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
The properties of force pairs are determined by a number of theorems, which are given without proof:
· Two pairs are equivalent if their vector moments are equal in magnitude and have the same direction.
· The action of the pair on the body will not change if it is transferred to any place in the plane of action.
· The action of the pair on the body will not change if it is transferred from the plane of action to a plane parallel to it.
· The effect of a couple on the body will not change if you increase (decrease) the magnitude of the force of the couple, while simultaneously decreasing (increasing) the shoulder of the couple by the same amount.
Conclusion: the vector moment of a pair of forces acting on a rigid body is a free vector, i.e. it can be applied at any point of the rigid body.
Let us consider the addition of pairs arbitrarily located in space. Let's prove the theorem:
A system of pairs arbitrarily located in space is equivalent to one pair with a moment equal to the geometric sum of the moments of the terms of the pairs.
Let's take two pairs () and (), located on planes intersecting at an arbitrary angle. Let's assume the pairs' shoulders are equal to and respectively. On the line of intersection of the planes, mark an arbitrary segment AB and bring each of the summand pairs to arm AB. By adding the corresponding forces (see figure) c and c, we obtain a new pair (), the moment of which will be equal to
Fig. 2.18 Resultant force pair
A system of pairs of forces acting on a body can, in accordance with the theorem just proven, be replaced by one pair equal to the sum of the moment vectors of the pairs. Consequently, equilibrium of a system of pairs is possible only if the condition is met
Projecting the reduced vector condition for the equilibrium of pairs onto any three axes that do not lie in the same plane and are not parallel to each other, we obtain scalar equations for the equilibrium of a system of pairs
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.
Solution results: |
|||
M = - 50 kNm |
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M = - 80 kNm |
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Rice. 5. 10, e |
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P3 "E |
||||||||||
M1 = 10kNm |
M2 = 20kNm |
|||||||||
M2 = 40kNm |
||||||||||
M3 = 40kNm |
||||||||||
M1 = 10kNm |
M4 = 80kNm |
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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 built 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
Ministry of Education and Science of the Russian Federation
Federal state budget educational
institution of higher professional education
Transbaikal State University
Department of Theoretical Mechanics
ABSTRACT
On the topic: “Equivalence of pairs of forces in space and on a plane, their addition and equilibrium conditions”
Student: Sadilov I.A.
Group: SUS-13-2
Teacher: Geller Yu.A.
Chita, 2014
What is a couple of forces……………………………………………………………3
Theorem on the sum of moments of a pair of forces…………………………….3
Theorem on the equivalence of pairs of forces……………………………4
Theorem on the transfer of a pair of forces into a parallel plane…….5
Theorem on the addition of pairs of forces…………………………………….8
Conditions for equilibrium of pairs of forces…………………………………..8
Conclusions……………………………………………………….9
List of references……………………………10
PAIR OF FORCE
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.
The plane of action of a pair of forces The plane in which these forces are located is called.
Shoulder of a couple of forces d is the shortest distance between the lines of action of the forces of the pair.
moment
pairs of forces
is called a vector whose modulus is equal to the product of the modulus of one of the forces of the pair and its shoulder and which is directed perpendicular to the plane of action of the forces of the pair in the direction from which the pair is visible trying to turn the body counterclockwise. 
Sum of moments theorem pairs of forces. The sum of the moments of the forces included in the pair relative to any point does not depend on the choice of this point and is equal to the moment of this pair of forces.
Proof: Let us arbitrarily choose point O. Draw radius vectors from it to points A and B (See Fig. 4.2).
, 
H 
that was what needed to be proven.
Two pairs of forces are said to be equivalent , if their effect on a solid body is the same, other things being equal.
Theorem on the equivalence of pairs of forces. A pair of forces acting on a rigid body can be replaced by another pair of forces located in the same plane of action and having the same moment as the first pair.
.
P 
let's bring back the power 
exactly 
, and strength 
exactly 
. Let's draw through the points 
any two parallel straight lines intersecting the lines of action of the forces of the pair. Let's connect the dots 
straight line segment and expand the forces 
at the point 
And 
at the point 
according to the parallelogram rule.
Because 
, That
And 
That's why 
is equivalent to the system 
, and this system is equivalent to the system 
, because 
is equivalent to zero.
Thus we have a given pair of forces 
replaced by another pair of forces 
. Let us prove that the moments of these pairs of forces are the same.
Moment of the initial force pair 
, and the moment of a couple of forces 
numerically equal to the area of the parallelogram 
. But the areas of these parallelograms are equal, since the area of the triangle is 
equal to the area of the triangle 
.
Q.E.D.
Theorem on the transfer of a pair of forces into a parallel plane . The action of a pair of forces on a rigid body will not change if this pair is transferred to a parallel plane.
Proof: Let a pair of forces act on a rigid body 
in the plane 
. From the points of application of forces A and B, we lower perpendiculars to the plane 
and at the points of their intersection with the plane 
let's apply two systems of forces 
And 
, each of which is equivalent to zero.
WITH 
we apply two equal and parallel forces 
And 
. Their resultant 
at point O.
Let's add two equal and parallel forces 
And 
. Their resultant 
parallel to these forces, equal to their sum and applied in the middle of the segment 
at point O.
Because 
, then the system of forces 
is equivalent to zero and can be discarded.
So a couple of forces 
equivalent to a couple of forces 
, but lies in a different, parallel plane. Q.E.D.
Consequence: The moment of a pair of forces acting on a rigid body is a free vector.
Two pairs of forces acting on the same rigid body are equivalent if they have moments of the same magnitude and direction.
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. Couple of forces 
in the plane 
characterized by moment 
, and a couple of forces 
in the plane 
characterized by moment 
.
Let us arrange the pairs of forces so that the shoulder 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 
.
Q.E.D.
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.
Conditions for the equilibrium of a system of forces
Vector shape
For the equilibrium of an arbitrary system of forces applied to a rigid body, it is necessary and sufficient that the main vector of the force system is equal to zero and the main moment of the force system relative to any center of reduction is also equal to zero.
Algebraic form.
For the equilibrium of an arbitrary system of forces applied to a solid body, it is necessary and sufficient that the three sums of the projections of all forces on the Cartesian coordinate axes are equal to zero and the three sums of the moments of all forces relative to the three coordinate axes are also equal to zero.
Conditions for the equilibrium of a spatial system
parallel forces
A system of parallel forces acts on the body. Let's place the Oz axis parallel to the forces.
Equations 
For the equilibrium of a spatial system of parallel forces acting on a solid body, it is necessary and sufficient that the sum of the projections of these forces be equal to zero and the sum of the moments of these forces relative to two coordinate axes perpendicular to the forces are also equal to zero.
- projection of force onto the Oz axis.
Conclusions:
A couple of forces as a rigid figure can be rotated and transferred in any way in its plane of action.
The leverage and forces of a force couple can be changed while maintaining the couple's moment and plane of action.
3. The moment of the couple is a free vector and completely determines the action of the couple on an absolutely rigid body. For deformable bodies, the theory of pairs is not applicable.
LITERATURE:
1. Kirsanov M.N. Theoretical mechanics. Self-study textbook.
2. Targ S.M. Course on Theoretical Mechanics.
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.
