Equilibrium of a material point.
We have a problem to determine, at what conditions the material point, that is the body, which sizes may not be taken into account, is in the equilibrium in some inertial frame of reference. The material point acceleration is equal to zero because the material point is motionless in this frame of reference. Therefore we have, according to the second Newton's law, that.
Thus, the condition of the equilibrium of the material point in some inertial point of reference will be the equality of the sum of all forces, acting to the material point, to zero: (1).
Because all forces are applied in a single point, they can be added with the help of the parallelogram rule. We should obtain the resultant force , which is equal to the sum of these forces and is applied to the material point. Then if is possible to say that the condition of the equilibrium of the material point should be the equality of the resultant force to zero, that is .
The vector equality (1) may be written in the projections to every axis in space. If we write (1) in the projections to three coordinate axes x, y, and z in space, we will obtain the system of three scalar equalities, equivalent to the single equality (1):
Here , Fix, Fiy,Fiz , are the projections of the force to the coordinate axes x, y, and z. Each of the equalities of the system (2) declares that in the equilibrium of the material point the sum of the projections of all forces acting to the material point to the corresponding axis should be equal to zero.
Let's note, that the coordinate system x, y, z is not necessarily rectangular. Equations (1) and (2) will be anyway equivalent, excluding the case when the axes x, y, z are parallel to each other, or when they are situated in the same plane. The equivalence means that if the system (1) is valid, the conditions (2) will be also fulfilled, and if all three equations (3) are simultaneously valid, then the vector equation will consequently be valid.
You may often meet in the technology a significant number of practical problems when all forces are situated in the same plane (a planar case). Let's superpose x and y axis with this plane. Then last equality in the system (2) became the strict identity and the conditions of the equilibrium of the material point in a planar case can be written in the form.
Conditions (1) are necessary, but not sufficient conditions of the equilibrium. It means that the validity of the system (1) is the obligatory (necessary) consequence of the fact of the equilibrium of the material point. But the equilibrium of the material point is not the strict consequence of the validity of the equality (1). According to the second Newton's law the equality of the material point acceleration to zero is the consequence of the equality of the sum of all forces to zero. And it shows that the material point may not only be motionless, but also can to move straight-line and uniformly.
Let's emphasize once more that the condition of the equilibrium (1) is in fact the equation of the second Newton's law for the material point written for the particular case when the acceleration is equal to zero. Therefore the methods and trips used for the solution of the problems of the equilibrium of the material point are analogous to those which are used in the problems connected with the use of the equations of the second Newton's law.
The ball of mass m = 0.2 kg is suspended on two wires attached to the ceiling and to the wall (see Figure 3). The first wire is directed along the horizontal and the angle between the second wire and the ceiling is equal to 60° . Let's find the tension forces of the wires.
The force of gravity and the wire tension forces act to the ball (see Figure 3). Let's write the condition of the equilibrium of the ball considering it as the material point:
The following solution of the will be realized by three methods and it is very helpful for the obtaining of practical experience. In fact it will be three mathematical methods; which should help to find interesting for us values from the written vector equality.
1-st method of the solution : Let's add any two forces, for example . We will obtain the force , which substitutes these two forces. Because of the equality of the sum of forces to zero, the modulus of the force is equal to the modulus of the force , and their directions are opposite. We can find from the obtained parallelogram of forces that
2-nd method of the solution : Let's write the vector equality (3) in the projections to the horizontal axis x and the vertical axis y (see Figure 4):
3-rd method of the solution : Sometimes the necessity (or desire) arises to obtain from the vector equality such scalar equality which should not contain the projection of the concrete (usually unknown) vector. It is not difficult to guess that it is necessary for this purpose to write the vector equality in the projection to the axis perpendicular to this vector.
Let's direct axis in our problem (see Figure 4) perpendicular to , force and axis perpendicular to force. The condition of the equilibrium (3) written in the projections to these axes has the form:
We obtained in each equation only one unknown value and it is very convenient. We can find from the first equation and from the second equation.
The analysis of the third method of the solution of Problem 1 allows to give two useful practical advises for the solution of problems.
- §1. Introduction.
- §2. Force. Equivalence of forces. Resultant force. Addition of forces. Decomposition of force.
- §3. Equilibrium of a material point.
- §4. Equilibrium of a body in the absence of the rotation.
- §5. Equilibrium of a body with a fixed axis of the rotation in the planar case. Moment of the force.
- Test questions.