Acceleration of any point in a moving flat figure can be defined in two ways: 1) as the geometric sum of the accelerations of this point in the translational and rotational motions of the figure, and 2) as the acceleration of this point in the rotational motion around the instantaneous center of accelerations, and the instantaneous center of accelerations is such a point of the Flat figure, the acceleration of which at the given moment is equal to zero.

If the acceleration of some point A of the figure (the acceleration of the pole), as well as the angular velocity and angular acceleration of the figure, are known, then the acceleration of any point B of it is determined by the formula

Here the vector is the acceleration of point B in rotational motion around the pole, the tangent and normal components of this acceleration.

Hence,

in this case, the vector is directed along AB (from point B to point A), and the vector is perpendicular to AB.

The angle between the vectors and VA is determined by the formula

moreover, in the case of accelerated rotation of the figure, the vectors (the rotational speed of point B around the pole A) lie on one side of the straight line AB, otherwise these vectors are located on opposite sides of this straight line.

If the angular velocity of the figure is constant, i.e., then, and therefore, and, i.e., the vector coincides in direction with the vector BA. If at the moment , then the vector is perpendicular to the vector BA.

Based on equality (78), the acceleration of point B can be found by constructing a polygon of accelerations and then applying the projection method, projecting vector equality (78) onto the selected axes.

If the instantaneous acceleration center Q is taken as a pole, then for the acceleration of an arbitrarily chosen point M of the figure we have:

but, therefore

i.e. acceleration of any point M of a flat figure is defined as acceleration in rotational motion around the instantaneous center of acceleration (Fig. 108).

In this case, the acceleration is directed along the straight line MQ from the point M to the center Q, and the acceleration is perpendicular to MQ and

The acceleration of the point M is modulo

and will make an angle with the MQ direction

(84)

It follows from this: 1) the angle a for all points of the figure has the same value at the given moment; 2) accelerations of points of a flat figure are proportional to the distances of these points from the instantaneous center of accelerations.

To determine the position of the instantaneous center of acceleration for a given moment, you need:

1) find the acceleration of some point A of the figure [usually, when solving problems of the type under consideration, the acceleration of one point of the figure (mechanism) is either given or can be easily found];

2) rotate the half-line along which the vector is directed around point A by an acute angle or in the direction of rotation of the figure, if this rotation is accelerated, or in the opposite direction otherwise;

3) on the half-line obtained after this turn, set aside the segment

We note two special cases:

1) let , then , therefore, the acceleration of any point M of the moving figure is directed, i.e. passes through the center Q. Therefore, the instantaneous center of accelerations Q in this case can be found as the intersection point of the lines along which the accelerations of any two points of the figure are directed ;

2) let , then therefore, the acceleration of any point M of the figure is perpendicular to MQ. Therefore, the instantaneous center of accelerations Q in this case can be found as the point of intersection of the perpendiculars erected from any two points of the moving figure to the accelerations of these points.

The tasks related to this paragraph can be divided into the following four groups:

1) problems in which the vectors of velocity and acceleration of one point and the rectilinear trajectory of the second point of a plane figure are given, the acceleration of which is required to be found (problems 566-571, 573-579);

2) problems in which the velocity and acceleration vectors of one point and the curvilinear trajectory of the second point of a plane figure are given, the acceleration of which is required to be found (problems 572, 573, 575);

3) tasks in which it is required to determine the acceleration of a point of a wheel rolling without slipping (problems 556-563);

4) problems in which the accelerations of two points of a plane figure are given, but it is required to determine the acceleration of the third point of this figure (problems 564, 574, 576-578).

The acceleration of an arbitrary point of a rigid body participating in a plane motion can be found as the geometric sum of the acceleration of the pole and the acceleration of the given point in rotational motion around the pole.

To prove this position, we use the theorem of addition of estrus accelerations in a compound motion. Let's take a point as a pole. We will move the moving coordinate system translationally along with the pole (Fig. 1.15 a). Then the relative motion will be rotation around the pole. It is known that the Coriolis acceleration in the case of a portable forward movement is zero, so

Because in translational motion, the accelerations of all points are the same and equal to the acceleration of the pole, we have .

