Addition of force pairs in space. Reducing a system of force pairs to its simplest form or adding force pairs Adding force pairs is a condition for equilibrium of force pairs

Theorem: a system of pairs of forces acting on an absolutely rigid body in one plane is equivalent to a pair of forces with a moment equal to the algebraic sum of the moments of the pairs of the system.

A resultant pair is a pair of forces that replaces the action of these pairs of forces applied to a solid body in one plane.

Condition for the equilibrium of a system of pairs of forces: for the equilibrium of a plane system of pairs of forces, it is necessary and sufficient that the sum of their moments be equal to 0.

Moment of force about a point.

The moment of a force relative to a point is the product of the modulus of force and its shoulder relative to a given point, taken with a plus or minus sign. The arm of a force relative to a point is the length of the perpendicular drawn from a given point to the line of action of the force. The following sign rule is accepted: the moment of a force about a given point is positive if the force tends to rotate the body around this point counterclockwise, and negative in the opposite case. If the line of action of a force passes through a certain point, then relative to this point the leverage of the force and its moment are equal to zero. The moment of force relative to a point is determined by the formula.

Properties of the moment of force relative to a point:

1. The moment of force relative to a given point does not change when the force is transferred along its line of action, because in this case, neither the force modulus nor its leverage changes.

2. The moment of force relative to a given point is equal to zero if the line of action of the force passes through this point, because in this case the force arm is zero: a=0

Poinsot's theorem on bringing a force to a point.

A force can be transferred parallel to the line of its action; in this case, it is necessary to add a pair of forces with a moment equal to the product of the modulus of the force and the distance over which the force is transferred.

The operation of parallel transfer of force is called bringing the force to a point, and the resulting pair is called an attached pair.

The opposite effect is also possible: a force and a pair of forces lying in the same plane can always be replaced by one force equal to a given force transferred parallel to its initial direction to some other point.

Given: force at a point A(Fig. 5.1).

Add at point IN balanced system of forces (F"; F"). A couple of forces is formed (F; F"). Let's get the force at the point IN and the moment of the pair m.

Bringing a plane system of arbitrarily located forces to one center. The main vector and the main moment of the system of forces.

The lines of action of an arbitrary system of forces do not intersect at one point, therefore, to assess the state of the body, such a system should be simplified. To do this, all the forces of the system are transferred to one arbitrarily selected point - the point of reduction (PO). Apply Poinsot's theorem. Whenever a force is transferred to a point not lying on the line of its action, a couple of forces are added.

The pairs that appear during transfer are called attached pairs.

The SSS obtained at the point O is folded according to the force polygon method and we obtain one force at the point O - this is the main vector.

The resulting system of attached pairs of forces can also be added and one pair of forces is obtained, the moment of which is called the main moment.

The main vector is equal to the geometric sum of the forces. The main moment is equal to the algebraic sum of the moments of the attached pairs of forces or the moments of the original forces relative to the reduction point.

Definition and properties of the main vector and main moment of a plane system of forces.

Properties of the main vector and main moment

1 The module and direction of the main vector do not depend on the choice of the reduction center, because at the center of reduction, the force polygon constructed from these forces will be the same)

2. The magnitude and sign of the main moment depend on the choice of the reduction center, because when the center of adduction changes, the shoulders of the forces change, but their modules remain unchanged.

3. The main vector and the resultant of the force system are vectorially equal, but in the general case they are not equivalent, because there is still a moment

4. The main vector and the resultant are equivalent only in the special case when the main moment of the system is equal to zero, and this is in the case when the center of reduction is on the line of action of the resultant

Consider a flat system of forces ( F 1 ,F 2 , ...,F n), acting on a solid body in the Oxy coordinate plane.

The main vector of the force system called a vector R, equal to the vector sum of these forces:

R = F 1 + F 2 + ... + F n= F i.

For a plane system of forces, its main vector lies in the plane of action of these forces.

The main point of the system of forces relative to the center O is called a vector L O, equal to the sum of the vector moments of these forces relative to point O:

L O= M O( F 1) +M O( F 2) + ... +M O( F n) = M O( F i).

Vector R does not depend on the choice of center O, and the vector L When the position of the center changes, O can generally change.

For a plane system of forces, instead of a vector main moment, the concept of an algebraic main moment is used. Algebraic main point L O of a plane system of forces relative to the center O lying in the plane of action of the forces is called the sum of algebraic moments uh quiet forces relative to the center O.

The main vector and the main moment of a plane system of forces are usually calculated by analytical methods.

