Task 3.2.1

Determine the resultant of two forces F 1 \u003d 50N and F 2 \u003d 30N, forming an angle of 30 ° between them (Fig. 3.2a).

Figure 3.2

We transfer the force vectors F 1 and F 2 to the point of intersection of the action lines and add them according to the parallelogram rule (Fig. 2.2b). The point of application and the direction of the resultant is shown in the figure. The module of the resulting resultant is determined by the formula:

Answer: R=77.44N

Task 3.2.2

Determine the resultant of the system of converging forces F 1 =10N, F 2 =15N, F 3 =20N, if the angles formed by the vectors of these forces with the Ox axis are known: α 1 =30 °, α 2 =45 ° and α 3 =60 ° ( fig.3.3a)

Figure 3.3

We project the forces on the Ox and Oy axes:

Resultant modulus

Based on the obtained projections, we determine the direction of the resultant (Fig. 3.3b)

Answer: R=44.04N

Task 3.2.3

At the point of connection of two threads, a vertical force P = 100N is applied (Fig. 3.4a). Determine the forces in the threads, if in equilibrium the angles formed by the threads with the OY axis are equal to α=30°, β=75°.

Figure 3.4

The tension forces of the threads will be directed along the threads from the connection node (Fig. 3.4b). The system of forces T 1 , T 2 , P is a system of converging forces, because the lines of action of the forces intersect at the junction of the threads. The equilibrium condition for this system:

We compose analytical equations for the equilibrium of a system of converging forces, projecting a vector equation onto the axis.

We solve the system of equations obtained. From the first we express T 2 .

Substitute the resulting expression into the second and determine T 1 and T 2 .

H,

Let's check the solution from the condition that the modulus P'of the sum of forces T 1 and T 2 must be equal to P (Fig. 3.4c).

Answer: T 1 \u003d 100N, T 2 \u003d 51.76N.

Task 3.2.4

Determine the resultant of the system of converging forces if their modules F 1 =12N, F 2 =10N, F 3 =15N and the angle α=60 ° are given (Fig. 3.5a).

Figure 3.5

We determine the projections of the resultant

Resultant modulus:

Based on the obtained projections, we determine the direction of the resultant (Fig. 3.5b)

Answer: R=27.17N

Task 3.2.6

Three rods AC, BC, DC are pivotally connected at point C. Determine the forces in the rods if the force F=50N, the angle α=60° and the angle β=75° are given. The force F is in the Oyz plane. (fig.3.6)

Figure 3.6

Initially, we assume that all the rods are stretched, respectively, we direct the reactions in the rods from node C. The resulting system N 1 , N 2 , N 3 , F is a system of converging forces. The equilibrium condition for this system.

To answer this question, it is necessary to draw some conclusions from the condition of the problem:

1. The direction of these forces;
2. Modular value of forces F1 and F2;
3. Can these forces create such a resultant force to move the cart from its place.

Direction of forces

In order to determine the main characteristics of the movement of a cart under the influence of two forces, it is necessary to know their direction. For example, if a cart is pulled to the right by a force equal to 5 N and the same force is pulling the cart to the left, then it is logical to assume that the cart will stand still. If the forces are co-directed, to find the resultant force, it is only necessary to find their sum. If any force is directed at an angle to the plane of movement of the cart, then the value of this force must be multiplied by the cosine of the angle between the direction of the force and the plane. Mathematically it will look like this:

F = F1 * cosa; where

F is the force directed parallel to the surface of motion.

Cosine theorem for finding the resulting force vector

If two forces have their origin at one point and there is a certain angle between their direction, then it is necessary to complete the triangle with the resulting vector (that is, the one that connects the ends of the vectors F1 and F2). We find the resulting force using the cosine theorem, which states that the square of any side of a triangle is equal to the sum of the squares of the other two sides of the triangle minus twice the product of these sides by the cosine of the angle between them. Let's write this in mathematical form:

F \u003d F 1 2 + F 2 2 - 2 * F 1 * F 2 * cosa.

Substituting all the known values, you can determine the magnitude of the resulting force.

The content of the article

STATICS, branch of mechanics, the subject of which are material bodies that are at rest under the action of external forces on them. In the broad sense of the word, statics is the theory of equilibrium of any bodies - solid, liquid or gaseous. In a narrower sense, this term refers to the study of the equilibrium of rigid bodies, as well as non-stretching flexible bodies - cables, belts and chains. The equilibrium of deforming solids is considered in the theory of elasticity, and the equilibrium of liquids and gases - in hydroaeromechanics.
Cm. HYDROAEROMECHANICS.

