A body is at rest (or moves uniformly and rectilinearly) if the vector sum of all forces acting on it is equal to zero. They say that forces balance each other. When we are dealing with a body of a certain geometric shape, when calculating the resultant force, all forces can be applied to the center of mass of the body.
Condition for equilibrium of bodies
For a body that does not rotate to be in equilibrium, it is necessary that the resultant of all forces acting on it be equal to zero.
F → = F 1 → + F 2 → + . . + F n → = 0 .
The figure above shows the equilibrium of a rigid body. The block is in a state of equilibrium under the influence of three forces acting on it. The lines of action of the forces F 1 → and F 2 → intersect at point O. The point of application of gravity is the center of mass of the body C. These points lie on the same straight line, and when calculating the resultant force F 1 →, F 2 → and m g → are brought to point C.
The condition that the resultant of all forces be equal to zero is not enough if the body can rotate around a certain axis.
The arm of force d is the length of the perpendicular drawn from the line of action of the force to the point of its application. The moment of force M is the product of the force arm and its modulus.
The moment of force tends to rotate the body around its axis. Those moments that turn the body counterclockwise are considered positive. The unit of measurement of moment of force in the international SI system is 1 Newtonmeter.
Definition. Rule of Moments
If the algebraic sum of all moments applied to a body relative to a fixed axis of rotation is equal to zero, then the body is in a state of equilibrium.
M 1 + M 2 + . . +Mn=0
Important!
In the general case, for bodies to be in equilibrium, two conditions must be met: the resultant force must be equal to zero and the rule of moments must be observed.
In mechanics there are different types of equilibrium. Thus, a distinction is made between stable and unstable, as well as indifferent equilibrium.
A typical example of indifferent equilibrium is a rolling wheel (or ball), which, if stopped at any point, will be in a state of equilibrium.
Stable equilibrium is such an equilibrium of a body when, with its small deviations, forces or moments of forces arise that tend to return the body to an equilibrium state.
Unstable equilibrium is a state of equilibrium, with a small deviation from which forces and moments of forces tend to throw the body out of balance even more.
In the figure above, the position of the ball is (1) - indifferent equilibrium, (2) - unstable equilibrium, (3) - stable equilibrium.
A body with a fixed axis of rotation can be in any of the described equilibrium positions. If the axis of rotation passes through the center of mass, indifference equilibrium occurs. In stable and unstable equilibrium, the center of mass is located on a vertical straight line that passes through the axis of rotation. When the center of mass is below the axis of rotation, the equilibrium is stable. Otherwise, it's the other way around.
A special case of balance is the balance of a body on a support. In this case, the elastic force is distributed over the entire base of the body, rather than passing through one point. A body is at rest in equilibrium when a vertical line drawn through the center of mass intersects the area of support. Otherwise, if the line from the center of mass does not fall into the contour formed by the lines connecting the support points, the body tips over.
An example of body balance on a support is the famous Leaning Tower of Pisa. According to legend, Galileo Galilei dropped balls from it when he conducted his experiments on studying the free fall of bodies.
A line drawn from the center of mass of the tower intersects the base approximately 2.3 m from its center.
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DEFINITION
Stable balance- this is an equilibrium in which a body, removed from a position of equilibrium and left to its own devices, returns to its previous position.
This occurs if, with a slight displacement of the body in any direction from the original position, the resultant of the forces acting on the body becomes non-zero and is directed towards the equilibrium position. For example, a ball lying at the bottom of a spherical depression (Fig. 1 a).
DEFINITION
Unstable equilibrium- this is an equilibrium in which a body, taken out of an equilibrium position and left to itself, will deviate even more from the equilibrium position.
In this case, with a slight displacement of the body from the equilibrium position, the resultant of the forces applied to it is non-zero and directed from the equilibrium position. An example is a ball located at the top point of a convex spherical surface (Fig. 1 b).
DEFINITION
Indifferent Equilibrium- this is an equilibrium in which a body, taken out of an equilibrium position and left to its own devices, does not change its position (state).
In this case, with small displacements of the body from the original position, the resultant of the forces applied to the body remains equal to zero. For example, a ball lying on a flat surface (Fig. 1c).
Fig.1. Different types of body balance on a support: a) stable balance; b) unstable equilibrium; c) indifferent equilibrium.
