The system of forces called. balanced, if under the action of this system the body remains at rest.
Equilibrium conditions:
The first condition for the equilibrium of a rigid body:
For the equilibrium of a rigid body, it is necessary that the sum of external forces applied to the body be equal to zero.
The 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 about any axis is equal to zero.
General equilibrium condition for a rigid body:
For the equilibrium of a rigid body, the sum of external forces and the sum of the moments of forces acting on the body must be equal to zero. The initial velocity of the center of mass and the angular velocity of rotation of the body must also be equal to zero.
Theorem. Three forces balance a rigid body only if they all lie in the same plane.
11. Flat system of forces are forces in the same plane.
Three forms of equilibrium equations for a flat system:
The center of gravity of the body.
center of gravity bodies of finite size is called the point, relative to which the sum of the moments of gravity of all particles of the body is equal to zero. At this point, the force of gravity of the body is applied. The center of gravity of a body (or system of forces) usually coincides with the center of mass of the body (or system of forces).
Center of gravity of a flat figure:
A practical way to find the center of mass of a plane figure: hang the body in the field of gravity so that it can freely rotate around the point of suspension O1 . Center of mass in balance With is on the same vertical with the suspension point (below it), since it is equal to zero
the moment of gravity, which can be considered applied at the center of mass. By changing the suspension point, in the same way we find another straight line About 2 C , passing through the center of mass. The position of the center of mass is given by the point of their intersection.
Center of mass speed:
The momentum of a system of particles is equal to the product of the mass of the entire system M= Σmi to the speed of its center of mass V :
The center of mass characterizes the movement of the system as a whole.
15. Sliding friction- friction during the relative motion of the contacting bodies.
Friction of rest– friction in the absence of relative movement of the contacting bodies.
sliding friction force Ftr between the surfaces of contacting bodies during their relative motion depends on the force of the normal reaction N , or from the force of normal pressure PN , and Ftr=kN or Ftr=kPn , where k – coefficient of sliding friction , which depends on the same factors as the static friction coefficient k0 , as well as on the speed of the relative motion of the contacting bodies.
16. Rolling friction is the rolling of one body over another. The sliding friction force does not depend on the size of the rubbing surfaces, but only on the quality of the surfaces of the rubbing bodies and on the force that reduces the rubbing surfaces and is directed perpendicular to them. F=kN, where F- friction force, N is the value of the normal reaction and k is sliding friction coefficient.
17. Equilibrium of bodies in the presence of friction- this is the maximum adhesion force proportional to the normal pressure of the body on the plane.
The angle between the total reaction, built on the greatest frictional force for a given normal reaction, and the direction of the normal reaction is called friction angle.
A cone with a vertex at the point of application of the normal reaction of a rough surface, the generatrix of which makes a friction angle with this normal reaction, is called friction cone.
Dynamics.
1. AT dynamics the influence of interactions between bodies on their mechanical motion is considered.
Weight- this is the painting characteristic of a material point. The mass is constant. Weight is additive (adds up)
Force - this is a vector that completely characterizes the interaction of a material point on it with other material points.
Material point- a body, the dimensions and shape of which are insignificant in the considered movement. (ex: in translational motion, a rigid body can be considered a material point)
system of material points called set of material points interacting with each other.
1 Newton's law: any material point maintains a state of rest or uniform rectilinear motion until external influences change this state.
Newton's 2nd law: acceleration acquired by a material point in an inertial frame of reference, which is directly proportional to the force acting on the point, inversely proportional to the mass of the point and coincides in direction with the force: a=F/m
Definition
The equilibrium of the body is called such a state when any acceleration of the body is equal to zero, that is, all actions on the body of forces and moments of forces are balanced. In this case, the body can:
- be in a state of calm;
- move evenly and in a straight line;
- rotate uniformly around an axis that passes through its center of gravity.
Body equilibrium conditions
If the body is in equilibrium, then two conditions are simultaneously satisfied.
- The vector sum of all forces acting on the body is equal to the zero vector : $\sum_n((\overrightarrow(F))_n)=\overrightarrow(0)$
- The algebraic sum of all moments of forces acting on the body is equal to zero: $\sum_n(M_n)=0$
The two equilibrium conditions are necessary but not sufficient. Let's take an example. Consider a wheel rolling uniformly without slipping on a horizontal surface. Both equilibrium conditions are met, but the body is moving.