History reference.

Statics is the oldest branch of mechanics; some of its principles were already known to the ancient Egyptians and Babylonians, as evidenced by the pyramids and temples they built. Among the first creators of theoretical statics was Archimedes (c. 287–212 BC), who developed the theory of leverage and formulated the basic law of hydrostatics. The ancestor of modern statics was the Dutchman S. Stevin (1548–1620), who in 1586 formulated the law of addition of forces, or the parallelogram rule, and applied it in solving a number of problems.

Basic laws.

The laws of statics follow from the general laws of dynamics as a special case when the velocities of rigid bodies tend to zero, but for historical reasons and pedagogical considerations, statics is often presented independently of dynamics, building it on the following postulated laws and principles: a) the law of addition of forces, b) the principle of equilibrium and c) the principle of action and reaction. In the case of rigid bodies (more precisely, ideally rigid bodies that do not deform under the action of forces), another principle is introduced based on the definition of a rigid body. This is the principle of force transferability: the state of a rigid body does not change when the point of application of the force moves along the line of its action.

Force as a vector.

In statics, a force can be considered as a pulling or pushing force that has a certain direction, magnitude, and point of application. From a mathematical point of view, this is a vector, and therefore it can be represented as a directed straight line segment, the length of which is proportional to the magnitude of the force. (Vector quantities, unlike other quantities that have no direction, are denoted in bold letters.)

Parallelogram of forces.

Consider the body (Fig. 1, a) on which the forces act F 1 and F 2 applied at the point O and represented in the figure by directed segments OA and OB. As experience shows, the action of forces F 1 and F 2 is equivalent to one strength R, represented by a segment OC. The magnitude of the force R is equal to the length of the diagonal of the parallelogram built on the vectors OA and OB how its sides; its direction is shown in Fig. one, a. Strength R called the resultant force F 1 and F 2. Mathematically, this is written as R = F 1 + F 2 , where addition is understood in the geometric sense of the word indicated above. This is the first law of statics, called the rule of the parallelogram of forces.

Balanced force.

Instead of building a parallelogram OACB, to determine the direction and magnitude of the resultant R one can construct the triangle OAC by translating the vector F 2 parallel to itself until its starting point (former point O) coincides with the end point (point A) of the vector OA. The trailing side of the triangle OAC will obviously have the same magnitude and the same direction as the vector R(Fig. 1, b). This method of finding the resultant can be generalized to a system of many forces F 1 , F 2 ,..., F n applied at the same point O of the considered body. So, if the system consists of four forces (Fig. 1, in), then you can find the resultant of forces F 1 and F 2, fold it with force F 3 , then add the new resultant with the force F 4 and, as a result, obtain the total resultant R. Resultant R, found by such a graphical construction, is represented by the closing side of the OABCD force polygon (Fig. 1, G).

The definition of the resultant given above can be generalized to the system of forces F 1 , F 2 ,..., F n applied at points O 1 , O 2 ,..., O n of the rigid body. A point O is chosen, called the reduction point, and a system of parallel transferred forces is built in it, equal in magnitude and direction to the forces F 1 , F 2 ,..., F n. Resultant R these parallel transferred vectors, i.e. the vector represented by the closing side of the polygon of forces is called the resultant of the forces acting on the body (Fig. 2). It is clear that the vector R does not depend on the chosen reduction point. If the magnitude of the vector R(segment ON) is not equal to zero, then the body cannot be at rest: in accordance with Newton's law, any body on which a force acts must move with acceleration. Thus, a body can be in equilibrium only if the resultant of all forces applied to it is zero. However, this necessary condition cannot be considered sufficient - the body can move when the resultant of all forces applied to it is equal to zero.