Static and dynamic balance of bodies
If, as a result of the action of forces, the body does not receive acceleration, it can be at rest or move uniformly in a straight line. Therefore, we can talk about static and dynamic equilibrium.
DEFINITION
Static balance- this is an equilibrium when, under the influence of applied forces, the body is at rest.
Dynamic balance- this is an equilibrium when, due to the action of forces, the body does not change its movement.
A lantern suspended on cables, or any building structure, is in a state of static equilibrium. As an example of dynamic equilibrium, consider a wheel that rolls on a flat surface in the absence of friction forces.
« Physics - 10th grade"
Remember what a moment of force is.
Under what conditions is the body at rest?
If a body is at rest relative to the chosen frame of reference, then this body is said to be in equilibrium. Buildings, bridges, beams with supports, machine parts, a book on a table and many other bodies are at rest, despite the fact that forces are applied to them from other bodies. The task of studying the conditions of equilibrium of bodies is of great practical importance for mechanical engineering, construction, instrument making and other fields of technology. All real bodies, under the influence of forces applied to them, change their shape and size, or, as they say, are deformed.
In many cases encountered in practice, the deformations of bodies when they are in equilibrium are insignificant. In these cases, deformations can be neglected and calculations can be carried out, considering the body absolutely hard.
For brevity, we will call an absolutely rigid body solid body or just body. Having studied the equilibrium conditions of a solid body, we will find the equilibrium conditions of real bodies in cases where their deformations can be ignored.
Remember the definition of an absolutely rigid body.
The branch of mechanics in which the conditions of equilibrium of absolutely rigid bodies are studied is called static.
In statics, the size and shape of bodies are taken into account; in this case, not only the value of the forces is significant, but also the position of the points of their application.
Let us first find out, using Newton's laws, under what condition any body will be in equilibrium. For this purpose, let us mentally divide the entire body into a large number of small elements, each of which can be considered as a material point. As usual, we will call the forces acting on the body from other bodies external, and the forces with which the elements of the body itself interact internal (Fig. 7.1). So, a force of 1.2 is a force acting on element 1 from element 2. A force of 2.1 acts on element 2 from element 1. These are internal forces; these also include forces 1.3 and 3.1, 2.3 and 3.2. It is obvious that the geometric sum of internal forces is equal to zero, since according to Newton’s third law
12 = - 21, 23 = - 32, 31 = - 13, etc.
Statics is a special case of dynamics, since the rest of bodies, when forces act on them, is a special case of motion ( = 0).
In general, several external forces can act on each element. By 1, 2, 3, etc. we will understand all external forces applied respectively to elements 1, 2, 3, .... In the same way, through "1, "2, "3, etc. we denote the geometric sum of internal forces applied to elements 2, 2, 3, ... respectively (these forces are not shown in the figure), i.e.
" 1 = 12 + 13 + ... , " 2 = 21 + 22 + ... , " 3 = 31 + 32 + ... etc.
If the body is at rest, then the acceleration of each element is zero. Therefore, according to Newton’s second law, the geometric sum of all forces acting on any element will also be equal to zero. Therefore, we can write:
1 + "1 = 0, 2 + "2 = 0, 3 + "3 = 0. (7.1)
Each of these three equations expresses the equilibrium condition of a rigid body element.
The first condition for the equilibrium of a rigid body.
Let us find out what conditions external forces applied to a solid body must satisfy in order for it to be in equilibrium. To do this, we add equations (7.1):
(1 + 2 + 3) + ("1 + "2 + "3) = 0.
In the first brackets of this equality the vector sum of all external forces applied to the body is written, and in the second - the vector sum of all internal forces acting on the elements of this body. But, as is known, the vector sum of all internal forces of the system is equal to zero, since according to Newton’s third law, any internal force corresponds to a force equal to it in magnitude and opposite in direction. Therefore, on the left side of the last equality only the geometric sum of external forces applied to the body will remain:
1 + 2 + 3 + ... = 0 . (7.2)
In the case of an absolutely rigid body, condition (7.2) is called the first condition for its equilibrium.
It is necessary, but not sufficient.
So, if a rigid body is in equilibrium, then the geometric sum of external forces applied to it is equal to zero.