It is convenient to represent the acceleration of a point when moving along a circle as the sum of the centripetal and rotational components:

.

Hence

The directions of the acceleration components are shown in Figure 1.15 a.

The normal (centripetal) component of the relative acceleration is given by

Its value is equal to The vector is directed along the segment AB to the pole A (the center of rotation around is ).

Rice. 1. 15. The theorem on the addition of accelerations (a) its corollaries (b)

The tangential (rotational) component of the relative acceleration is determined by the formula

.

The modulus of this acceleration is found in terms of angular acceleration. The vector is directed perpendicular to AB in the direction of angular acceleration (in the direction of angular velocity, if the movement is accelerated and in the opposite direction of rotation, if the movement is slow).

The value of the total relative acceleration is determined by the Pythagorean theorem:

.

The vector of relative acceleration of any point of a flat figure is deviated from the straight line connecting the point under consideration with the pole by an angle defined by the formula

Figure 1.15 b shows that this angle is the same for all points of the body.

Corollary from the acceleration theorem.

The ends of the acceleration vectors of the points of a rectilinear segment on a flat figure lie on one straight line and divide it into parts proportional to the distances between the points.

The proof of this assertion follows from the figure:

.

The methods for determining the accelerations of the points of a body during its plane motion are identical to the corresponding methods for determining the velocities.

Instant acceleration center

At any moment in the plane of the moving figure, there is one single point whose acceleration is zero. This point is called the instantaneous center of acceleration (MCC).

The proof follows from the way in which the position of this point is determined. Let us take point A as the pole, assuming its acceleration to be known. We decompose the movement of a flat figure into translational and rotational. Using the acceleration addition theorem, we write down the acceleration of the desired point and equate it to zero.

It follows from here that, i.e., the relative acceleration of the point Q is equal to the acceleration of the pole A in magnitude and is directed in the opposite direction. This is possible only if the angles of inclination of the relative acceleration and the acceleration of the pole A to the straight line connecting the point Q with the pole A are the same.

, , .

Examples of finding the MCU.

Consider ways to find the position of the MCC.

Example No. 1: , , are known (Fig. 1.16 a).

Determine the angle . We set aside the angle in the direction of angular acceleration (that is, in the direction of rotation with accelerated rotation and against - with slow rotation), from the direction of the known acceleration of the point and build a ray. On the constructed ray, we set aside a segment of length AQ.

Rice. 1. 16. Examples of finding the MCC: example No. 1 (a), example No. 2 (b)

Example No. 2. Accelerations of two points A and B are known: and (Fig. 1.16 b).

We take one of the points with a known acceleration as a pole and determine the relative acceleration of the other point using geometric constructions. By measuring we find the angle and draw rays from known accelerations at this angle. The point of intersection of these beams is the MCU. The angle is plotted from the acceleration vectors in the same direction as the angle from the relative acceleration vector to the straight line VA.

It should be noted that the MCC and MCC are different points of the body, and the acceleration of the MCC is not equal to zero and the speed of the MCC is not equal to zero (Fig. 1.17).

Rice. 1. 17. The position of the MCC and MCC in the case of a rolling skating rink without slipping

In those cases where the accelerations of the points are parallel to each other, the following particular cases of finding the MCC are possible (Fig. 1.17)

Rice. 1. 18. Special cases of finding the MCU:
a) the accelerations of two points are parallel and equal; b) the accelerations of two points are antiparallel; c) accelerations of two points are parallel but not equal

STATICS

INTRODUCTION TO STATICS

Basic concepts of statics, their scope

Statics is a branch of mechanics that studies the conditions for the equilibrium of material bodies and includes the doctrine of forces.

Speaking of balance, we must remember that "any rest, any balance is relative, they make sense only in relation to one or another specific form of movement." For example, bodies resting on the Earth move with it around the Sun. It is more accurate and correct to speak of relative equilibrium. Equilibrium conditions are different for solid, liquid and gaseous, deformable bodies.

