Laws of addition of forces in mechanics. Concept of force How to find the value of the resultant force

Newton's laws are a mathematical abstraction. In reality, the cause of movement or rest of bodies, as well as their deformation, is caused by several forces at once. Therefore, an important addition to the laws of mechanics will be the introduction of the concept of resultant force and its application.

About the reasons for the changes

Classical mechanics is divided into two sections - kinematics, which uses equations to describe the trajectory of bodies, and dynamics, which deals with the reasons for changes in the position of objects or the objects themselves.

The cause of the changes is a certain force, which is a measure of the action of other bodies or force fields on the body (for example, an electromagnetic field or gravity). For example, the force of elasticity causes the deformation of a body, the force of gravity causes bodies to fall to the Earth.

Force is a vector quantity, that is, its action is directed. In the general case, the modulus of force is proportional to a certain coefficient (for the deformation of a spring, this is its rigidity), as well as to the action parameters (mass, charge).

For example, in the case of the Coulomb force, this is the magnitude of both charges taken modulo, the squared distance between the charges and the coefficient k, in the SI system determined by the expression: $k = (1 \over 4 \pi \epsilon)$, where $\epsilon$ – dielectric constant.

Addition of forces

In the case when n forces act on a body, we speak of a resultant force, and the formula of Newton’s second law takes the form:

$m\vec a = \sum\limits_(i=1)^n \vec F_i$.

Rice. 1. Resultant of forces.

Since F is a vector quantity, the sum of forces is called geometric (or vector). This addition is performed according to the rule of a triangle or parallelogram, or by components. Let's explain each method with an example. To do this, we write the formula for the resultant force in general form:

$F = \sum\limits_(i=1)^n \vec F_i$

And let's represent the force $F_i$ in the form:

$F = (F_(xi), F_(yi), F_(zi))$

Then the sum of the two forces will be a new vector $F_(ab) = (F_(xb) + F_(xa), F_(yb) + F_(ya), F_(zb) + F_(za))$.

Rice. 2. Componentwise addition of vectors.

The absolute value of the resultant can be calculated as follows:

$F = \sqrt((F_(xb) + F_(xa))^2 + (F_(yb) + F_(ya))^2 + (F_(zb) + F_(za))^2)$

Now let's give a strict definition: the resultant force is the vector sum of all forces influencing the body.

Let's look at the rules of triangles and parallelograms. Graphically it looks like this:

Rice. 3. Rule of triangle and parallelogram.

Outwardly, they seem different, but when it comes to calculations, they come down to finding the third side of a triangle (or, which is the same thing, the diagonal of a parallelogram) using the cosine theorem.

If there are more than two forces, sometimes it is more convenient to use the polygon rule. At its core, this is still the same triangle, only repeated in one picture a certain number of times. If the resulting contour is closed, the total action of forces is zero and the body is at rest.

Tasks

• A box placed at the center of a Cartesian rectangular coordinate system is subject to two forces: $F_1 = (5, 0)$ and $F_2 = (3, 3)$. Calculate the resultant using two methods: using the triangle rule and using component-wise addition of vectors.

Solution

The resultant force will be the vector sum of $F_1$ and $F_2$.

Therefore, let's write:

$\vec F = \vec F_1 + \vec F_2 = (5+3, 0+3) = (8, 3)$
Absolute value of the resultant force:

$F = \sqrt(8^2 + 3^2) = \sqrt(64 + 9) = 8.5 N$

Now we get the same value using the triangle rule. To do this, we first find the absolute values ​​of $F_1$ and $F_2$, as well as the angle between them.

$F_1 = \sqrt(5^2 + 0^2) = 5 Н$

$F_2 = \sqrt(3^2 + 3^2) = 4.2 N$

The angle between them is 45˚, since the first force is parallel to the Ox axis, and the second divides the first coordinate plane in half, that is, it is the bisector of a rectangular angle.

Now, having placed the vectors according to the triangle rule, we calculate the resultant using the cosine theorem:

$F = \sqrt(F_1^2 + F_2^2 - 2F_1F_2 cos135) = \sqrt(F_1^2 + F_2^2 + 2F_1F_2 sin45) = \sqrt(25 + 18 + 2 \cdot 5 \cdot 4,2 \ cdot sin45) = 8.5 N$

• Three forces act on the machine: $F_1 = (-5, 0)$, $F_2 = (-2, 0)$, $F_1 = (7,0)$. What is their resultant?

Solution

It is enough to add the X components of the vectors:

$F = -5 – 2 + 7 = 0$

What have we learned?

During the lesson, the concept of resultant forces was introduced and various methods for calculating it were considered, as well as the entry of Newton's second law for the general case when the number of forces is unlimited.

Test on the topic

Evaluation of the report

Average rating: 4.7. Total ratings received: 175.

