensionless and depends on the nature of the surfaces in contact.
Fs max = μs N
If a particle is in dynamic equilibrium, movement already takes place and static friction no longer intervenes. If any friction force opposing the movement is present, the dynamic friction acts, whose magnitude is constant and is given by:
Fk = μk N
Where μk is he dynamic friction coefficient, It also depends on the type of surfaces in contact. Like the static friction coefficient, it is dimensionless and its value is determined experimentally.
The value of the coefficient of dynamic friction is usually less than that of static friction.
The book in Figure 3 is at rest and has a mass of 1.30 kg. The plane has an angle of inclination of 30º. Find the coefficient of static friction between the book and the plane surface.
It is important to select a suitable reference system, see the following figure:
The weight of the book has magnitude W = mg, however it is necessary to break it down into two components: Wx Y WY, since it is the only force that does not fall just above any of the coordinate axes. The decomposition of the weight is observed in the figure on the left.
WY = mg.cosθ = 1.30 x 9.8 x cos 30º N = 11.03 N
Wx = mg sinθ = 1.30 x 9.8 x sin 30º = 6.37 N
The 2nd. Newton's law for the vertical axis is:
N - Wy = 0
N = mg. cos θ = 11.03 N.
Applying the 2nd. Newton's law for the x-axis, choosing the direction of the possible motion as positive:
Wx - Fs = 0
The maximum friction is Fs max= μsN, Thus:
Wx - μsN = 0
μs = Wx / N = 6.37 / 11.03 = 0.58
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