If the sum of external forces is zero, then the sum of the projections of these forces on the coordinate axes is also zero. In particular, for the projections of external forces on the OX axis, we can write:
F 1x + F 2x + F 3x + ... = 0. (7.3)
The same equations can be written for the projections of forces on the OY and OZ axes.
The second condition for the equilibrium of a rigid body.
Let us make sure that condition (7.2) is necessary, but not sufficient for the equilibrium of a rigid body. Let us apply two forces equal in magnitude and oppositely directed to the board lying on the table at different points, as shown in Figure 7.2. The sum of these forces is zero:
+ (-) = 0. But the board will still rotate. In the same way, two forces of equal magnitude and opposite directions turn the steering wheel of a bicycle or car (Fig. 7.3).
What other condition for external forces, besides their sum being equal to zero, must be satisfied for a rigid body to be in equilibrium? Let's use the theorem about the change in kinetic energy.
Let us find, for example, the equilibrium condition for a rod hinged on a horizontal axis at point O (Fig. 7.4). This simple device, as you know from the basic school physics course, is a lever of the first kind.
Let forces 1 and 2 be applied to the lever perpendicular to the rod.
In addition to forces 1 and 2, the lever is acted upon by a vertically upward normal reaction force 3 from the side of the lever axis. When the lever is in equilibrium, the sum of all three forces is zero: 1 + 2 + 3 = 0.
Let's calculate the work done by external forces when turning the lever through a very small angle α. The application points of forces 1 and 2 will travel along the paths s 1 = BB 1 and s 2 = CC 1 (arcs BB 1 and CC 1 at small angles α can be considered straight segments). The work A 1 = F 1 s 1 of force 1 is positive, because point B moves in the direction of the force, and the work A 2 = -F 2 s 2 of force 2 is negative, because point C moves in the direction opposite to the direction of force 2. Force 3 does not do any work, since the point of its application does not move.
The traveled paths s 1 and s 2 can be expressed in terms of the angle of rotation of the lever a, measured in radians: s 1 = α|VO| and s 2 = α|СО|. Taking this into account, let us rewrite the expressions for work as follows:
A 1 = F 1 α|BO|, (7.4)
A 2 = -F 2 α|CO|.
The radii BO and СО of the circular arcs described by the points of application of forces 1 and 2 are perpendiculars lowered from the axis of rotation on the line of action of these forces
As you already know, the arm of a force is the shortest distance from the axis of rotation to the line of action of the force. We will denote the force arm by the letter d. Then |VO| = d 1 - arm of force 1, and |СО| = d 2 - arm of force 2. In this case, expressions (7.4) will take the form
A 1 = F 1 αd 1, A 2 = -F 2 αd 2. (7.5)
From formulas (7.5) it is clear that the work of each force is equal to the product of the moment of force and the angle of rotation of the lever. Consequently, expressions (7.5) for work can be rewritten in the form
A 1 = M 1 α, A 2 = M 2 α, (7.6)
and the total work of external forces can be expressed by the formula
A = A 1 + A 2 = (M 1 + M 2)α. α, (7.7)
Since the moment of force 1 is positive and equal to M 1 = F 1 d 1 (see Fig. 7.4), and the moment of force 2 is negative and equal to M 2 = -F 2 d 2, then for work A we can write the expression
A = (M 1 - |M 2 |)α.
When a body begins to move, its kinetic energy increases. To increase kinetic energy, external forces must do work, i.e. in this case A ≠ 0 and, accordingly, M 1 + M 2 ≠ 0.
If the work of external forces is zero, then the kinetic energy of the body does not change (remains equal to zero) and the body remains motionless. Then
M 1 + M 2 = 0. (7.8)
Equation (7 8) is second condition for the equilibrium of a rigid body.
When a rigid body is in equilibrium, the sum of the moments of all external forces acting on it relative to any axis is equal to zero.
So, in the case of an arbitrary number of external forces, the equilibrium conditions for an absolutely rigid body are as follows:
1 + 2 + 3 + ... = 0, (7.9)
M 1 + M 2 + M 3 + ... = 0.