Let a system be applied to an absolutely rigid body N strength ( F 1, F 2, … F N), located in space so that their lines of action intersect at one point ABOUT(picture 1).

Such a system of forces is called a system of converging forces. Let us simplify the system of converging forces, i.e. Let's solve the first problem of statics.

Reduction to resultant

Let us prove that this system of forces is equivalent to one force, i.e. is reduced to a resultant force.

Picture 1

In fact, since force is a sliding vector, all forces of a given system can be transferred along the lines of their action to the point ABOUT.

Further, according to the fourth axiom, forces F 1 And F 2 you can replace them with the resultant R 1.2(Figure 1), which is determined by the diagonal of the parallelogram, built on these forces as on the sides, and directed along this diagonal, i.e.

(F 1, F 2) ~ R 1,2,

(R 1,2 F 3) ~ (F 1, F 2, F 3) ~ R 1,2,3,

Where R 1,2,3 =F 1 +F 2 +F 3 etc.

For system N we will finally have the strength

(F 1 F 2 … F N) ~ R *,

R * = F 1 + F 2 + … + F N = ∑ F i . (1)

Figure 2, a shows the construction of the resultant in the indicated way using the example of a system consisting of four forces. However, it is more convenient to determine the resultant in a different way, by constructing a so-called force polygon.

Force polygon

From the end of the force vector F 1(dots IN) draw the vector Sun, geometrically equal to the force F 2 WITH) draw the vector CD equal to strength F 3. From the end of this vector (points D) draw the vector DE, equal to strength F 4.

Figure 2

The resulting polygon ABCDE called force polygon. The process of its construction is clearly visible in Figure 2, b. The sides of a force polygon are called component forces.

Vector AE, connecting the beginning A the first force with the end E the last force and directed towards the component forces is called the closing side of the force polygon.

Consequently, the resultant of the system of converging forces is depicted on a selected scale as the closure of a force polygon built on the component forces.

Finding the resultant system of converging forces according to the force polygon rule is called vector or geometric addition of forces.

Thus, we have proven that a system of converging forces is generally equivalent to a single force, i.e. the resultant, which is applied at the point of intersection of the lines of action of all forces and is equal to their geometric sum.

Calculation of the resultant

To analytically determine the resultant, we find its projections Rx, Ry, Rz on the axis of the Cartesian coordinate system. We have

R x =∑ F kx ,
R y =∑ F ky ,
R z =∑ F kz . (2)

Then the magnitude of the resultant will be determined by the following formula:

To determine the direction of the resultant R* Let's use the usual expressions for direction cosines:

cos α = Rx/R, cos β = Ry/R, cos γ = Rz/R. (5)

Here α, β, γ are the angles between the positive direction of the coordinate axes and the resultant.

This is the vector sum of all forces acting on the body.

The cyclist leans towards the turn. The force of gravity and the reaction force of the support from the earth provide a resultant force that imparts the centripetal acceleration necessary for motion in a circle

Relationship with Newton's second law

Let's remember Newton's law:

The resultant force can be equal to zero in the case when one force is compensated by another, the same force, but opposite in direction. In this case, the body is at rest or moving uniformly.

If the resultant force is NOT zero, then the body moves with uniform acceleration. Actually, it is this force that causes the uneven movement. Direction of resultant force Always coincides in direction with the acceleration vector.

When it is necessary to depict the forces acting on a body, while the body moves with uniform acceleration, it means that in the direction of acceleration the acting force is longer than the opposite one. If the body moves uniformly or is at rest, the length of the force vectors is the same.

Finding the resultant force

In order to find the resultant force, it is necessary: ​​firstly, to correctly designate all the forces acting on the body; then draw coordinate axes, select their directions; in the third step it is necessary to determine the projections of the vectors on the axes; write down the equations. Briefly: 1) identify the forces; 2) select the axes and their directions; 3) find the projections of forces on the axis; 4) write down the equations.

How to write equations? If in a certain direction the body moves uniformly or is at rest, then the algebraic sum (taking into account signs) of the projections of forces is equal to zero. If a body moves uniformly accelerated in a certain direction, then the algebraic sum of the projections of forces is equal to the product of mass and acceleration, according to Newton’s second law.

Examples

A body moving uniformly on a horizontal surface is subject to the force of gravity, the reaction force of the support, the force of friction and the force under which the body moves.

Let us denote the forces, choose the coordinate axes

Let's find the projections

Writing down the equations

A body that is pressed against a vertical wall moves downward with uniform acceleration. The body is acted upon by the force of gravity, the force of friction, the reaction of the support and the force with which the body is pressed. The acceleration vector is directed vertically downwards. The resultant force is directed vertically downwards.