Consider the case when the body does not rotate. In order for the body not to rotate and be in balance, it is necessary that the sum of the projections of all forces on an arbitrary axis be equal to zero, that is, the resultant of the forces. Then the body is either at rest, or moves uniformly and rectilinearly.
A body that has an axis of rotation will be in equilibrium if the rule of moments of forces is followed: the sum of the moments of forces that rotate the body clockwise must be equal to the sum of the moments of forces that rotate it counterclockwise.
To get the right moment with the least effort, you need to apply force as far as possible from the axis of rotation, increasing the same arm of the force and, accordingly, reducing the value of the force. Examples of bodies that have an axis of rotation are: a lever, doors, blocks, a brace, and the like.
Three types of balance of bodies that have a fulcrum
- stable equilibrium, if the body, being removed from the equilibrium position to the neighboring nearest position and left in peace, returns to this position;
- unstable equilibrium, if the body, being removed from the equilibrium position to a neighboring position and left at rest, will deviate even more from this position;
- indifferent equilibrium - if the body, being brought to a neighboring position and left in peace, remains in its new position.
Balance of a body with a fixed axis of rotation
- stable, if in the equilibrium position the center of gravity C occupies the lowest position of all possible near positions, and its potential energy will have the smallest value of all possible values in neighboring positions;
- unstable if the center of gravity C occupies the highest of all nearby positions, and the potential energy has the greatest value;
- indifferent if the center of gravity of the body C in all nearby possible positions is at the same level, and the potential energy does not change during the transition of the body.
Task 1
A body A with mass m = 8 kg is placed on a rough horizontal table surface. A thread is tied to the body, thrown over block B (Figure 1, a). What weight F can be tied to the end of the thread hanging from the block so as not to disturb the balance of the body A? Friction coefficient f = 0.4; ignore the friction on the block.
Let's define body weight ~A: ~G = mg = 8$\cdot $9.81 = 78.5 N.
We assume that all forces are applied to body A. When the body is placed on a horizontal surface, only two forces act on it: the weight G and the oppositely directed reaction of the support RA (Fig. 1, b).
If we apply some force F acting along a horizontal surface, then the reaction RA, which balances the forces G and F, will begin to deviate from the vertical, but the body A will be in equilibrium until the modulus of the force F exceeds the maximum value of the friction force Rf max , corresponding to the limit value of the angle $(\mathbf \varphi )$o (Fig. 1, c).
Having decomposed the reaction RA into two components Rf max and Rn, we obtain a system of four forces applied to one point (Fig. 1, d). Projecting this system of forces onto the x and y axes, we obtain two equilibrium equations:
$(\mathbf \Sigma )Fkx = 0, F - Rf max = 0$;
$(\mathbf \Sigma )Fky = 0, Rn - G = 0$.
We solve the resulting system of equations: F = Rf max, but Rf max = f$\cdot $ Rn, and Rn = G, so F = f$\cdot $ G = 0.4$\cdot $ 78.5 = 31.4 H; m \u003d F / g \u003d 31.4 / 9.81 \u003d 3.2 kg.
Answer: Mass of cargo m = 3.2 kg
Task 2
The system of bodies shown in Fig. 2 is in a state of equilibrium. Weight of cargo tg=6 kg. Angle between vectors $\widehat((\overrightarrow(F))_1(\overrightarrow(F))_2)=60()^\circ $. $\left|(\overrightarrow(F))_1\right|=\left|(\overrightarrow(F))_2\right|=F$. Find the mass of weights.
The resultant force $(\overrightarrow(F))_1and\ (\overrightarrow(F))_2$ is equal in absolute value to the weight of the load and opposite to it in direction: $\overrightarrow(R)=(\overrightarrow(F))_1+(\overrightarrow (F))_2=\ -m\overrightarrow(g)$. By the law of cosines, $(\left|\overrightarrow(R)\right|)^2=(\left|(\overrightarrow(F))_1\right|)^2+(\left|(\overrightarrow(F) )_2\right|)^2+2\left|(\overrightarrow(F))_1\right|\left|(\overrightarrow(F))_2\right|(cos \widehat((\overrightarrow(F)) _1(\overrightarrow(F))_2)\ )$.
