The homogeneous lever is balanced using five weights. Balance of bodies

The part of dynamics that studies the conditions of equilibrium of bodies is called static(gr. statos - standing).

The equilibrium of a body is a position that is maintained without additional influences. Based on the equations of dynamics of translational and rotational motions, we can formulate the following conditions for the equilibrium of a rigid body.

A body will not go into rotational motion if for any axis the sum of the moments of forces acting on it is equal to zero:

A body will not begin to move translationally if the sum of the forces acting on it is zero:

Equality (7.8) is called the rule of moments.

The conditions for equilibrium of a body at rest are the simultaneous equality to zero sum of forces And sum of moments of forces, acting on the body.

Let us find out what position the axis of rotation must occupy so that the body fixed to it remains in equilibrium under the action

According to rule of moments for equilibrium it is necessary that the sum of the moments of all these forces about the axis equals zero.

It can be shown that for each body there is a single point where the sum of the moments of gravity about any axis passing through this point is equal to zero. This point is called center of gravity (usually coincides with the center of mass).

Body center of gravity (CG) is the point relative to which the sum of the moments of gravity acting on all particles of the body is equal to zero.

Thus, the force of gravity does not cause the body to rotate around the center of gravity. Therefore, all gravity forces could be replaced by a single force, which attached to this point and is equal to the force of gravity.

A common center of gravity (GCG) is often introduced for an athlete's body.

Basic properties of the center of gravity:

if the body is fixed on an axis passing through the center of gravity, then the force of gravity will not cause it to rotate;

the center of gravity is the point of application of gravity;

in a uniform field of gravity, the center of gravity coincides with the center of mass.

Equilibrium is a body position in which it can remain at rest for as long as desired. At

When a body deviates from its equilibrium position, the forces acting on it change and the balance of forces is disrupted. There are different types of equilibrium (Fig. 7.11) for a body resting on one point:

stable equilibrium (Fig. 7.11, a) - with a small deviation of the body from the equilibrium position, a force arises that tends to return the body to its original state;

indifferent equilibrium (Fig. 7.11, b) - with a small deviation the body remains in the equilibrium position;

unstable equilibrium (Fig. 7.11, c) - with a small deviation of the body from the equilibrium position, forces arise that tend to increase this deviation.

Example indifferent equilibrium is the equilibrium of a body fixed on an axis passing through its center of gravity. If the axis passes through another point and the center of gravity is located higher axis, then it is only possible unstable equilibrium. There will be balance sustainable, if the center of gravity is located below axes.

In a stable equilibrium position, a body has minimal potential energy.

Let us now consider the equilibrium of a body resting not on one point, as in the example with a ball, but on a whole area. In these cases, the condition for stability is as follows: for balance, it is necessary that the vertical drawn through the center of gravity pass inside the area of ​​the body’s support.

Violation of this condition leads to the impossibility of maintaining balance. For example, the cylinder shown in Fig. 7.12, a, must capsize, because the plumb line drawn through the CG passes outside its base.

A standing person maintains balance as long as the plumb line from the GCT is inside the area limited by the edges of his feet, Fig. 7.12, b.

A person sitting on a chair holds his torso upright, Fig. 7.12, c. The GCT of the torso is located inside the body (near the spine, approximately 20 cm above the level of the navel). A plumb line drawn down from the GCP passes through the area of ​​support limited by the feet and legs of the chair. You can sit in this position. However, in order to stand up, a person must move the line of gravity inside the area limited by the feet. To do this, you need to tilt your torso forward and at the same time move your legs back (you can stand up without changing the position of your legs if you tilt forward sharply).

The simplest mechanisms

The action of the simplest mechanisms used to change the magnitude or direction of force is based on the use of the laws of statics.

Lever - a solid body, often in the form of a rod, which can rotate (rotate) around a fixed axis.

Let the axis divide the lever in the ratio L,:L, and two parallel forces F act on it, and F 2 (Fig. 7.13). We will also assume that the force of gravity acting on the lever can be neglected.

Let us determine the position of the rotation axis (O) at which the lever will remain in balance.

When the lever is in balance under the action of two parallel forces, the axis of rotation divides the distance between the points of application of forces into segments inversely proportional to the magnitudes of the forces.