Axiom on the condition of equivalence of pairs of forces in space. Instead of the moment vector of each pair of forces perpendicular to the plane of the drawing, only the direction in which the pair of forces tends to rotate this plane is indicated.

Pairs of forces in space are equivalent if their moments are geometrically equal. Without changing the action of a pair of forces on a rigid body, a pair of forces can be transferred to any plane parallel to the plane of action of the pair, and also change its forces and leverage, keeping the modulus and direction of its moment constant. Thus, the moment vector of a pair of forces can be transferred to any point, i.e., the moment of a pair of forces is a free vector. The moment vector of a pair of forces describes all three of its elements: the position of the plane of action of the pair, the direction of rotation and the numerical value of the moment. Let's look at the addition of two pairs of forces located in intersecting planes and prove the following axiom: the geometric sum of the moments of the constituent pairs of forces is equal to the moment of the pair equivalent to them. Let it be required to add two pairs of forces located in intersecting planes I and II having moments

Rice. 34 Having chosen the forces of these pairs to be equal in magnitude

Let's define the shoulders of these pairs:

Let us arrange these pairs of forces in such a way that the forces are oriented along the strip of intersection of the KL planes in opposite directions and are balanced. The remaining forces form a force pair equivalent to the given two force pairs. This pair of forces has a shoulder BC = d and a moment perpendicular to the plane of action of the pair of forces, equal in magnitude to M = Pd.

The geometric sum of the moments of the constituent force pairs is equal to the moment of the equivalent pair. Because the moment of a pair of forces is a free vector, let us transfer the moments of the constituent pairs of forces to point B and add them, constructing a parallelogram on these moments. The diagonal of this parallelogram

represents the moment of an equivalent pair. It follows that the vector, i.e., the geometric sum of the moments of the constituent pairs of forces is equal to the moment of the equivalent pair of forces:

This method of adding the moments of pairs of forces is called the moment parallelogram rule. The construction of a parallelogram of moments can be replaced by the construction of a triangle of moments.

Using the construction of a parallelogram or triangle of moments, you can also solve the inverse problem, i.e., decompose any pair of forces into two components. Let it be necessary to add several pairs of forces located arbitrarily in space (Fig. 35). Having determined the moments of these pairs, they can be transferred to any point O of the place. By adding the moments of these pairs of forces one by one, it is possible to construct a polygon of the moments of the pairs, the closing side of which will determine the moment of the equivalent pair of forces. (Fig. 35) shows the construction of a moment polygon when adding 3 pairs.

The moment of a pair of forces, forces equivalent to a given system of pairs of forces in space, is equal to the geometric sum of the moments of the constituent pairs of forces:
or

Plane I of the action of a given pair of forces is perpendicular to the direction of its moment

If the moment of an equivalent pair of forces is zero, then the pairs of forces are mutually balanced:

Thus, the equilibrium condition for pairs of forces arbitrarily located in space can be constructed as follows: pairs of forces arbitrarily located in space are mutually balanced in this case if the geometric sum of their moments is zero. If pairs of forces are placed in the same plane (Fig. 36), then the moments of these pairs of forces, directed along one straight line, add up algebraically.

A system of force pairs acting on a body is equivalent to one force pair, the moment of which is equal to the algebraic sum of the moments of the component pairs.

Let three pairs of forces (P1, P1 ′), (P2, P2 ′), (P3, P3 ′) act on a solid body (Fig. 5.9), located in the same plane. Moments of these couples:

M 1 = P 1. d 1, M 2 = P 2. d 2, M 3 = - P 3. d 3

Let us choose an arbitrary segment AB of length d in the same plane and replace the given pairs with equivalent ones (Q1, Q1 ′), (Q2, Q2 ′), (Q3, Q3 ′) with a common arm d.

Let us find the moduli of forces of equivalent pairs from the relations

M1 = P1. d1 = Q1 . d, M2 = P2. d2 = Q2 . d, M3 = - P3. d3 = - Q3 . d.

Let's add up the forces applied to the ends of the segment AB and find the modulus of their resultant:

R = Q1 + Q2 - Q3

R′ = - R = (-Q′ 1 - Q′ 2 + Q′ 3 )

The resultants R and R′ form a resulting pair equivalent to the system of given pairs.

This couple's moment:

M = R. d = (Q1 + Q2 - Q3) d = Q1. d + Q2 . d - Q3 . d = M1 + M2 + M3

If “n” pairs act on a body, then the moment of the resulting pair is equal to the algebraic sum of the moments of the constituent pairs:

M = ∑ Mi

A pair is called balancing, the moment of which is equal in absolute value to the moment of the resulting pair, but opposite in direction.