Most engineering structures can be considered low-deformable or rigid. By abstraction, one can introduce the concept of an absolutely rigid body: the distances between the points of which do not change over time.

In the statics of an absolutely rigid body, two problems are solved:

addition of forces and reduction of the system of forces to the simplest form;

determination of equilibrium conditions.

Forces have a different physical nature, often unclear to the end and at the present time. Following Newton, we will understand force as a quantitative model, a measure of the interaction of material bodies.

Newton's model of force is defined by three main characteristics: magnitude, direction of action and point of its application. It has been experimentally established that the value introduced in this way has vector properties. They are considered in more detail in the axioms of statics. In the international SI system of units used in accordance with GOST, the unit of force is the newton (N). The image and designation of forces is shown in Fig. 2.1 a

The totality of forces acting on a body (or system of bodies) is called a system of forces.

A body that is not fastened to other bodies, which can be given movement in any direction, is called free.

A system of forces that completely replaces another system of forces acting on a free body without changing the state of motion or rest is called equivalent.

Rice. 2. 1. Basic concepts of forces

The system of forces, under the action of which the body can be at rest, is called equivalent to zero or balanced.

One force equivalent to a system of forces is called its resultant. The resultant does not always exist, for example, in the case shown in the figure, it does not exist.

One force, equal in magnitude to the resultant, but directed oppositely to it, is called balancing for the original system of forces (Fig. 2.1 b).

The forces acting between the particles of one body are called internal, and those acting from other bodies are called external.

Axioms of statics

Considering the plane motion of a flat figure as the sum of translational motion, in which all points of the figure move with acceleration a A pole A , and rotational

movement around this pole, we obtain a formula for determining the acceleration of any point B of a flat figure in the form

a B =		a A +		a BA =		a A + a BA +	a BAc .
Here a						acceleration	poles A ; a		Acceleration

rotational motion of the point B around the pole A, which, as in the case of rotation of the body around a fixed axis, is a vector

is the sum of the rotational acceleration a BA in and centric

rapid acceleration a BA c . The modules of these accelerations are determined by the formulas

angular acceleration module. Rotational acceleration a BA в is directed perpendicular to the segment AB in the direction of the arc arrow ε, and centripetal acceleration a BA c is directed along the line AB from point B to pole A (Fig. 12). The total acceleration module a BA of point B relative to the pole A due to the condition a BA in a BA u is calculated by the formula

Fig 12. Determining the acceleration of point B

using pole A.

To find the acceleration a B by the formula (2.18)

recommended to use analytical method. This method introduces a rectangular cartesian system coordinates (system Bxy in Fig. 12) and the projections a Bx , a By

of the desired acceleration as the algebraic sums of the projections of the accelerations included in the right side of equality (2.18):

(a in

(a c

a cosα

c;

(a in

(a c

sinα

where α is the angle between the vector a A

and Bx axis. Found

The described method for determining the accelerations of points of a plane figure is applicable to solving problems in which the movement of the pole A and the angle of rotation of the figure are given

equations (2.14). If the dependence of the angle of rotation on time is unknown, then for a given position of the figure it is necessary to determine the instantaneous angular velocity and instantaneous angular acceleration. Methods for determining them are discussed further in the examples of task 2.

We also note that when determining the accelerations of the points of a plane figure, one can use instant center of acceleration is a point whose acceleration is zero at a given time. However, the use of an instantaneous acceleration center is associated with rather laborious methods for finding its position, therefore, it is recommended to determine the accelerations of the points of a plane figure using the formula

2.4 Task 2. Determination of speeds and accelerations of points of a flat mechanism

Mechanisms (see p. 5) are called flatif all its points move in one or parallel planes, otherwise the mechanisms are called spatial

nym.

V task 2.1 are consideredplanetary gears,

in task 2.2 - crank mechanisms, and in task

2.3, in addition to the two types mentioned, the movement of mechanisms of other types is studied. Most of the mechanisms considered are mechanisms with one degree of freedom,

in which, to determine the motion of all links, it is necessary to set the law of motion of one link.