Equilibrium of the lever occurs under the condition that the ratio of the parallel forces applied to its ends is inverse to the ratio of the arms

and the moments of these forces are opposite in sign. Therefore, by applying a small force to the long end of the lever, it is possible to balance the much larger force applied to the short end of the lever. Depending on the relative position of the points of application of forces and the axis, levers of the 1st and 2nd kind are distinguished (Fig. 7.13):

A) Lever of the 1st kind. The forces are located on both sides of the axis. Similar levers are a long pole, with the help of which a heavy stone is lifted (Fig. 7.14.).

b) Lever of the 2nd kind. The forces are located on one side of the support. This type includes, for example, a wheelbarrow (Fig. 7.15), in which the hand force is applied at the “maximum” distance from the wheel axis (maximum shoulder), which makes it possible to transport large loads.

The use of leverage in mechanisms gives a gain in strength, while the same amount is lost in movement. The lever does not give any gain in work.

Many joints operate on the principle of a second-class lever. At the same time, the muscles act on a smaller shoulder lever, fig. 7.16. This leads to loss in power and to gain in movement And speed. As a result, with a relatively small movement, a muscle, link or limb describes a significantly larger trajectory.

This feature in the structure of the musculoskeletal nodes should cause additional complications in central regulation

movements, since an increase in the trajectory of movement of links is combined with a large number of degrees of freedom of mobility inherent in the human body as a kinematic chain.

Balance(French balancier - rocker) - a double-arm lever that performs rocking (oscillating) movements around a fixed axis. It is used in a balancing pendulum used in mechanotherapy.

Block, like the lever, it belongs to the simplest mechanisms, Fig. 7.17. It is made in the form of a disk that rotates freely on an axis. Along the circumference of the disk there is a groove for a chain (rope, thread). Equal tension is used at all points of the chain, which moves without friction.

Fixed block(Fig. 7.17, a) does not give a gain in strength, but allows you to change it direction. Yes, you can lift the load up, acting on the rope with a downward force, which is less tiring: F- R.

Movable block(Fig. 7.17, b) gives two-

For ease of use, a movable block is often used in combination with a fixed one (Fig. 7.17, c).

Block-type devices are used in mechanotherapy for training to facilitate (restore) movements in joints and strengthen muscles.

The simplest mechanisms include inclined plane. When describing the position of the body in this case, a rectangular coordinate system is used, the OX axis of which is directed parallel to the plane, and the OU axis is directed perpendicular to it. On a body located on an inclined plane, Fig. 7.18, gravity acts mg, ground reaction force - N and friction force F The projections of gravity onto the coordinate axes are equal to mg-sina (rolling force) and mgcosa.

When moving down an inclined plane, the rolling force assists the movement and contributes to a significant increase in speed. For a given length of an inclined plane, the rolling force is directly proportional to the height, Fig. 7.19.

An inclined surface is often used in training when performing various exercises, Fig. 7.20.

When recovering from injuries, exercises on a special table, the design of which allows you to change the angle of inclination of its plane to the horizon, are effective, Fig. 7.21.

Changing the angle of inclination and the location of fastening of the fixing belts (at the level of large joints of the legs, lumbar and thoracic spine) allows you to dose the load on the musculoskeletal, cardiovascular and vestibular systems.

Elements of mechanics of the human musculoskeletal system

The human musculoskeletal system consists of interconnected skeletal bones. The bones of the skeleton act as levers that have a fulcrum in the joints or in the external environment and are driven by the traction force generated by the contraction of the muscles attached to the bones.

Lever of the first kind, providing movement or balance of the head in the sagittal plane.

In Fig. Figure 7.22 shows a skull and the forces acting on it.

The axis of rotation (O) passes through the articulation of the skull with the first vertebra. The skull is acted upon by two forces applied on opposite sides of the axis.

The force of gravity (/?) applied to the center of gravity of the skull. The leverage of this force is indicated by the letter b.

Traction strength of muscles and ligaments (F), attached to the occipital bone. The leverage of this force is indicated by the letter A.

Lever equilibrium condition: F- a = Rb. In this case a > b, hence, F < R. Therefore, the lever gives a gain in strength, but a loss in movement.

The human forearm works on the principle of a second-class lever.

In Fig. Figure 7.24 shows the forearm and hand with a load, as well as the forces acting on them.

The axis of rotation (O) is located in the elbow joint. The lever is acted upon by two forces applied on one side of the axis.

The force of gravity (/?), equal to the weight of the load. The leverage of this force is indicated by the letter b.

Muscle traction force (F), transmitted by the biceps. The leverage of this force is indicated by the letter A.