Example 5.1

Determine the moment of the resulting pair for three given pairs (Fig. 5.

10, a), if P1 = 10 kN, P2 = 15 kN, P3 = 20 kN, d1 = 4 m, d2 = 2 m, d3 = 6 m.

We determine the moment of each pair of forces:

M1 = 10 N. 4 m = 40 Nm M2 = - 15 N. 2 m = - 30 Nm M3 = - 20 N. 6 m = - 120 Nm

М = ∑ Мi = М1 + М2 + М3 = 40 – 30 – 120 = - 110 Nm

Example 5.2

The frame (Fig. 5. 10, b) is affected by three pairs of forces (P1, P1 ′), (P2, P2 ′), (P3, P3 ′), applied at points A1, A2, A3, respectively. Define the moment

resultant pair, if P1 = 10 N, P2 = 15 N, P3 = 20 N, and the arms of the force pairs d1 =

0.4 m, d2 = 0.2 m, d3 = 0.6 m.

We determine the moments of force pairs:

M1 = P1. d1 = 10 . 0.4 = 4 Nm M2 = - P2. d2 = - 15 . 0.2 = - 3 Nm M3 = - P3. d3 = - 20 . 0.6 = - 12 Nm

We determine the moment of the resulting pair:

М = ∑ Мi = М1 + М2 + М3 = 4 – 3 – 120 = - 11 Nm

Example 5.3

The beam (Fig. 5. 10, c) is affected by three pairs of forces (P1, P1 ′), (P2, P2 ′), (P3, P3 ′), applied at points A1, A2, A3. Determine the moment of the resulting pair,

if P1 = 2 kN, P2 = 3 kN, P3 = 6 kN, and the arms of the force pairs d1 = 0.2 m, d2 = 0.4 m, d3 = 0.3 m.

We determine the moments of force pairs:

M1 = - P1. d1 = - 2 . 0.2 = - 0.4 kNm M2 = - P2. d2 = - 3 . 0.4 = - 1.2 kNm M3 = P3. d3 = 6 . 0.3 = 1.8 kNm

We determine the moment of the resulting pair:

М = ∑ Мi = М1 + М2 + М3 = - 0.4 – 1.2 + 1.8 = 0.2 kNm

Example 5.4

Determine the moments of the resulting pairs acting on the frames (Fig. 5. 10, d, e, f) independently.

5. 5. Addition of force pairs in space

Theorem. A system of pairs of forces acting on a rigid body is equivalent to one pair of forces, the moment of which is equal to the geometric sum of the moments of the constituent pairs.

Proof

Let us prove the theorem for two pairs of forces, the planes of action of which are I and II, and the moments M1 and M2 (Fig. 5. 11, a). Let us transform the pairs of forces so that their shoulders are the segment AB lying on the line of intersection of the planes. We obtain two pairs of forces (Р1, Р1 ′) and (Q2, Q2 ′) having identical shoulders and correspondingly modified force modules, which we find from the relations

M 1 = P1. AB

M2 = Q1. AB

Adding up the forces applied at points A and B, we find their resultants

R = P1 + Q1

R′ = Р1 ′ + Q1 ′

The parallelograms of forces are equal and lie in parallel planes. Consequently, the resultants R and R′ are equal in magnitude, parallel and directed in opposite directions, i.e. form the resulting pair (R, R′ ).

Let's find the moment of this couple:

M = r x R = AB x R = AB x (P1 + Q1) = AB x P1 + AB x Q1 = M1 + M 2

Consequently, the moment of a pair M is equal to the geometric sum of the moments M1 and M2 and is depicted by the diagonal of a parallelogram constructed on the vectors M1 and M2.

If a rigid body is acted upon by “n” pairs of forces with moments M1, M2 ... Mn, then the resulting pair will have a moment equal to the geometric sum of the moments of these pairs

M = ∑ Mi

5. 6. Conditions for equilibrium of a system of pairs of forces

For equilibrium of pairs of forces on a plane, it is necessary and sufficient that the algebraic sum of the moments of all pairs be equal to zero

∑ Mi = 0

For equilibrium of pairs of forces in space, it is necessary and sufficient that the geometric sum of the moments of all pairs be equal to zero

∑ Mi = 0

Example 5.5

Determine the support reactions RA and RB of the beam (Fig. 5. 11, b) under the action of two pairs of forces, using the conditions of equilibrium of pairs of forces on the plane.