Task 2.1

In the planetary mechanism (Fig. 13), the crank 1 with the length OA = 0.8 (m) rotates around the fixed axis O, perpendicular to the plane drawing, by law

ϕ OA (t ) = 6t − 2t 2 (rad). At point A, the crank is pivotally connected

with the center of disk 2 of radius r = 0.5 (m), which is in internal engagement with a fixed wheel 3, coaxial with

crank OA . On disk 2, at time t 1 = 1 (s), point B is set, the position of which is determined by distance AB = 0.5 (m) and angle α = 135°. (At a given time, the angle α is measured from the Ax axis in the counterclockwise direction for α > 0 or in the opposite direction for

α < 0).

Fig 13. Planetary mechanism and method of setting the position of point B.

Determine at time t 1

1) the speed of point B in two ways: using the instantaneous center of velocities (MCS) of disk 2 and using pole A;

2) acceleration of point B using pole A .

1) Determining the speed of point B.

First you need to make a graphic image

mechanism in the selected scale (for example, in 1 cm of the figure - 0.1 m of segment OA and radius r) and show the specified position of point B (Fig. 14).

Figure 14. Determining the speed of point B using the instantaneous center of velocities P and pole A.

According to the given law of rotation of the crank OA, we find the speed of the center A of the disk 2. We determine the angular velocity of the crank at a given time t 1 \u003d 1 (c):

ω OA = ϕ ! OA = (6 t −				6 − 4 t ;	ω OA (t 1) \u003d 2 (rad / s).

The resulting value ω OA (t 1 ) is positive, so we direct the arc arrow ω OA counterclockwise, that is, in the positive direction of the angle ϕ.

Calculate the modulus of speed

v A \u003d ω OA (t 1) OA \u003d 2 0.8 \u003d 1.6 (m / s)

and build the velocity vector v A perpendicular to OA in the direction of the arc arrow ω OA .

the arc arrow ω OA and the vector v A are drawn in the opposite direction, and the modulus is used to calculate v A

ω OA (t 1 ) .

The instantaneous center of velocities (point P) of disk 2 is located at the point of its contact with wheel 3 (see item 5 on p. 34). Let us determine the instantaneous angular velocity ω of the disk from the found value of the velocity v A :

ω \u003d v A / AP \u003d v A / r \u003d 1.6 / 0.5 \u003d 3.2 (rad / s)

and depict in the figure its arc arrow (Fig. 14).

To determine the speed of point B using the MCS, we find the distance BP using the cosine theorem from the triangle ABP:

BP = AB2 + AP2 - 2 AB AP cos135 " =

0.5 2 + 0.52 − 2 0.52 (− 2 / 2) ≈ 0.924 (m).

The speed v B is modulo

v B = ω PB = 3.2 0.924 ≈ 2.956 (m / s)

and is directed perpendicular to the segment РВ in the direction of the arc arrow ω.

The same vector v B can be found using the pole A by formula (2.15): v B = v A + v BA . We move the vector v A to point B and construct a vector v BA perpendicular to the segment AB and directed towards the arc arrow ω . Module

that the angle between the vectors v A and v BA is 45°. Then by formula (2.16) we find

vB = vA 2 + vBA 2 + 2 vA vBA cos 45" =

1.62 + 1.62 + 2 1.62(2/2) ≈ 2.956(m/s).

In the figure, the vector v B must coincide with the diagonal of the parallelogram whose sides are the vectors v A and v BA . This is achieved by constructing vectors v A , v B and v BA in the selected

scale (for example, 1 cm in the figure corresponds to 0.5 m/s). Note that the scales given in the considered example can be changed and assigned independently.

2). Determination of the acceleration of point B.

The acceleration of point B is determined by the formula (2.18) using the pole A, the acceleration of which is a vectorial sum of the tangential and normal accelerations:

a B = a A + a BA c + a BA c = a τ A + a A n + a BA c + a BA c .