The body moves uniformly along a wedge whose slope is alpha. The body is acted upon by the force of gravity, the reaction force of the support, and the force of friction.

The main thing to remember

1) If the body is at rest or moving uniformly, then the resultant force is zero and the acceleration is zero;
2) If the body moves uniformly accelerated, then the resultant force is not zero;
3) The direction of the resultant force vector always coincides with the direction of acceleration;
4) Be able to write equations of projections of forces acting on a body

A block is a mechanical device, a wheel that rotates around its axis. Blocks can be mobile And motionless.

Fixed block used only to change the direction of force.

Bodies connected by an inextensible thread have equal accelerations.

Movable block designed to change the amount of effort applied. If the ends of the rope clasping the block make equal angles with the horizon, then lifting the load will require a force half as much as the weight of the load. The force acting on a load is related to its weight as the radius of a block is to the chord of an arc encircled by a rope.

The acceleration of body A is half the acceleration of body B.

In fact, any block is lever arm, in the case of a fixed block - equal arms, in the case of a movable one - with a ratio of shoulders of 1 to 2. As for any other lever, the following rule applies to the block: the number of times we win in effort, the same number of times we lose in distance

A system consisting of a combination of several movable and fixed blocks is also used. This system is called a polyspast.

2.3. Resultant force

2.3.1. Resultant force

A force that replaces the action of several forces on a body is called resultant; the resultant force is equal to the vector sum of the forces applied to a given body:

F → = F → 1 + F → 2 + ... + F → N,

where F → 1, F → 2, ..., F → N are the forces applied to a given body.

It is convenient to find the resultant of two forces graphically using parallelogram rule(Fig. 2.14, a) or triangle (Fig. 2.14, b).

Rice. 2.14

To add several forces (calculate the resultant), use the following algorithm:

1) introduce a coordinate system and record the projections of all forces on the coordinate axes:

F 1 x , F 2 x , ..., F Nx ,

F 1 y, F 2 y, ..., F Ny;

2) calculate the projections of the resultant as the algebraic sum of the projections of forces:

F x = F 1 x + F 2 x + ... + F Nx ,

F y = F 1 y + F 2 y + ... + F Ny ;

3) the modulus of the resultant is calculated using the formula

F = F x 2 + F y 2 .

Let us consider special cases of the resultant.

The force of interaction between the body and the horizontal support, along which the body can move, is calculated as the resultant of the friction force and the support reaction force (Fig. 2.15):

Rice. 2.15

F rise = F tr 2 + N 2,

where F → tr is the force of sliding or static friction; N → - ground reaction force.

The force of interaction between the body and combined support(for example, a seat in a car, airplane, etc.) is calculated as the resultant of the pressure forces on the vertical and horizontal parts of the support (Fig. 2.16):

F → up = F → up + F → up,

where F → hor is the pressure force acting on the body from the horizontal part of the support (numerically equal to the weight of the body); F → vert - pressure force acting on the body from the vertical part of the support (numerically equal to the inertial force).

Rice. 2.16

Special cases of the resultant:

The resultant force of gravity and the Archimedes force is called lifting force (Fig. 2.17):

its module is calculated by the formula

F under = F A − m g ,

where F → A is the Archimedes force (buoyant force); m g → - gravity.

Rice. 2.17

Special cases of the resultant:

If, under the influence of several forces, a body moves uniformly in a circle, then the resultant of all forces applied to the body is centripetal force(Fig. 2.18):

F → c.c = F → 1 + F → 2 + ... + F → N.

where F → 1, F → 2, ..., F → N are the forces applied to the body.

The modulus of the centripetal force directed radially to the center of the circle can be calculated using one of the formulas:

F c.s = m v 2 R, F c.s = m ω 2 R, F c.s = m v ω,

where m is body weight; v is the module of the linear velocity of the body; ω is the magnitude of the angular velocity; R is the radius of the circle.

Rice. 2.18

Example 21. A body weighing 10 kg, completely submerged in water, begins to slide along the bottom of a reservoir, inclined at an angle of 60° to the horizontal. Find the modulus of the resultant of all forces applied to the body if there is no water between the body and the bottom of the reservoir and the friction coefficient is 0.15.

Solution. Since there is no water layer between the body and the bottom, the Archimedes force does not act on the body.

The required quantity is the modulus of the vector sum of all forces applied to the body:

F → = F → tr + m g → + N → ,

where N → is the normal ground reaction force; m g → - gravity; F → tr - friction force. The indicated forces and coordinate system are shown in the figure.

We will calculate the module of the resulting force F in accordance with the algorithm.