The second equilibrium condition can be derived from the basic equation of the dynamics of the rotational motion of a rigid body. According to this equation where M is the total moment of forces acting on the body, M = M 1 + M 2 + M 3 + ..., ε is the angular acceleration. If the rigid body is motionless, then ε = 0, and, therefore, M = 0. Thus, the second equilibrium condition has the form M = M 1 + M 2 + M 3 + ... = 0.
If the body is not absolutely solid, then under the action of external forces applied to it it may not remain in equilibrium, although the sum of external forces and the sum of their moments relative to any axis are equal to zero.
Let us, for example, apply two forces to the ends of a rubber cord, equal in magnitude and directed along the cord in opposite directions. Under the influence of these forces, the cord will not be in equilibrium (the cord is stretched), although the sum of external forces is equal to zero and the sum of their moments relative to the axis passing through any point of the cord is equal to zero.
This lecture covers the following issues:
1. Conditions for equilibrium of mechanical systems.
2. Stability of balance.
3. An example of determining equilibrium positions and studying their stability.
The study of these issues is necessary to study the oscillatory movements of a mechanical system relative to the equilibrium position in the discipline “Machine Parts”, to solve problems in the disciplines “Theory of Machines and Mechanisms” and “Strength of Materials”.
An important case of motion of mechanical systems is their oscillatory motion. Oscillations are repeated movements of a mechanical system relative to some of its positions, occurring more or less regularly over time. The course work examines the oscillatory motion of a mechanical system relative to an equilibrium position (relative or absolute).
A mechanical system can oscillate for a sufficiently long period of time only near a stable equilibrium position. Therefore, before composing the equations of oscillatory motion, it is necessary to find equilibrium positions and study their stability.
Equilibrium conditions for mechanical systems.
According to the principle of possible displacements (the basic equation of statics), in order for a mechanical system on which ideal, stationary, restraining and holonomic constraints are imposed to be in equilibrium, it is necessary and sufficient that all generalized forces in this system be equal to zero:
Where - generalized force corresponding j- oh generalized coordinate;
s- the number of generalized coordinates in the mechanical system.
If differential equations of motion have been compiled for the system under study in the form of Lagrange equations of the second kind, then to determine possible equilibrium positions it is sufficient to equate the generalized forces to zero and solve the resulting equations with respect to the generalized coordinates.
If the mechanical system is in equilibrium in a potential force field, then from equations (1) we obtain the following equilibrium conditions:
Therefore, in the equilibrium position, the potential energy has an extreme value. Not every equilibrium determined by the above formulas can be realized practically. Depending on the behavior of the system when it deviates from the equilibrium position, one speaks of stability or instability of this position.
Equilibrium stability
The definition of the concept of stability of an equilibrium position was given at the end of the 19th century in the works of the Russian scientist A. M. Lyapunov. Let's look at this definition.
To simplify the calculations, we will further agree on generalized coordinates q 1 , q 2 ,...,q s count from the equilibrium position of the system:
Where
An equilibrium position is called stable if for any arbitrarily small numbercan you find another number? , that in the case when the initial values of generalized coordinates and velocities will not exceed:
the values of generalized coordinates and velocities during further movement of the system will not exceed .
In other words, the equilibrium position of the system q 1 = q 2 = ...= q s = 0 is called sustainable, if it is always possible to find such sufficiently small initial values, at which the movement of the systemwill not leave any given, arbitrarily small, neighborhood of the equilibrium position. For a system with one degree of freedom, the stable motion of the system can be clearly depicted in the phase plane (Fig. 1).For a stable equilibrium position, the movement of the representing point, starting in the region [ ] , will not go beyond the region in the future.
Fig.1
The equilibrium position is called asymptotically stable , if over time the system approaches the equilibrium position, that is
Determining the conditions for the stability of an equilibrium position is a rather complex task, so we will limit ourselves to the simplest case: studying the stability of the equilibrium of conservative systems.
Sufficient conditions for the stability of equilibrium positions for such systems are determined Lagrange-Dirichlet theorem : the equilibrium position of a conservative mechanical system is stable if in the equilibrium position the potential energy of the system has an isolated minimum .
The potential energy of a mechanical system is determined to within a constant. Let us choose this constant so that in the equilibrium position the potential energy is equal to zero:
P (0)=0.
Then, for a system with one degree of freedom, a sufficient condition for the existence of an isolated minimum, along with the necessary condition (2), will be the condition
Since in the equilibrium position the potential energy has an isolated minimum and P (0)=0 , then in some finite neighborhood of this position
P(q)=0.
