Translational equilibrium: conditions, examples, exercises
We explain the translational equilibrium with conditions, examples and exercises. It is stated that an object is in translational equilibrium when the sum of the forces acting on it is zero. This does not mean that there is necessarily rest, but the movement, if it exists, would be uniform rectilinear or exclusively rotational, in the case of a large object. translational equilibrium
The conditions of mechanical equilibrium are based on Newton’s Laws of Mechanics. Indeed, the first law tells us that an object is at rest or moving with uniform rectilinear motion MRU, provided that no net force acts on it.
Now, the net force or resultant force is simply the vector sum of all the forces acting on the object. According to Newton’s second law, this sum must be equal to the product of the mass and the acceleration, but if the object is not accelerated, this sum vanishes. translational equilibrium
And in the absence of acceleration, there are the two aforementioned possibilities: the body is at rest, that is, it does not move, or if it does, it must be with MRU. In the first case, we speak of static transnational equilibrium, and in the second, dynamic. translational equilibrium
Translational equilibrium is an important factor in many aspects of engineering, for example in construction. The elements that make up a building: beams, cables, frames and more, must be in balance to ensure the stability of the enclosure.
Translational balance is also sought in mobile structures, such as escalators, conveyor belts and in the practice of numerous sports.
Translational equilibrium condition
Let us suppose that several forces act on a body, which we denote as F 1 , F 2 , F 3 …. F n , using bold type to highlight the fact that forces are vectors and must be added as such.
The vector sum of all these forces is called the resultant force or the net force . If this summation results in the null vector, the condition for translational equilibrium is fulfilled:
F 1 + F 2 + F 3 …. + F n = 0
This condition can be written compactly using summation notation:
∑ F i = 0
In terms of the components of the resultant force, the above equation, which is vector, can be broken down into three scalar equations, one for each component of the resultant force:
∑ F ix = 0; ∑ F y = 0 and ∑ F z = 0
In practice, it is not easy to cancel the sum of forces, because friction is a contact force between surfaces that is hardly completely canceled by any other force.
This is the reason that real objects are almost never exempt from external forces, and as a consequence it is difficult to obtain translational equilibrium. translational equilibrium
So engineers use mechanisms to reduce friction, such as bearings and the use of lubricating oils.
The free-body diagram is a diagram in which the forces acting on the body are drawn. When seeking translational equilibrium, these forces must be balanced. For example, if a downward vertical force is acting, such as weight , then there must be an upward vertical force of exactly the same magnitude.
This force can be supplied by the hand that supports the object so that it does not fall, a rope or simply the surface of a table.
If there is a force tangential to the surface, such as kinetic or static friction, there must be another opposing force for balance to exist. For example, let’s look at the weight hanging from the strings shown in the following figure.
The weight is balanced translatory and not moving, thanks to the vertical rope that holds exerting a tension T that compensates the weight W . Each force has been represented on the weight by an arrow, each one of equal size and with the same direction, but in the opposite direction. translational equilibrium
The balancing force
Suppose that a set of forces acts on an object. This is called a system of forces from which the resultant can be found as explained above: by vectorly adding each of the forces in the system.
Well, the force opposite to this resultant is called the balancing force . If the resultant force is F R and the balancing force is E , then:
E + F R = 0
E = – F R
Examples of translational equilibrium
Many objects that we find on a daily basis, inside and outside the home, are in translational equilibrium:
Buildings and roads
Buildings and roads are built to remain stable and not tip over or collapse. However, in skyscrapers and in general very tall buildings, some flexibility is necessary to resist the action of the wind. translational equilibrium
Books and objects on shelves
Books in a library and products on store shelves are objects that remain in translational equilibrium and do not move.
The furniture, the flat-screen TV and the pictures on the wall, as well as the lamps hanging from the ceiling, to mention a few objects, are in translational balance.
The traffic lights are fastened by poles and cables, so that they do not fall. However, we know that the wind makes them sway.
The streetlights are also in translational balance, fixed on the lamp posts, like the lamppost in the main image.
What magnitude must the static friction force f s have for the box in the figure to remain at rest in the middle of the inclined plane at an angle α of 37º? The mass of the box is m = 8 kg.
The figure shows the free-body diagram of the box on the plane. There are three forces acting on it: the weight W , directed vertically downwards, the normal N , which is the perpendicular force exerted by the surface of the plane on the box, and finally the static friction force f s that opposes the box slipping downhill.
The translational equilibrium condition states that:
W + N + f s = 0
But it must be remembered that this is a vector sum and to carry it out it is necessary to decompose the forces into components along the coordinate axes. translational equilibrium
In the figure a coordinate system has been drawn in which the x-axis runs parallel to the surface of the inclined plane. With this choice, the static friction falls on said axis, while the normal is on the y axis. Weight is the only force that is inclined and must be broken down with the help of trigonometry:
W x = W. sin α
W y = W. cos α
The sum of forces along each axis is:
∑ F y = N – W y = 0
∑ F x = f s – W x = 0
From this last equation it follows that:
f s = W x
And since W x = W. sin α and the magnitude of the weight in turn is W = mg, where g is the value of gravity, then the magnitude of the static friction is simply: translational equilibrium
f s = m⋅g⋅sen α = 8 kg × 9.8 m / s 2 × sin 37º = 47.2 N.