Hence $(\left(mg\right))^2=$; $F=\frac(mg)(\sqrt(2\left(1+(cos 60()^\circ \ )\right)))$;
Since the blocks are movable, $m_g=\frac(2F)(g)=\frac(2m)(\sqrt(2\left(1+\frac(1)(2)\right)))=\frac(2 \cdot 6)(\sqrt(3))=6.93\ kg\ $
Answer: The mass of each weight is 6.93 kg.
A body is at rest (or moves uniformly and in a straight line) if the vector sum of all forces acting on it is zero. The forces are said to 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.
The condition for the equilibrium of bodies
In order 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 action of three forces acting on it. The lines of action of the forces F 1 → and F 2 → intersect at the point O. The point of application of gravity is the center of mass of the body C. These points lie on one straight line, and when calculating the resultant force F 1 → , F 2 → and m g → are reduced to point C .
The condition that the resultant of all forces be equal to zero is not enough if the body can rotate around some axis.
The shoulder of the 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 arm of the force and its modulus.
The moment of force tends to rotate the body around its axis. Those moments that rotate the body counterclockwise are considered positive. The unit of measurement of the moment of force in the international SI system is 1 Newton meter.
Definition. moment rule
If the algebraic sum of all the moments applied to the body relative to the fixed axis of rotation is equal to zero, then the body is in equilibrium.
M1 + M2 + . . + M n = 0
Important!
In the general case, for the equilibrium of bodies, two conditions must be met: the resultant force is equal to zero and the rule of moments is observed.
There are different types of equilibrium in mechanics. Thus, a distinction is made between stable and unstable, as well as indifferent equilibrium.
A typical example of an 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 - a state of equilibrium, with a small deviation from which the forces and moments of forces tend to bring 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, there is an indifferent equilibrium. In stable and unstable equilibrium, the center of mass is located on a vertical line that passes through the axis of rotation. When the center of mass is below the axis of rotation, the equilibrium is stable. Otherwise, vice versa.
A special case of equilibrium is the equilibrium of a body on a support. In this case, the elastic force is distributed over the entire base of the body, and does not pass 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 overturns.
An example of the balance of a body 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 the study of 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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Equilibrium conditions for a material point and a rigid body.
All forces acting on a material point are applied at one point. The resulting force is defined as the geometric sum of all forces acting on a material point. If the resulting force is zero, then according to Newton's 2nd law, the acceleration of a material point is zero, the speed is constant or equal to zero, the material point is in equilibrium.
The equilibrium condition of a material point: . (6.1)
A much more important issue in statics is the question of the equilibrium of an extended body, since in practice it is precisely such bodies that have to be dealt with. It is clear that for the balance of the body it is extremely important that the resulting force acting on the body is zero. But fulfilling this condition is not enough. Consider a horizontal rod that can rotate about a horizontal axis O(Fig. 6.2). The following acts on the rod: gravity, the reaction force of the axis, two external forces and, equal in magnitude and opposite in direction. The resultant of these forces is zero:
however, our practical experience tells us that the rod will begin to rotate, ᴛ.ᴇ. will not be in equilibrium. Please note that the moments of forces and relative to the axis O are zero, the moments of forces and are not equal to zero and both are positive, the forces try to rotate the rod clockwise about the axis O.
In Fig. 6.3, the forces and are equal in magnitude and are directed in the same way. The resultant of all forces acting on the rod is equal to zero (in this case, the force is greater than in the first case, it balances the resultant of three forces -, and). The resulting moment of all forces is zero, the rod is in equilibrium. We come to the conclusion that two conditions are extremely important for the equilibrium of the body.
Equilibrium conditions for an extended body:
Let us write down important rules that can be used when considering the equilibrium conditions of a body.
1. The vectors of forces applied to the body can be moved along the line of their action. The resulting force and the resulting moment do not change.