1. Let us determine the projections of forces applied to the body onto the coordinate axes:

• on the Ox axis:

friction force projection

F tr x = − F tr = − μ N ;

gravity projection

(m g) x = m g sin 60 ° = 0.5 3 m g ;

ground reaction force projection

N x = 0;

• to the Oy axis:

friction force projection

F tr y = 0 ;

gravity projection

(m g) y = − m g cos 60 ° = − 0.5 m g ;

ground reaction force projection

Ny = N,

where m is body weight; g - free fall acceleration module; µ - friction coefficient.

2. Let us calculate the projections of the resultant onto the coordinate axes, summing up the corresponding projections of the indicated forces:

F x = F tr x + (m g) x = − μ N + 0.5 3 m g ;

F y = (m g) y + N y = − 0.5 m g + N .

There is no movement along the Oy axis, i.e. F y = 0, or, explicitly:

− 0.5 m g + N = 0 .

It follows that

N = 0.5 mg,

which allows us to obtain a formula for calculating the friction force:

F tr = μ N = 0.5 μ m g.

3. The required value of the resultant:

F = F x 2 + F y 2 = | F x | = − 0.5 μ m g + 0.5 3 m g = 0.5 m g (3 − μ) .

Let's do the calculation:

F = 0.5 ⋅ 10 ⋅ 10 (3 − 0.15) = 79 N.

Example 22. A body with a mass of 2.5 kg moves horizontally under the influence of a force equal to 45 N and directed at an angle of 30° to the horizontal. Determine the magnitude of the force of interaction between the body and the surface if the coefficient of sliding friction is 0.5.

Solution. We find the force of interaction between the body and the support as the resultant of the friction force F → tr and the normal reaction force of the support N →:

F → vz = F → tr + N → ,

F rise = F tr 2 + N 2.

The forces applied to the body are shown in the figure.

The modulus of the normal ground reaction force is determined by the formula

N = m g − F sin 30 ° ,

and the modulus of sliding friction force is

F tr = µN,

where m is body weight; g - free fall acceleration module; µ - friction coefficient; F is the modulus of the force causing the movement of the body.

Taking into account the expressions for N and F tr, the formula for calculating the required force takes the form:

F in = (μ N) 2 + N 2 = N μ 2 + 1 = (m g − F sin 30 °) μ 2 + 1.

Let's do the calculation:

F in = (2.5 ⋅ 10 − 45 ⋅ 0.5) (0.5) 2 + 1 ≈ 2.8 N.

Example 23. How many times will the lifting force change if ballast equal to half its mass is dropped from the balloon? The air density is assumed to be 1.3 kg/m 3, the mass of the balloon with ballast is 50 kg. The volume of the balloon is 50 m 3 .

Solution. The lifting force acting on the balloon is the resultant of the Archimedes force F → A and the force of gravity m g →:

F → sub = F → A + m g → ,

whose modulus is determined by the formula

F under = F A − mg ,

where F A = ​​ρ air gV - module of the Archimedes force; ρ air - air density; g - free fall acceleration module; V is the volume of the balloon; m is the mass of the balloon (with or without ballast).

The lift modulus can be calculated using the formulas:

• for balloons with ballast

F under 1 = ρ air g V − m 1 g ,

• for balloons without ballast

F under 2 = ρ air g V − m 2 g,

where m 1 is the mass of the balloon with ballast; m 2 is the mass of the balloon without ballast.

The required ratio of the lift force modules is

F under 2 F under 1 = ρ air V − m 2 ρ air V − m 1 = 1.3 ⋅ 50 − 25 1.3 ⋅ 50 − 50 ≈ 2.7.

Example 24. The modulus of the resultant of all forces acting on the body is equal to 2.5 N. Determine in degrees the angle between the velocity and acceleration vectors if it is known that the modulus of velocity remains constant.

Solution. The speed of the body does not change in magnitude. Consequently, the body has only a normal acceleration component a → n ≠ 0. This case occurs when the body moves uniformly in a circle.

The resultant of all forces applied to the body is the centripetal force and is shown in the figure.

The force, velocity and acceleration vectors have the following directions:

• centripetal force F → c.c is directed towards the center of the circle;
• the normal acceleration vector a → n is directed in the same way as the force;
• the velocity vector v → is directed tangentially to the trajectory of the body.

Therefore, the required angle between the velocity and acceleration vectors is 90°.

>> Resultant force

Submitted by readers from Internet sites

Lesson content lesson notes supporting frame lesson presentation acceleration methods interactive technologies Practice tasks and exercises self-test workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures, graphics, tables, diagrams, humor, anecdotes, jokes, comics, parables, sayings, crosswords, quotes Add-ons abstracts articles tricks for the curious cribs textbooks basic and additional dictionary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in a textbook, elements of innovation in the lesson, replacing outdated knowledge with new ones Only for teachers perfect lessons calendar plan for the year; methodological recommendations; discussion programs Integrated Lessons