Functions that have a constant sign and are equal to zero only when all their arguments are zero are called definite. Consequently, in order for the equilibrium position of a mechanical system to be stable, it is necessary and sufficient that in the vicinity of this position the potential energy is a positive definite function of generalized coordinates.
For linear systems and for systems that can be reduced to linear for small deviations from the equilibrium position (linearized), the potential energy can be represented in the form of a quadratic form of generalized coordinates
Where - generalized stiffness coefficients.
Generalized coefficientsare constant numbers that can be determined directly from the series expansion of potential energy or from the values of the second derivatives of potential energy with respect to generalized coordinates at the equilibrium position:
From formula (4) it follows that the generalized stiffness coefficients are symmetrical with respect to the indices
For that In order for sufficient conditions for the stability of the equilibrium position to be satisfied, the potential energy must be a positive definite quadratic form of its generalized coordinates.
In mathematics there is Sylvester criterion , which gives necessary and sufficient conditions for the positive definiteness of quadratic forms: quadratic form (3) will be positive definite if the determinant composed of its coefficients and all its principal diagonal minors are positive, i.e. if the odds will satisfy the conditions
.....
In particular, for a linear system with two degrees of freedom, the potential energy and the conditions of the Sylvester criterion will have the form
In a similar way, it is possible to study the positions of relative equilibrium if, instead of potential energy, we introduce into consideration the potential energy of the reduced system.
P An example of determining equilibrium positions and studying their stability
Fig.2
Consider a mechanical system consisting of a tube AB, which is the rod OO 1 connected to the horizontal axis of rotation, and a ball that moves along the tube without friction and is connected to a point A tubes with a spring (Fig. 2). Let us determine the equilibrium positions of the system and evaluate their stability under the following parameters: tube length l 2 = 1 m , rod length l 1 = 0,5 m . undeformed spring length l 0 = 0.6 m spring stiffness c= 100 N/m. Tube weight m 2 = 2 kg, rod - m 1 = 1 kg and the ball - m 3 = 0.5 kg. Distance O.A. equals l 3 = 0.4 m.
Let us write down an expression for the potential energy of the system under consideration. It consists of the potential energy of three bodies located in a uniform field of gravity and the potential energy of a deformed spring.
The potential energy of a body in a gravity field is equal to the product of the body's weight and the height of its center of gravity above the plane in which the potential energy is considered equal to zero. Let the potential energy be zero in the plane passing through the axis of rotation of the rod O.O. 1, then for gravity
For the elastic force, the potential energy is determined by the magnitude of the deformation
Let us find possible equilibrium positions of the system. The coordinate values at the equilibrium positions are the roots of the following system of equations.
A similar system of equations can be compiled for any mechanical system with two degrees of freedom. In some cases, it is possible to obtain an exact solution of the system. For system (5) such a solution does not exist, so the roots must be sought using numerical methods.
Solving the system of transcendental equations (5), we obtain two possible equilibrium positions:
To assess the stability of the obtained equilibrium positions, we will find all the second derivatives of the potential energy with respect to the generalized coordinates and from them we will determine the generalized rigidity coefficients.
The branch of mechanics in which the conditions of equilibrium of bodies are studied is called statics. The easiest way is to consider the equilibrium conditions of an absolutely rigid body, that is, a body whose dimensions and shape can be considered unchanged. The concept of an absolutely rigid body is an abstraction, since all real bodies, under the influence of forces applied to them, are deformed to one degree or another, that is, they change their shape and size. The magnitude of the deformations depends both on the forces applied to the body and on the properties of the body itself - its shape and the properties of the material from which it is made. In many practically important cases, deformations are small and the use of concepts of an absolutely rigid body is justified.
Model of an absolutely rigid body. However, the smallness of deformations is not always a sufficient condition for a body to be considered absolutely solid. To illustrate this, consider the following example. A board lying on two supports (Fig. 140a) can be considered as an absolutely rigid body, despite the fact that it bends slightly under the influence of gravity. Indeed, in this case, the conditions of mechanical equilibrium make it possible to determine the reaction forces of the supports without taking into account the deformation of the board.