2. The second equilibrium condition is satisfied with respect to any axis of rotation. It is convenient to choose such an axis of rotation with respect to which equation (6.3) will be the simplest. For example, about the axis O in fig. 6.2 the moments of forces and are equal to zero.
sustainable balance. In stable equilibrium, the potential energy of the body is minimal. When a body is displaced from a position of stable equilibrium, the potential energy increases, and a resulting force arises, directed towards the equilibrium position.
Unstable equilibrium. When a body is displaced from a position of unstable equilibrium, the potential energy decreases, and a resulting force arises, directed away from the equilibrium position.
Center of gravity of the body- the point of application of the resulting all gravity forces acting on individual elements of the body.
A sign of balance. The body maintains balance if the vertical line passing through the center of gravity intersects the body's support area.
Topic 7. (4 hours)
MOLECULAR PHYSICS. Atomistic hypothesis of the structure of matter and its experimental evidence. Gas pressure. Absolute temperature as a measure of the average kinetic energy of the thermal motion of particles of matter. The equation of state for an ideal gas. Isoprocesses of an ideal gas. Structure and properties of liquids and solids. Water vapor.
Hosted on ref.rf
Air humidity.
Equilibrium conditions for a material point and a rigid body. - concept and types. Classification and features of the category "Equilibrium conditions of a material point and a rigid body." 2017, 2018.
The static calculation of engineering structures in many cases is reduced to the consideration of the equilibrium conditions for a structure from a system of bodies connected by some kind of connections. The connections connecting the parts of this construction will be called internal Unlike external connections that fasten the structure with bodies that are not included in it (for example, with supports).
If, after discarding the external bonds (supports), the structure remains rigid, then the problems of statics are solved for it as for an absolutely rigid body. However, there may be such engineering structures that, after discarding the external links, do not remain rigid. An example of such a design is a three-hinged arch. If supports A and B are discarded, then the arch will not be rigid: its parts can rotate around the hinge C.
Based on the principle of solidification, the system of forces acting on such a structure must, at equilibrium, satisfy the equilibrium conditions of a solid body. But these conditions, as has been pointed out, while necessary, will not be sufficient; therefore, it is impossible to determine all unknown quantities from them. To solve the problem, it is necessary to additionally consider the balance of one or more parts of the structure.
For example, compiling the equilibrium conditions for the forces acting on a three-hinged arch, we get three equations with four unknowns X A, Y A, X B, Y B . Having additionally considered the equilibrium conditions for the left (or right) half of it, we obtain three more equations containing two new unknowns X C, Y C, in fig. 61 not shown. Solving the resulting system of six equations, we find all six unknowns.
14. Particular cases of reducing the spatial system of forces
If, when the system of forces is reduced to a dynamic screw, the main moment of the dynamo turned out to be equal to zero, and the main vector is different from zero, then this means that the system of forces is reduced to the resultant, and the central axis is the line of action of this resultant. Let us find out under what conditions, relating to the main vector Fp and the main moment M 0 , this can be. Since the main moment of the dynamo M * is equal to the component of the main moment M 0 directed along the main vector, then the case under consideration M * \u003d O means that the main moment M 0 is perpendicular to the main vector, i.e. / 2 \u003d Fo * M 0 \u003d 0. This directly implies that if the main vector F 0 is not equal to zero, and the second invariant is equal to zero, Fo≠O, / 2 = F 0 *M 0 =0, (7.9) then the considered the system is reduced to a resultant.
In particular, if for any center of reduction F 0 ≠0, and M 0 = 0, then this means that the system of forces is reduced to the resultant, passing through this center of reduction; in this case, condition (7.9) will also be satisfied. Let us generalize the theorem on the moment of the resultant (Varignon's theorem) presented in Chapter V to the case of a spatial system of forces. If the spatial system.
forces is reduced to the resultant, then the moment of the resultant with respect to an arbitrary point is equal to the geometric sum of the moments of all forces with respect to the same point. P 
let the system of forces have a resultant R and a point O lies on the line of action of this resultant. If we bring the given system of forces to this point, then we get that the main moment is equal to zero. 
Let us take some other reference center O1; (7.10)C 
on the other hand, based on formula (4.14) we have Mo1=Mo+Mo1(Fo), (7.11) because М 0 = 0. Comparing expressions (7.10) and (7.11) and taking into account that in this case F 0 = R, we get (7.12).
