WHAT IS DYNAMIC EQUILIBRIUM? (WITH EXAMPLE) - PHYSICAL - 2025

The dynamic equilibrium is the state in which a moving object represented ideally as a particle when its motion is rectilinear uniform lies. This phenomenon occurs when the sum of external forces acting on it is canceled.

It is often believed that if there is no net or resultant force on an object, rest is the only possible consequence. Or also that for a body to be in balance there must be no force acting.

Figure 1. This cat moves in dynamic equilibrium if it moves with constant speed. Source: Pixabay.

In reality, equilibrium is the absence of acceleration, and therefore constant speed is perfectly possible. The cat in the figure may be moving without acceleration.

An object with uniform circular motion is not in dynamic equilibrium. Although its speed is constant, there is an acceleration directed towards the center of the circumference that keeps it on the path. This acceleration is responsible for changing the velocity vector appropriately.

The null velocity is a particular situation of the equilibrium of a particle, equivalent to affirming that the object is at rest.

As for considering objects as particles, this is a very useful idealization when describing their global motion. In reality, the moving objects that surround us are made up of a large number of particles whose individual study would be cumbersome.

The principle of superposition

This principle allows substituting the action of multiple forces on an object by an equivalent called resultant force FR or net force FN, which in this case is null:

F1 + F2 + F3 +…. = FR = 0

Where the forces F1, F2, F3…., Fi are the different forces that act on the body. The summation notation is a compact way of expressing it:

As long as an unbalanced force does not intervene, the object can keep moving indefinitely with constant speed, since only a force can alter this panorama.

In terms of the components of the resultant force, the condition of dynamic equilibrium of a particle is expressed as follows: Fx = 0; Fy = 0; Fz = 0.

Rotation and equilibrium conditions

For the particle model, the condition FR = 0 is sufficient guarantee of equilibrium. However, when taking into account the dimensions of the mobile under study, there is the possibility that the object can rotate.

The rotational movement implies the existence of an acceleration, therefore the rotating bodies are not in dynamic equilibrium. The rotation of a body requires not only the participation of a force, but it is necessary to apply it in the appropriate place.

To check this, a thin rod in length can be placed on a friction-free surface, such as a frozen surface or a highly polished mirror or glass. The normal balances the weight vertically, and by applying two forces F1 and F2 of the same magnitude horizontally, according to the diagram in the following figure, what happens is verified:

Figure 2. A rod on a friction-free surface may or may not be in equilibrium, depending on how forces 1 and 2 are applied. Source: own elaboration.

If F1 and F2 are applied as shown to the left, with a common line of action, the rod will remain at rest. But if F1 and F2 are applied as shown on the right, with different lines of action, although parallel, a clockwise rotation occurs, around the axis that passes through the center.

In this case, F1 and F2 constitute a couple of forces or simply a couple.

Torque or moment of a force

The effect of torque is to produce a rotation on an extended object such as the rod in the example. The charged vector magnitude is called torque or also moment of a force. It is denoted as τ and is calculated by:

τ = rx F

In this expression F is the applied force and r is the vector that goes from the axis of rotation to the point of application of the force (see figure 2). The direction of τ is always perpendicular to the plane where F and r lie and its units in the international system are Nm

For the example, the direction of the moments produced by F1 and F2 is towards the paper, according to the rules of the vector product.

Although the forces cancel each other out, their torques do not. And the result is the displayed rotation.

Equilibrium conditions for an extended object

There are two conditions that must be met to guarantee the balance of an extended object:

There is a box or trunk that weighs 16 kg-f, which slides down an inclined plane with constant speed. The angle of inclination of the wedge is θ = 36º. Answer:

a) What is the magnitude of the dynamic friction force necessary for the trunk to slide with constant speed?

b) How much is the coefficient of kinetic friction?

c) If the height h of the inclined plane is 3 meters, find the speed of descent of the trunk knowing that it takes 4 seconds to reach the ground.

Solution

The trunk can be treated as if it were a particle. Therefore, the forces will be applied at a point located approximately in its center, on which all its mass can be assumed to be concentrated. It is at this point that it will be tracked.

Figure 3. Free-body diagram for trunk sliding downhill and weight breakdown (right). Source: self made.

The weight W is the only force that does not fall on one of the coordinate axes and must be decomposed into two components: Wx and Wy. This decomposition is shown in the scheme (figure 3).

It is also convenient to transfer the weight to units of the international system, for which it is enough to multiply by 9.8:

Wy = W. cosθ = 16 x 9.8 x cos 36º N = 126.9 N

Wx = W. sinθ = 16 x 9.8 x sin 36º = 92.2 N

Paragraph a

Along the horizontal axis are the horizontal component of the weight Wx and the dynamic or kinetic friction force fk, which opposes the movement.

By choosing the positive direction in the direction of movement, it is easy to see that Wx is responsible for the block going downhill. And since the friction is opposed, instead of sliding rapidly, the block has the possibility of sliding with constant speed downhill.

The first equilibrium condition is sufficient, since we are treating the trunk as a particle, which is assured in the statement that it is in dynamic equilibrium:

Wx - fk = 0 (no acceleration in the horizontal direction)

fk = 92.2 N

Section b

The magnitude of the dynamic friction is constant and is given by fk = μk N. This means that the dynamic friction force is proportional to the normal and the magnitude of this is required to know the coefficient of friction.

Observing the free body diagram it can be seen that on the vertical axis we have the normal force N, which the wedge exerts on the trunk and is directed upwards. She is balanced with the vertical component of the weight Wy. Selecting up as a positive sense and making use of Newton's second law and the equilibrium condition results:

N - Wy = 0 (there is no movement along the vertical axis)

Thus:

N = Wy = 126.9 N

fk = μk N

μk = fk / N = 92.2 /126.9= 0.73

Section c

The total distance traveled by the trunk from the top of the wedge to the ground is found by trigonometry:

d = h / sin 36º = 3 / sin 36º m = 5.1 m.

To calculate the velocity, the definition for uniform rectilinear motion is used:

v = d / t = 5.1 m / 4 s = 1.3 m / s

References

1. Rex, A. 2011. Fundamentals of Physics. Pearson. 76 - 90.
2. Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1. 7th. Ed. Cengage Learning. 120-124.
3. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9th Ed. Cengage Learning. 99-112.
4. Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. MacGraw Hill. 71 - 87.
5. Walker, J. 2010. Physics. Addison Wesley. 148-164.