As a simple but important example to clarify what has been said, consider a thin rigid rod of length l, the weight of which is negligible compared to the magnitude of the forces applied to it. Let two forces act on the rod F and -F applied to its ends, equal in magnitude but oppositely directed, as shown in Fig. 3, a. In this case, the resultant R is equal to F – F= 0, but the rod will not be in equilibrium; obviously, it will rotate around its midpoint O. The system of two equal, but oppositely directed forces, acting not in one straight line, is a “pair of forces”, which can be characterized by the product of the magnitude of the force F on the shoulder" l. The significance of such a product can be shown by the following reasoning, which illustrates the lever rule derived by Archimedes and leads to the conclusion about the condition of rotational equilibrium. Consider a lightweight homogeneous rigid rod that can rotate around an axis at point O, on which the force acts F 1 applied at a distance l 1 from the axis, as shown in fig. 3, b. Under the force F 1 the rod will rotate around the point O. As you can easily see from experience, the rotation of such a rod can be prevented by applying some force F 2 at that distance l 2 to satisfy the equality F 2 l 2 = F 1 l 1 .

Thus rotation can be prevented in countless ways. It is only important to choose the force and the point of its application so that the product of the force on the shoulder is equal to F 1 l one . This is the rule of leverage.

It is not difficult to derive the equilibrium conditions for the system. Action of forces F 1 and F 2 per axis causes a reaction in the form of a reaction force R, applied at the point O and directed opposite to the forces F 1 and F 2. According to the law of mechanics about action and reaction, the magnitude of the reaction R equal to the sum of forces F 1 + F 2. Therefore, the resultant of all forces acting on the system is equal to F 1 + F 2 + R= 0, so that the above necessary equilibrium condition is satisfied. Strength F 1 creates a clockwise torque, i.e. moment of power F 1 l 1 about point O, which is balanced by a counterclockwise moment F 2 l 2 strength F 2. Obviously, the equilibrium condition of the body is the equality to zero of the algebraic sum of the moments, which excludes the possibility of rotation. If strength F acts on the rod at an angle q, as shown in fig. four, a, then this force can be represented as the sum of two components, one of which ( F p), value F cos q, acts parallel to the rod and is balanced by the reaction of the support - F p , and the other ( F n) F sin q directed at right angles to the lever. In this case, the torque is Fl sin q; it can be balanced by any force that creates an equal moment acting counterclockwise.

To make it easier to take into account the signs of the moments in cases where a lot of forces act on the body, the moment of force F relative to any point O of the body (Fig. 4, b) can be considered as a vector L equal to the vector product r ґ F position vector r for strength F. In this way, L = rґ F. It is easy to show that if a system of forces applied at the points O 1 , O 2 ,..., O n (Fig. 5) acts on a rigid body, then this system can be replaced by the resultant R forces F 1 , F 2 ,..., F n applied at any point Oў of the body, and a pair of forces L, the moment of which is equal to the sum [ r 1 ґ F 1 ] + [r 2 ґ F 2 ] +... + [r n ґ F n]. To verify this, it is enough to mentally apply at the point Oў a system of pairs of equal but oppositely directed forces F 1 and - F 1 ; F 2 and - F 2 ;...; F n and - F n , which obviously does not change the state of the rigid body.

Wore F 1 applied at the point O 1 , and the force - F 1 , applied at the point Oў, form a pair of forces, the moment of which relative to the point Oў is equal to r 1 ґ F one . Just the same strength F 2 and - F 2 applied at the points O 2 and Oў, respectively, form a pair with moment r 2 ґ F 2 , etc. Total moment L of all such pairs with respect to the point Oў is given by the vector equality L = [r 1 ґ F 1 ] + [r 2 ґ F 2 ] +... + [r n ґ F n]. Remaining forces F 1 , F 2 ,..., F n , applied at the point Oў, in total give the resultant R. But the system cannot be in equilibrium if the quantities R and L are different from zero. Consequently, the condition of equality to zero at the same time of the quantities R and L is a necessary condition for equilibrium. It can be shown that it is also sufficient if the body is initially at rest. Thus, the equilibrium problem is reduced to two analytical conditions: R= 0 and L= 0. These two equations represent the mathematical notation of the equilibrium principle.

Theoretical provisions of statics are widely used in the analysis of forces acting on structures and structures. In the case of a continuous distribution of forces, the sums that give the resulting moment L and resultant R, are replaced by integrals and in accordance with the usual methods of integral calculus.

Often, not one, but several forces act simultaneously on the body. Consider the case when two forces ( and ) act on the body. For example, a body resting on a horizontal surface is affected by gravity () and the surface support reaction () (Fig. 1).