Thus, the theorem is proved.
Let at any choice of the center of reduction Fo=O, M ≠0. Since the main vector does not depend on the center of reduction, it is equal to zero for any other choice of the center of reduction. Therefore, the main moment also does not change with a change in the center of reduction, and, therefore, in this case, the system of forces is reduced to a pair of forces with a moment equal to M0.
Let us now make a table of all possible cases of reduction of the spatial system of forces:
If all forces are in the same plane, for example, in the plane Ohu then their projections on the axis G and moments about the axes X and at will be equal to zero. Therefore, Fz=0; Mox=0, Moy=0. Introducing these values into formula (7.5), we find that the second invariant of the plane system of forces is equal to zero. We will obtain the same result for the spatial system of parallel forces. Indeed, let all forces be parallel to the axis z. Then their projections on the axes X and at and the moments about the z-axis will be equal to 0. Fx=0, Fy=0, Moz=0
On the basis of what has been proved, it can be argued that a flat system of forces and a system of parallel forces are not reduced to a dynamic screw.
1
1.
Body equilibrium in the presence of sliding friction If two bodies / and // (Fig. 6.1) interact with each other, touching at a point BUT, then always the reaction R A, acting, for example, from the side of the body // and applied to the body /, can be decomposed into two components: N.4, directed along the common normal to the surface of the contacting bodies at point L, and T 4, lying in the tangent plane . Component N.4 is called normal response, force T l is called sliding friction force - it prevents the "sliding of the body / over the body //. In accordance with the axiom 4
(3 Newton's law) on the body // from the side of the body / there is an equal in magnitude and oppositely directed reaction force. Its component perpendicular to the tangent plane is called force of normal pressure. As mentioned above, the force of friction T BUT
=
Oh, if the mating surfaces are perfectly smooth. Under real conditions, the surfaces are rough and in many cases the friction force cannot be neglected. 6.2, a. To the body 5, located on a fixed plate D, is attached a thread thrown over the block C, the free end of which is provided with a support platform BUT. If pad BUT gradually load, then with an increase in its total weight, the tension of the thread will increase S,
which tends to move the body to the right. However, as long as the total load is not too large, the friction force T will hold the body AT at rest. On fig. 6.2, b depicted acting on the body AT forces, and P is the force of gravity, and N is the normal reaction of the plate D.
If the load is insufficient to break the rest, the following equilibrium equations are valid: N-
P
= 0,
(6.1) S-T = 0. (6.2). It follows from here that N
=
Pand T = S. Thus, while the body is at rest, the friction force remains equal to the tension force of the thread S. Denote by Tmax
friction force at the critical moment of the loading process, when the body AT loses balance and starts to slide on the slab D.
Therefore, if the body is in equilibrium, then T≤Tmax.Maximum friction force T max
depends on the properties of the materials from which the bodies are made, their condition (for example, on the nature of the surface treatment), as well as on the magnitude of the normal pressure N. As experience shows, the maximum friction force is approximately proportional to the normal pressure, i.e. e. there is an equality Tmax=
fN.
(6.4). This relation is called Amonton-Coulomb law. The dimensionless coefficient / is called coefficient of sliding friction. As follows from experience, it the value in a wide range does not depend on the area of the contacting surfaces, but depends on the material and the degree of roughness of the contacting surfaces. The values of friction coefficients are established empirically and can be found in reference tables. Inequality" (6.3) can now be written as T≤fN
(6.5). The case of strict equality in (6.5) corresponds to the maximum value of the friction force. This means that the friction force can be calculated by the formula T
=
fN
only in those cases where it is known in advance that there is a critical case. In all other cases, the friction force should be determined from the equilibrium equations. Consider a body located on a rough surface. We will assume that as a result of the action of active forces and reaction forces, the body is in limiting equilibrium. On fig. 6.6, a
the limit reaction R and its components N and T max are shown (in the position shown in this figure, the active forces tend to move the body to the right, the maximum friction force T max is directed to the left). Injection f between limit reaction R and the normal to the surface is called the angle of friction. Let's find this corner. From fig. 6.6, but we have tgφ \u003d Tmax / N or, using the expression (6.4), tgφ \u003d f (6-7) 
both quantities are given).