According to the given law of rotation of the crank OA, we find its angular acceleration:

ε OA = ω ! OA = (6 − 4t ! ) = − 4 (rad / s 2 ).

The resulting value ε OA is negative, so we direct the arc arrow ε OA clockwise, then

is in the negative direction, and in the further calculation we will take this value modulo.

The modules of the tangential and normal accelerations of the pole A at a given time t 1 are found by the formulas (2.11):

a τ A = ε OA OA = 4 0.8 = 3.2 (m / s 2 ); a n A \u003d ω OA 2 OA \u003d 22 0.8 \u003d 3.2 (m / s 2).

The tangential acceleration a τ A is directed perpendicular to the crank OA in the direction of the arc arrow ε OA , and the normal acceleration a A n - from the anguish A to the point O for any direction of the angular velocity of the crank (Fig. 15). The total acceleration a A need not be determined.

Fig 15. Determining the acceleration of point B using pole A.

ω = v A / r = ω OA (OA / r ) .
	by definition angular	acceleration	disk (with
OA/r = const) equals
ε = ω ! =	w! OA (OA / r ) = ε OA (OA / r ) = −	4 (0.8 / 0.5) =	− 6.4 (rad / s 2 ).

we direct the angular arrow ε in the opposite direction to the arc arrow ω.

We calculate the modules of rotational and centripetal accelerations of point B relative to pole A using the formulas

a BA						AB=	6.4 0.5 \u003d 3.2 (m / s 2);
a BAC						2AB=	3.22 0.5 \u003d 5.12 (m / s 2).

The vector a BA в is directed perpendicular to the segment AB to the side

arc arrow ε, and the vector a BA c - from point B to pole A

We find the acceleration of point B by its projections on the axes of the coordinate system Axy:

a Bx = (a τ A ) x +					(a An) x + (a BAв) x + (a BAц) x =
	0 − a n A −			a BA in cos 45" +			a BAC	cos 45" =
		3.2 −				/ 2 + 5.12		2 / 2 ≈	− 1.84 (m / s 2 );
a By = (a τ A ) y +					(a An) y + (a BAв) y + (a BAц) y =
		a τ A +	0 −		a BA	cos45"	− a BA c cos 45" =
		3.2 −				/ 2 − 5.12		2 / 2 ≈	− 9.08 (m / s 2 ).
Module a B =		a Bx2			a By2	≈ 9.27 (m / s 2 ).
					acceleration		a τ A ,	a A n ,	a BA in , a BA c is required

draw on the selected scale and construct the vector a B on the same scale using the found projections (Fig. 15).

The initial data for independent performance of task 2.1 are given in the table on p. 44.

				Rigid Body Kinematics
		ϕ OA (t), rad				α , deg	t1, c
		t2 + 3t
		8t-3t2
		t2-4t
		3t-2t2
		2t2-t
		4t-t2
		2t2 - 6t
		2t-3t2
		3t2 - 4t
		8t-2t2
		4t2 - 6t
		3t-4t2
		4t2 - 2t
		6t-t2
		2t2 - 4t
		4t-3t2
		2t2 + t
		4t-2t2
		3t2 - 10t
		t-2t2
		3t2 + 2t
		6t-3t2
		3t2 - 8t
		2t-4t2

( the answer is taken from question 16, it's just that in all formulas you need to express instead of the distance to the MCS - the acceleration of the point)

When determining the velocities of the points of a flat figure, it was found that at each moment of time there is such a point P of the figure (MCP), the speed of which is zero. Let us show that at each moment of time there is a point of the figure whose acceleration is equal to zero. Such a point is called instantaneous center of acceleration (MCC). Let's denote it by Q.

Consider a flat figure moving in the plane of the drawing (Fig.). Let us take as a pole any point A, the module and direction of acceleration aA of which are known at the considered moment of time. Let the angular velocity and angular acceleration of the figure be known at this point in time. It follows from the formula that the point Q will be the MCC if , i.e. when . Since the vector aQA makes an angle "alpha" with the line AQ , then the vector aA parallel to it is directed to the line connecting the pole A with the point Q, also at an angle "alpha" (see Fig.).