These two forces can be replaced by one, which is called the resultant force (). Find it as a vector sum of forces and:

Determination of the resultant of two forces

DEFINITION

The resultant of two forces called a force that produces an effect on a body similar to the action of two separate forces.

Note that the action of each force does not depend on whether there are other forces or not.

Newton's second law for the resultant of two forces

If two forces act on the body, then we write Newton's second law as:

The direction of the resultant always coincides in direction with the direction of acceleration of the body.

This means that if two forces () act on a body at the same time, then the acceleration () of this body will be directly proportional to the vector sum of these forces (or proportional to the resultant forces):

M is the mass of the considered body. The essence of Newton's second law is that the forces acting on the body determine how the speed of the body changes, and not just the magnitude of the speed of the body. Note that Newton's second law holds exclusively in inertial frames of reference.

The resultant of two forces can be equal to zero if the forces acting on the body are directed in different directions and are equal in absolute value.

Finding the value of the resultant of two forces

To find the resultant, it is necessary to depict on the drawing all the forces that must be taken into account in the problem acting on the body. The forces must be added according to the rules of vector addition.

Let's assume that two forces act on the body, which are directed along one straight line (Fig. 1). It can be seen from the figure that they are directed in different directions.

The resultant of forces () applied to the body will be equal to:

To find the modulus of the resultant forces, we choose an axis, denote it X, direct it along the direction of the forces. Then, projecting expression (4) onto the X axis, we get that the value (modulus) of the resultant (F) is equal to:

where are the modules of the corresponding forces.

Imagine that two forces act on the body and directed at some angle to each other (Fig. 2). The resultant of these forces is found by the parallelogram rule. The value of the resultant will be equal to the length of the diagonal of this parallelogram.

Examples of problem solving

EXAMPLE 1

Exercise	A body of mass 2 kg is moved vertically upwards by a thread, while its acceleration is 1. What is the magnitude and direction of the resultant force? What forces are applied to the body?
Solution	The force of gravity () and the reaction force of the thread () are applied to the body (Fig. 3). The resultant of the above forces can be found using Newton's second law: In projection onto the X axis, equation (1.1) takes the form: Let's calculate the magnitude of the resultant force:
Answer	H, the resultant force is directed in the same way as the acceleration of the movement of the body, that is, vertically upwards. There are two forces acting on the body.

Resultant. You already know that two forces balance each other when they are equal in magnitude and directed oppositely. Such, for example, are the force of gravity and the force of normal reaction acting on a book lying on the table. In this case, the resultant of the two forces is said to be zero. In the general case, the resultant of two or more forces is the force that produces the same effect on the body as the simultaneous action of these forces.

Consider by experience how to find the resultant of two forces directed along one straight line.

Let's put experience

Let's put a light block on a smooth horizontal table surface (so that the friction between the block and the table surface can be neglected). We will pull the bar to the right using one dynamometer, and to the left - using two dynamometers, as shown in Fig. 16.3. Please note that the dynamometers on the left are attached to the bar so that the tension forces of the springs of these dynamometers are different.

Rice. 16.3. How can you find the resultant of two forces

We will see that a block is at rest if the modulus of the force pulling it to the right is equal to the sum of the moduli of the forces pulling the block to the left. The scheme of this experiment is shown in Fig. 16.4.

Rice. 16.4. Schematic representation of the forces acting on the bar

The force F 3 balances the resultant of the forces F 1 and F 2, that is, it is equal in absolute value and opposite in direction. This means that the resultant of the forces F 1 and F 2 is directed to the left (like these forces), and its module is equal to F 1 + F 2. Thus, if two forces are directed in the same way, their resultant is directed in the same way as these forces, and the modulus of the resultant is equal to the sum of the modules of the force terms.

Consider the force F 1 . It balances the resultant of the forces F 2 and F 3 , directed oppositely. This means that the resultant of the forces F 2 and F 3 is directed to the right (that is, towards the larger of these forces), and its module is equal to F 3 - F 2. Thus, if two forces that are not equal in absolute value are directed oppositely, their resultant is directed as the largest of these forces, and the module of the resultant is equal to the difference between the modules of the greater and lesser forces.

Finding the resultant of several forces is called the addition of these forces.

Two forces are directed along the same straight line. The modulus of one force is equal to 1 N, and the modulus of another force is equal to 2 N. Can the modulus of the resultant of these forces be equal to: a) zero; b) 1 N; c) 2 N; d) 3 N?