First equilibrium condition: explanation, examples, exercises
The first equilibrium condition requires that the vector sum of all forces acting on a body be null, so that it is at rest (static equilibrium) or with uniform rectilinear motion (dynamic equilibrium).
This sum of forces is none other than the net force acting on the body, expressed mathematically as follows:
net f =
∑ F =
In space, the first equilibrium condition gives rise to three equations, one for each dimension:
∑ F x = 0; ∑ F y = 0 y ∑ F z = 0
When these equations are satisfied, the object does not translate, or if it does, it will be at a constant speed.
Looking around us, we realize that we are continually seeking to satisfy the first balance condition so that things don’t fall apart.
For this reason, it tries to compensate for the Earth’s gravitational pull by means of supports, ropes or supports of some kind, so that, in this way, things stay in place and don’t end up on the ground.
At other times, it is necessary to prevent external electromagnetic fields from interfering with the operation of electrical circuits and communication devices. In this case, it is the electrical charges that must be in balance.
A large number of everyday objects satisfy the first equilibrium condition, it is a matter of carefully observing:
Builders seek stability in buildings so that users stay safe. The purpose of statics is to study the conditions for static equilibrium to occur in buildings, bridges, roads and all types of structures.
Traffic lights and suspension signals
These signaling devices must remain fixed to fulfill their functions, therefore, they are connected to cables, poles and rods so that the first equilibrium condition is met.
Conductors in the balanced electrostatic attic
When conductive materials, such as copper and other metals, acquire an electrical charge, electrostatic balance is soon established, leaving the excess charge on the conductive surface. Inside the electric field is null.
This effect is often used to isolate electrical and electronic equipment from external fields, using a so-called Faraday cage. The cage is made of conductive material and surrounds the equipment to be protected.
During storms, automobiles serve as Faraday cages, protecting occupants from electrical shocks.
In lighting systems such as lighting fixtures, the first equilibrium condition is used to fix them to the ceiling, floor or wall.
Books and objects on the tables
Objects placed on tables and shelves meet the first equilibrium condition. The normal force that the support exerts on the objects is responsible for compensating the weight.
Measuring the Viscosity of a Liquid
To determine the viscosity of a liquid, a spherical object of known diameter is thrown into it, which will be slowed down by resistance. The ball’s speed is constant, thus being in dynamic equilibrium.
The greater the viscosity of the liquid, the slower the speed at which the sphere moves within it.
Steps to Apply the First Equilibrium Condition
Draw a free-body diagram showing all the forces acting on the body (omit those exerted by the body on others).
-Select a Cartesian coordinate system, ensuring that, whenever possible, the forces are located in one of the axes. The positive direction is usually taken in the direction of movement or possible movement.
-Determine the Cartesian components of each force.
– Apply Newton’s second law to each component as established at the beginning, thus leaving a system of equations.
-Solve the system of equations proposed in the previous step.
– Exercise solved 1
The block in the figure, of mass m , moves downhill on the inclined plane at an angle θ with constant velocity. Calculate the value of the kinetic friction coefficient μ k , if the block mass is m = 5 kg and θ = 37º.
The first step is to draw the free-body diagram and choose a Cartesian coordinate system to express the vector of each force. The forces acting on the block are:
-The normal N exerted by the inclined plane is perpendicular to its surface.
-Weight W is vertically downwards.
-Kinetic friction f k opposing movement. If it didn’t exist, the body would move downhill with an acceleration equal to g.senθ .
As the weight W is skewed with respect to the selected coordinate axes, it must be decomposed into its Cartesian components:
W x = mg.sin 37º = 5 kg x 9.8 m / s 2 x sin 37º = 29. 5 N
W Y = mg.cos 37º = 5 kg x 9.8 m / s 2 x cos 37º = 39, 1 N
Newton’s second law is now applied, equating each sum to 0, since the block has no acceleration when moving at a constant velocity:
∑ F y = N – W y = 0
F x = W x – f k = 0
The magnitude of kinetic friction is proportional to the magnitude of normal, with the kinetic friction coefficient u k being the proportionality constant.
f k = μ k N
On your turn:
N = Wy = 39.1 N
f k = W x
- 5 N = μ k x N 39.1
μ k = 29.5 / 39.1 = 0.75
– Exercise solved 2
Calculate the magnitude of the voltages that support the semaphore with a mass of 33 kg, shown in the figure:
The free-body diagram is made for the traffic light and the knot that holds the cables:
They act on it: tension T 3 upwards and weight W downwards. Therefore:
∑ F y = W – T 3 = 0
T 3 = 33 kg x 9.8 m / s 2 = 323.4 N
Tensions break down into their Cartesian components:
∑ F y = T 1 sin 53º + T 2 sin 37º – T 3 = 0 x
F x = T2 2 cos 37º – T 1 cos 53º = 0
And the following system of linear equations with two unknowns T 1 and T 2, we obtain :
– 0.6 T 1 + 0.8 T 2 = 0
0.8 T 1 + 0.6 T 2 = 323.4
The solution to this system of equations is: T 1 = 258.7 N and T 2 = 194.0 N