1) Let's determine the moment of the resulting pair of forces

M = M1 + M2 = - 40 + 30 = - 30 kNm Since a pair of forces can only be balanced by a pair, then the reactions

RA and RB must form a pair of forces. The line of action of the reaction RB is defined (perpendicular to the supporting surface), the line of action of the reaction RA is parallel to the line of action of the reaction RB.

Let us accept the directions of reactions in accordance with Fig. 5. 11, b.

2) Let us determine the moment of the balancing pair of forces (R A, RB)

M (R A, RB) = МR = RА. AB = RB. AB

3) Let us determine the support reactions from the condition of equilibrium of pairs of forces

∑ Мi = 0 М + МR = 0

30 + RA. 6 = 0

RA = 5 kN; RВ = RA = 5 kN

With a couple of forces is a system of two forces equal in magnitude, parallel and directed in opposite directions, acting on an absolutely rigid body.

Theorem on the addition of pairs of forces. Two pairs of forces acting on the same solid body and lying in intersecting planes can be replaced by one equivalent pair of forces, the moment of which is equal to the sum of the moments of the given pairs of forces.

Proof: Let there be two pairs of forces located in intersecting planes. A pair of forces in a plane is characterized by a moment, and a pair of forces in a plane is characterized by a moment. Let us arrange the pairs of forces so that the arm of the pairs is common and located on the line of intersection of the planes. We add up the forces applied at point A and at point B. We get a couple of forces.

Conditions for equilibrium of pairs of forces.

If a solid body is acted upon by several pairs of forces, arbitrarily located in space, then by sequentially applying the parallelogram rule to each two moments of the pairs of forces, any number of pairs of forces can be replaced by one equivalent pair of forces, the moment of which is equal to the sum of the moments of the given pairs of forces.

Theorem. For equilibrium of pairs of forces applied to a solid body, it is necessary and sufficient that the moment of the equivalent pair of forces be equal to zero.

Theorem. For equilibrium of pairs of forces applied to a solid body, it is necessary and sufficient that the algebraic sum of the projections of the moments of pairs of forces onto each of the three coordinate axes is equal to zero.

20.dynamic differential equations regarding the motion of a material point. Dynamic Coriolis theorem

Differential equations of motion of a free material point.

To derive the equations, we will use the second and fourth axioms of dynamics. According to the second axiom ma = F (1)

where, according to the fourth axiom, F is the resultant of all forces applied to the point.

Taking into account the last remark, expression (1) is often called the basic equation of dynamics. In the form of writing, it represents Newton’s second law, where one force, according to the axiom of independence of the action of forces, is replaced by the resultant of all forces applied to a material point. Recalling that a = dV / dt = d2r / dt = r"", we obtain from (1) the differential equation of motion of a material point in vector form: mr"" = F (2)

differential equations of motion of a non-free material point.

According to the axiom of connections, replacing connections with their reactions, one can consider a non-free material point as free, under the influence of active forces and reactions of connections. According to the fourth axiom of dynamics, F will be the resultant of active forces and reactions of connections.

Therefore, the differential equations of motion of a free material point can be used to describe the motion of a non-free point, remembering that the projections of forces on the rectangular axes Fx, Fy, Fz in equations (4) and the projections of forces on the natural axes Fτ, Fn, Fb in equations (6 ) include not only projections of active forces, but also projections of bond reactions.

The presence of constraint reactions in the equations of motion of a point naturally complicates the solution of dynamics problems, since additional unknowns appear in them. To solve problems, you need to know the properties of bonds and have equations of bonds, of which there should be as many as the reactions of bonds.

The Coriolis force is equal to:

where m is a point mass, w is the vector of angular velocity of a rotating reference frame, v is the vector of the speed of motion of a point mass in this reference frame, square brackets indicate the vector product operation.

The quantity is called Coriolis acceleration.

The Coriolis force is one of the inertial forces that exists in a non-inertial reference frame due to rotation and the laws of inertia, manifesting itself when moving in a direction at an angle to the axis of rotation

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Any kinematic state of bodies that have a point or axis of rotation can be described by a moment of force that characterizes the rotational effect of the force.

Moment of force about the center- this is the vector product of the radius - the vector of the point of application of the force by the force vector.

Shoulder of power- the shortest distance from the center to the line of action of the force (perpendicular from the center to the line of action of the force).