But if the same board rests on the same supports (Fig. 1406), then the idea of an absolutely rigid body is inapplicable. In fact, let the outer supports be on the same horizontal line, and the middle one slightly lower. If the board is absolutely solid, that is, it does not bend at all, then it does not put pressure on the middle support at all. If the board bends, then it puts pressure on the middle support, and the greater the deformation, the stronger it is. Terms
The equilibrium of an absolutely rigid body in this case does not allow us to determine the reaction forces of the supports, since they lead to two equations for three unknown quantities.
Rice. 140. Reaction forces acting on a board lying on two (a) and three (b) supports
Such systems are called statically indeterminate. To calculate them, it is necessary to take into account the elastic properties of bodies.
The above example shows that the applicability of the model of an absolutely rigid body in statics is determined not so much by the properties of the body itself, but by the conditions in which it is located. So, in the example considered, even a thin straw can be considered an absolutely solid body if it lies on two supports. But even a very rigid beam cannot be considered an absolutely rigid body if it rests on three supports.
Equilibrium conditions. The equilibrium conditions of an absolutely rigid body are a special case of dynamic equations when there is no acceleration, although historically statics arose from the needs of construction equipment almost two millennia before dynamics. In an inertial reference frame, a rigid body is in equilibrium if the vector sum of all external forces acting on the body and the vector sum of the moments of these forces are equal to zero. When the first condition is met, the acceleration of the body's center of mass is zero. When the second condition is met, there is no angular acceleration of rotation. Therefore, if at the initial moment the body was at rest, then it will remain at rest further.
In the future, we will limit ourselves to the study of relatively simple systems in which all the acting forces lie in the same plane. In this case, the vector condition
reduces to two scalars:
if we position the axes of the plane of action of forces. Some of the external forces acting on the body included in the equilibrium conditions (1) can be specified, that is, their modules and directions are known. As for the reaction forces of connections or supports that limit the possible movement of the body, they, as a rule, are not predetermined and are themselves subject to determination. In the absence of friction, the reaction forces are perpendicular to the contact surface of the bodies.
Rice. 141. To determine the direction of reaction forces
Reaction forces. Sometimes doubts arise in determining the direction of the bond reaction force, as, for example, in Fig. 141, which shows a rod resting at point A on the smooth concave surface of a cup and at point B on the sharp edge of the cup.
To determine the direction of the reaction forces in this case, you can mentally move the rod slightly without disturbing its contact with the cup. The reaction force will be directed perpendicular to the surface along which the contact point is sliding. So, at point A the reaction force acting on the rod is perpendicular to the surface of the cup, and at point B it is perpendicular to the rod.
Moment of power. Moment M of force relative to some point
O is the vector product of the radius vector drawn from O to the point of application of the force and the force vector
The vector M of the moment of force is perpendicular to the plane in which the vectors lie
Equation of moments. If several forces act on a body, then the second equilibrium condition associated with the moments of forces is written in the form
In this case, the point O from which the radius vectors are drawn must be chosen to be common to all acting forces.
For a plane system of forces, the moment vectors of all forces are directed perpendicular to the plane in which the forces lie, if the moments are considered relative to a point lying in the same plane. Therefore, the vector condition (4) for moments is reduced to one scalar one: in the equilibrium position, the algebraic sum of the moments of all external acting forces is equal to zero. The modulus of the moment of force relative to point O is equal to the product of the modulus
forces at a distance from point O to the line along which the force acts. In this case, the moments tending to rotate the body clockwise are taken with the same sign, counterclockwise - with the opposite sign. The choice of the point relative to which the moments of forces are considered is made solely for reasons of convenience: the equation of moments will be simpler, the more forces have moments equal to zero.
An example of balance. To illustrate the application of the equilibrium conditions of an absolutely rigid body, consider the following example. A lightweight stepladder consists of two identical parts, hinged at the top and tied with a rope at the base (Fig. 142). Let us determine what the tension force of the rope is, with what forces the halves of the ladder interact in the hinge and with what forces they press on the floor, if a person weighing R stands in the middle of one of them.