Let us draw a straight line MN through the pole A, making the angle "alpha" with the vector of its acceleration, plotted from the vector aA in the direction of the arc arrow of the angular acceleration. Then there is a point Q on the ray AN for which . Because, according to , point Q (MTsU) will be separated from pole A at a distance .

In this way, at each moment of motion of a plane figure, if the angular velocity and angular acceleration are not equal to zero at the same time, there is a single point of this figure, the acceleration of which is equal to zero. At each subsequent moment of time, the MCC of a flat figure will be at its various points.

If MCC - point Q is chosen as a pole, then the acceleration of any point A of a flat figure
, since aQ = 0. Then . The acceleration aA makes with the segment QA, connecting this point with the MCU, the angle "alpha", laid off from QA in the direction opposite to the direction of the arc arrow of the angular acceleration. The accelerations of the points of the figure during the plane motion are proportional to the distances from the MCU to these points.

In this way, the acceleration of any point of the figure during its plane motion is determined at a given moment of time in the same way as during the rotational movement of the figure around the MCU.

Let us consider the cases when the position of the MCU can be determined using geometric constructions.

1) Let the directions of accelerations of two points of a flat figure, its angular velocity and acceleration be known. Then the MCC lies at the intersection of straight lines drawn to the acceleration vectors of the points of the figure at the same acute angle: , plotted from the acceleration vectors of the points in the direction of the arc arrow of the angular acceleration.

2) Let the acceleration directions of at least two points of a plane figure be known, its angular acceleration = 0, and the angular velocity not equal to 0.

3) Angular velocity = 0, angular acceleration is not 0. Right angle.

Let us show that the acceleration of any point M of a plane figure (as well as speed) is the sum of the accelerations that a point receives during the translational and rotational movements of this figure. Point position M in relation to the axes Oxy(see Fig.30) is determined by the radius vector where . Then

On the right side of this equality, the first term is the acceleration of the pole A, and the second term determines the acceleration that point m receives when the figure rotates around the pole A. hence,

The value of , as the acceleration of a point of a rotating rigid body, is defined as

where and - the angular velocity and angular acceleration of the figure, and - the angle between the vector and the segment MA(fig.41).

Thus, the acceleration of any point M plane figure is geometrically composed of the acceleration of some other point A, taken as a pole, and acceleration, which is a point M receives when the figure rotates around this pole. The module and direction of acceleration are found by constructing the corresponding parallelogram (Fig. 23).

However, the calculation using the parallelogram shown in Fig. 23 complicates the calculation, since it will first be necessary to find the value of the angle , and then the angle between the vectors and . Therefore, when solving problems, it is more convenient to replace the vector with its tangent and normal components and present it in the form

In this case, the vector is directed perpendicularly AM in the direction of rotation, if it is accelerated, and against rotation, if it is slow; the vector is always directed from the point M to the pole A(Fig. 42). Numerically

If the pole A does not move in a straight line, then its acceleration can also be represented as the sum of the tangent and normal components, then

Fig.41 Fig.42

Finally, when the point M moves curvilinearly and its trajectory is known, then it can be replaced by the sum .

Questions for self-examination

What motion of a rigid body is called flat? Give examples of links of mechanisms that make a plane movement.

What simple motions make up the plane motion of a rigid body?

How is the speed of an arbitrary point of a body determined in a plane motion?

What motion of a rigid body is called plane-parallel?

Complex point movement

This lecture covers the following questions:

1. Complicated movement of a point.

2. Relative, figurative and absolute movements.

3. Velocity addition theorem.

4. The theorem of addition of accelerations. Coriolis acceleration.

5. Complex motion of a rigid body.

6. Cylindrical gears.

7. Addition of translational and rotational movements.

8. Screw movement.

The study of these issues is necessary in the future for the dynamics of a plane motion of a rigid body, the dynamics of the relative motion of a material point, for solving problems in the disciplines "Theory of machines and mechanisms" and "Machine parts".