Lever equilibrium condition: F- a = Rb. In this case A< Ь, hence, F > R. Therefore, the lever gives a loss in strength (about 8 times). Is such a device advisable? At first glance, it seems not, since there is a loss in strength. However, according to the “golden rule” of mechanics, a loss in strength is rewarded with a gain in movement: the movement of the hand is 8 times greater

the magnitude of muscle contraction. At the same time, there is an improvement in the speed of movement: the hand moves 8 times faster than the muscle contracts.

Thus, the method of muscle attachment that exists in the human (animal) body provides the limbs with speed of movement, which is more important in the struggle for existence than strength. Man would be an extremely slow creature if his hands were not constructed according to this principle.

Bone traction systems for fractures

When healing broken bones, it is necessary to fix the damaged areas and remove the forces that normally act at the fracture site until it heals. To do this, various combinations of weights and blocks are used.

In Fig. 7.25, and shows an exhaust system using two identical weights and two blocks. In this case, the tension forces are 7", and G 2 are equal. The same conditions can be created in another way (Fig. 7.25, b), using one load and a combination of movable and fixed blocks. In this case, the total force acting on the leg is equal to the vector sum of two tension forces (Fig. 7.25, c).

9 = 20° to horizontal. The remaining angles are indicated in the figure. In this case, the vector sum of the three tension forces, indicated in Fig. 7.26, b, F, has the optimal direction.

In Fig. 7.26a shows the Russell traction system used to fix a fractured femur. This system was obtained by adding to the system shown in Fig. 7.25, two more blocks to ensure connection with the knee. The hip is set at an angle

A lever is a rigid body that can rotate around a fixed support.

Figure 149 shows how a worker uses it as a lifting tool lever crowbar In the first case (a) the worker presses the end of the crowbar B down with a force F, in the second (b) he lifts the end B.

The worker needs to overcome the weight of the load P - a force directed vertically downwards. To do this, he turns the crowbar around an axis passing through the only fixed point of the crowbar - the point of its support 0, Force F, with which the worker acts on lever in both cases, less force P, i.e., the worker is said to gain a gain in power. Thus, with the help of a lever you can lift such a heavy load that cannot be lifted without a lever.

Figure 153 shows a lever whose axis of rotation 0 (fulcrum) is located between the points of application of forces A and B; Figure 154 shows a diagram of this lever. Both forces F1 and F2 acting on the lever are directed in the same direction.

Shortest distance between a point support and a straight line along which The force acting on the lever is called the leverage.

To find the arm of the force, you need to lower the perpendicular from the fulcrum to the line of action of the force. The length of this perpendicular will be the arm of this force. Figure 154 shows that 0A is the arm of force F1, 0B is the arm of force F2.

The forces acting on the lever can rotate it around its axis in two directions: clockwise or counterclockwise. So, force F1 (Fig. 153) rotates the lever clockwise, and the forceF2 rotates it counterclockwise.

The condition under which the lever is in equilibrium under the influence of forces applied to it can be established experimentally. It must be remembered that the result of the action of a force depends not only on its numerical value (modulus), but also on , at what point it is applied to the body and how it is directed.

Various weights are suspended from the lever (Fig. 153) on both sides of the fulcrum so that the lever remains in balance each time. The forces acting on the lever are equal to the weights of these loads. For each case, the force modules and their shoulders are measured. Figure 153 shows that a 2N force balances a 4N force. In this case, as can be seen from the figure, the shoulder of the smaller force is 2 times larger than the shoulder of the greater force.

Based on such experiments, the condition (rule) of lever equilibrium was established: the lever is in equilibrium when the forces acting on it are inversely proportional to the arms of these forces.

This rule can be write it as a formula:

where F1 and F2 are the forces acting on the lever, l1 and l2 are the shoulders of these forces (Fig. 154).

The rule of lever equilibrium was established by Archimedes.

From this rule it is clear that with a smaller force you can balance a larger force with the help of a lever; you just need to select the shoulders of a certain length for this. For example, in Figure 149, and one lever arm is approximately 2 times larger another. This means that by applying a force of, for example, 400 N at point B, a worker can lift a stone of 800 N, i.e., weighing 80 kg. To lift an even heavier load, you need to increase the length of the lever arm on which the worker acts.

Example. What force is required (excluding friction) to lift a 240 kg stone using a lever? The force arm is 2.4 m, the gravity arm acting on the stone is 0.6 m.

Questions.

1. What is a lever?
2. What is called the shoulder of strength?
3. How to find leverage?
4. What effect do forces have on the lever?
5. What is the rule for lever equilibrium?
6. Who established the rule of lever equilibrium?

Exercise.