The system under consideration consists of two solid bodies - halves of the ladder, and equilibrium conditions can be applied both to the system as a whole and to its parts. Applying the equilibrium conditions to the entire system as a whole, one can find the floor reaction forces and (Fig. 142). In the absence of friction, these forces are directed vertically upward and the condition for the vector sum of external forces to be equal to zero (1) takes the form
The equilibrium condition for the moments of external forces relative to point A is written as follows:
where is the length of the stairs, the angle formed by the stairs with the floor. Solving the system of equations (5) and (6), we find
Rice. 142. The vector sum of external forces and the sum of the moments of external forces in equilibrium are equal to zero
Of course, instead of the equation of moments (6) about point A, one could write the equation of moments about point B (or any other point). This would result in a system of equations equivalent to the used system (5) and (6).
The tension force of the rope and the interaction forces in the hinge for the physical system under consideration are internal and therefore cannot be determined from the equilibrium conditions of the entire system as a whole. To determine these forces, it is necessary to consider the equilibrium conditions of individual parts of the system. At the same time
By successfully choosing the point with respect to which the equation of moments of forces is compiled, one can achieve simplification of the algebraic system of equations. So, for example, in this system we can consider the condition of equilibrium of the moments of forces acting on the left half of the staircase relative to point C, where the hinge is located.
With this choice of point C, the forces acting in the hinge will not be included in this condition, and we immediately find the tension force of the rope T:
where, given that we get
Condition (7) means that the resultant force T passes through point C, i.e., it is directed along the stairs. Therefore, equilibrium of this half of the ladder is possible only if the force acting on it at the hinge is also directed along the ladder (Fig. 143), and its modulus is equal to the modulus of the resultant forces T and
Rice. 143. The lines of action of all three forces acting on the left half of the staircase pass through one point
The absolute value of the force acting in the hinge on the other half of the ladder, based on Newton’s third law, is equal and its direction is opposite to the direction of the vector. The direction of the force could be determined directly from Fig. 143, taking into account that when a body is in equilibrium under the action of three forces, the lines along which these forces act intersect at one point. Indeed, let us consider the point of intersection of the lines of action of two of these three forces and construct an equation of moments about this point. The moments of the first two forces about this point are equal to zero; This means that the moment of the third force must also be equal to zero, which, in accordance with (3), is possible only if the line of its action also passes through this point.
The golden rule of mechanics. Sometimes the problem of statics can be solved without considering equilibrium conditions at all, but using the law of conservation of energy in relation to mechanisms without friction: no mechanism gives a gain in work. This law
called the golden rule of mechanics. To illustrate this approach, consider the following example: a heavy load of weight P is suspended on a weightless hinge with three links (Fig. 144). What tension force must the thread connecting points A and B withstand?
Rice. 144. To determine the tension force of a thread in a three-link hinge supporting a load of weight P
Let's try using this mechanism to lift the load P. Having untied the thread at point A, pull it up so that point B slowly rises to a distance. This distance is limited by the fact that the tension force of the thread T must remain unchanged during the movement. In this case, as will be clear from the answer, the force T does not depend at all on how much the hinge is compressed or stretched. The work done. As a result, the load P rises to a height which, as is clear from geometric considerations, is equal to Since in the absence of friction no energy losses occur, it can be argued that the change in the potential energy of the load is determined by the work done during lifting. That's why
Obviously, for a hinge containing an arbitrary number of identical links,
It is not difficult to find the tension force of the thread, and in the case when it is necessary to take into account the weight of the hinge itself, the work done during lifting should be equated to the sum of changes in the potential energies of the load and the hinge. For a hinge of identical links, its center of mass rises by Therefore
The formulated principle (“the golden rule of mechanics”) is also applicable when, during the process of movement, there is no change in potential energy, and the mechanism is used to convert force. Gearboxes, transmissions, gates, systems of levers and blocks - in all such systems, the converted force can be determined by equating the work of the converted and applied forces. In other words, in the absence of friction, the ratio of these forces is determined only by the geometry of the device.
Let us consider from this point of view the example with a stepladder discussed above. Of course, using a stepladder as a lifting mechanism, that is, lifting a person by bringing the halves of the stepladder closer together, is hardly advisable. However, this cannot prevent us from applying the described method to find the tension force of the rope. Equating the work done when the parts of the ladder come together with the change in the potential energy of the person on the ladder and, from geometric considerations, connecting the movement of the lower end of the ladder with a change in the height of the load (Fig. 145), we obtain, as one would expect, the previously given result:
As already noted, the movement should be chosen such that during the process the acting force can be considered constant. It is easy to see that in the example with a hinge this condition does not impose restrictions on movement, since the tension force of the thread does not depend on the angle (Fig. 144). On the contrary, in the stepladder problem the displacement should be chosen to be small, because the tension force of the rope depends on the angle a.