The vector is directed according to the vector product rule: the moment of the force relative to the center (point) as a vector is directed perpendicular to the plane in which the force and the center are located so that from its end it can be seen that the force is trying to rotate the body around the center counterclockwise.

Unit of measurement of moment of force there is 1

Moment of force relative to the center in the plane- an algebraic quantity that is equal to the product of the force modulus and the shoulder relative to the same center, taking into account the sign.

The sign of the moment of the force depends on the direction in which the force tries to rotate about the center:

• counterclockwise -„−” (negative)
• clockwise -„+” (positive);

Properties of the moment of force relative to the center (point).

1. The modulus of the moment of force relative to a point is equal to twice the area of ​​the triangle constructed on vectors.
2. The moment of a force relative to a point does not change when a force is transferred along its line of action, since the arm of the force remains unchanged.
3. The moment of force relative to the center (point) is equal to zero if:

• force is zero F = 0;
• arm of force h = 0, i.e. the line of action of the force passes through the center.

Varignon's theorem (about the moment of the resultant).

The moment of the resultant plane system of converging forces relative to any center is equal to the algebraic sum of the moments of the component forces of the system relative to the same center.

Force couple theory

The addition of two parallel forces directed in the same direction.

The resultant of a system of two parallel forces directed in one direction is equal in modulus to the sum of the moduli of the component forces, is parallel to them and directed in the same direction.

The line of action of the resultant passes between the points of application of the components at distances from these points inversely proportional to the forces

Addition of two parallel forces directed in different directions (the case of forces of different magnitudes)

The resultant of two parallel, unequal in magnitude, oppositely directed forces is parallel to them and directed in the direction of the greater force and is equal in magnitude to the difference in the component forces.

The line of action of the resultant passes outside the segment (on the side of the larger force) connecting the points of their application, and is spaced from them at distances inversely proportional to the forces.

Couple of forces- a system of two parallel forces, equal in magnitude and opposite in direction, applied to an absolutely rigid body.

Leverage of force couple- the distance between the lines of action of the forces of the pair, i.e. the length of a perpendicular drawn from an arbitrary point on the line of action of one of the forces of a pair to the line of action of the second force.

Plane of action of a couple of forces- this is the plane in which the lines of action of the pair’s forces are located.
The action of a pair of forces is reduced to rotational motion, which is determined by the moment of the pair.

Couple moment is called a vector with the following characteristics:

• it is perpendicular to the plane of the pair;
• directed in the direction from which the rotation performed by the pair is visible counterclockwise;
• its modulus is equal to the product of the modulus of one of the forces of the pair and the arm of the pair, taking into account the sign

Sign of the moment of a couple of forces:

• “+” - counterclockwise rotation
• „-„ - clockwise rotation

The moment of a pair of forces is equal to the product of the modulus of one of the forces of the pair and the arm of the pair.

The moment of a couple is a free vector - for it neither the point of application nor the line of action are designated, they can be arbitrary.

Property of the moment of a pair of forces: the moment of the pair is equal to the moment of one of the forces relative to the point of application of the second force.

Couple force theorems

Theorem 1. A pair of forces does not have a resultant, i.e. A pair of forces cannot be replaced by one force.

Theorem 2. A pair of forces is not a system of balanced forces.

Consequence: a pair of forces acting on an absolutely rigid body tries to rotate it.

Theorem 3. The sum of the moments of forces of a pair relative to an arbitrary center (point) in space is a constant quantity and represents the vector-moment of this pair.

Theorem 4. The sum of the moments of forces that make up a pair relative to an arbitrary center in the plane of action of the pair does not depend on the center and is equal to the product of the force by the arm of the pair, taking into account the sign, i.e. the very moment of the couple.

Theorem 5 - about the equivalence of pairs. Pairs of forces whose moments are equal in number and sign are equivalent. Those. a pair of forces can only be replaced or balanced by another equivalent pair of forces.

Theorem 6 is about the balance of a pair of forces. A pair of forces constitutes a balanced system of forces if and only if the moment of the pair is zero.

Theorem 7 - about the possibilities of moving a pair of forces in the plane of its action. The force pair obtained by moving the pair to any place in the plane of its action is equivalent to the provided pair.

Theorem 8 is about adding pairs of forces in the plane. The moment of a pair equivalent to the provided system of pairs in the plane is equal to the algebraic sum of the moments of the constituent pairs. Those. To add pairs of forces, you need to add their moments.

Conditions for the equilibrium of a system of pairs of forces.

Pairs of forces in a plane are balanced if the algebraic sum of their moments is equal to zero.

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