Stability of balance. Equilibrium can be stable, unstable and indifferent. Equilibrium is stable (Fig. 146a) if, with small movements of the body from the equilibrium position, the acting forces tend to return it back, and unstable (Fig. 1466) if the forces take it further from the equilibrium position.
Rice. 145. Movements of the lower ends of the ladder and movement of the load when the halves of the ladder come together
Rice. 146. Stable (a), unstable (b) and indifferent (c) equilibria
If, at small displacements, the forces acting on the body and their moments are still balanced, then the equilibrium is indifferent (Fig. 146c). In indifferent equilibrium, neighboring positions of the body are also equilibrium.
Let us consider examples of studying the stability of equilibrium.
1. Stable equilibrium corresponds to the minimum potential energy of the body in relation to its values in neighboring positions of the body. This property is often convenient to use when finding the equilibrium position and when studying the nature of equilibrium.
Rice. 147. Stability of body balance and position of the center of mass
A vertical free-standing column is in stable equilibrium, since at small inclinations its center of mass rises. This happens until the vertical projection of the center of mass goes beyond the support area, i.e., the angle of deviation from the vertical does not exceed a certain maximum value. In other words, the stability region extends from the minimum potential energy (in a vertical position) to the maximum closest to it (Fig. 147). When the center of mass is located exactly above the boundary of the support area, the column is also in equilibrium, but unstable. A horizontally lying column corresponds to a much wider range of stability.
2. There are two round pencils with radii and One of them is located horizontally, the other is balanced on it in a horizontal position so that the axes of the pencils are mutually perpendicular (Fig. 148a). At what ratio between the radii is equilibrium stable? At what maximum angle can the upper pencil be tilted from the horizontal? The coefficient of friction of pencils against each other is equal to
At first glance, it may seem that the balance of the upper pencil is generally unstable, since the center of mass of the upper pencil lies above the axis around which it can rotate. However, here the position of the rotation axis does not remain unchanged, so this case requires special study. Since the top pencil is balanced in a horizontal position, the centers of mass of the pencils lie on this vertical (Fig.).
Let's tilt the top pencil at a certain angle from the horizontal. In the absence of static friction, it would immediately slide down. In order not to think about possible slippage for now, we will assume that the friction is quite large. In this case, the upper pencil “rolls” over the lower one without slipping. The fulcrum from position A moves to a new position C, and the point at which the upper pencil rested on the lower one before the deviation
goes to position B. Since there is no slipping, the length of the arc is equal to the length of the segment
Rice. 148. The upper pencil is balanced horizontally on the lower pencil (a); to the study of equilibrium stability (b)
The center of mass of the upper pencil moves to position . If the vertical line drawn through passes to the left of the new fulcrum C, then gravity tends to return the upper pencil to its equilibrium position.
Let us express this condition mathematically. Drawing a vertical line through point B, we see that the condition must be met
Since from condition (8) we obtain
Since the force of gravity will tend to return the upper pencil to the equilibrium position only at Therefore, stable equilibrium of the upper pencil on the lower one is possible only when its radius is less than the radius of the lower pencil.
The role of friction. To answer the second question, you need to find out what reasons limit the permissible angle of deviation. Firstly, at large angles of deflection, the vertical drawn through the center of mass of the upper pencil can pass to the right of the fulcrum point C. From condition (9) it is clear that for a given ratio of the radii of the pencils the maximum angle of deflection
Are the equilibrium conditions of a rigid body always sufficient to determine reaction forces?
How can one practically determine the direction of reaction forces in the absence of friction?
How can you use the golden rule of mechanics when analyzing equilibrium conditions?
If in the hinge shown in Fig. 144, connect not points A and B with a thread, but points A and C, then what will its tension force be?
How is the stability of the equilibrium of a system related to its potential energy?
What conditions determine the maximum angle of deflection of a body resting on a plane at three points so that its stability is not lost?