Place a small support under the middle of the ruler so that the ruler is in balance. Balance coins of 5 and 1 k on the resulting lever. Measure the force arms and check the equilibrium condition of the lever. Repeat the work using 2 and 3 k coins.

Using this lever, determine the mass of the matchbox.

Note. Coins of 1, 2, 3 and 5 k. have masses of 1, 2, 3 and 5 g, respectively.

Lesson topic: Lever equilibrium condition. Problem solving.

Lesson objectives:

Educational: A) transfer of knowledge on the condition of lever equilibrium to solving problems, b) familiarization with the use of simple mechanisms in nature and technology; c) development of information and creative competencies.

Educational: A) education of ideological concepts: cause and effect relationships in the surrounding world, cognition of the surrounding world and man; b) moral education: a sense of comradely mutual assistance, ethics of group work.

Developmental: a) development of skills: classification and generalization, drawing conclusions based on the studied material; b) development of independent thinking and intelligence; V) development of competent oral speech.

Lesson plan:

I. Organizational part (1-2 minutes).

II. Activation of mental activity (7 min).

III. Solving problems of increased complexity (15 min)

IV. Differentiated work in groups (12 min)

V. Test of knowledge and skills (6 min).

VI. Summarizing and completing the lesson (2-3 min).

II.Activation of mental activity

Rice. 1 Fig. 2 Fig. 3

1. Will this lever be in equilibrium (Fig. 1)?

2. How to balance this lever (Fig. 2)?

3.How to balance this lever (Fig. 2)?

III. Solving problems of increased complexity

V.I. By whom No. 521*

Forces of 2 N and 18 N act at the ends of the lever. The length of the lever is 1 m. Where is the fulcrum if the lever is in equilibrium.

Given: Solution:

F 1 =2H F 1 d 1 =F 2 d 2

F 2 =18H d 1 +d 2 =L d 2 =L-d 1

L=1m F1d1=F2 (L-d 1) F 1 d 1 =F 2 L-F 2 d 1

M 1= M 2 F 1 d 1 +F 2 d 1 =F 2 L d 1 (F 1 +F 2) =F 2 L

Find: d 1 =F 2 L/(F 1 +F 2)

d 1 d 2 Answer: d 1 =0.9m; d 2 =0.1m

V.I.Kem No. 520*

Using a system of movable and fixed blocks, it is necessary to lift a load weighing 60 kg. How many movable and fixed blocks must the system consist of so that this load can be lifted by one person applying a force of 65 N?

Given: Solution:

m =60kg. F 1 =P/2 n =5-movable blocks

F =65H F =P/n*2 therefore fixed blocks

To find n P =mg you also need 5, but in general 10.

F=mg/2n

IV.Differentiated work in groups

Group 1

Task. The length of the smaller arm is 5 cm, the larger one is 30 cm. A force of 12 N acts on the smaller arm. What strength should it be applied to the larger arm to balance the lever? (Answer: 2H)

Message. Historical information.

The first simple machines (lever, wedge, wheel, inclined plane, etc.) appeared in ancient times. Man's first tool, the stick, is a lever. A stone ax is a combination of a lever and a wedge. The wheel appeared in the Bronze Age. Somewhat later, an inclined plane began to be used.

Group 2

Task. Forces of 100N and 140N act at the ends of a weightless lever. The distance from the fulcrum to the smaller force is 7 cm. Determine the distance from the fulcrum to the larger force. Determine the length of the lever. (Answer: 5cm; 12cm)

Message

Already in the 5th century BC, the Athenian army (Peloponnesian War) used battering rams - rams, throwing devices - ballistae and catapults. The construction of dams, bridges, pyramids, ships and other structures, as well as craft production, on the one hand, contributed to the accumulation of knowledge about mechanical phenomena, and on the other hand, required new knowledge about them.

Group 3

Task

Riddle: They work hard all the time, they are pressing for something. ??

Group 4

Riddle: Two sisters swayed, sought the truth, and when they achieved it, they stopped.

Group 5

Task

WITH
message. Levers in living nature.

In the skeleton of animals and humans, all bones that have some freedom of movement are levers. For example, in humans - the bones of the arms and legs, lower jaw, skull, fingers. In cats, levers are movable bones; many fish have dorsal fin spines. Lever mechanisms in the skeleton are mainly designed to gain speed while losing strength. Particularly large gains in speed are obtained in insects.

Let's consider the equilibrium conditions of a lever using the example of a skull (skull diagram). Here is the axis of rotation

lever ABOUT passes through the articulation of the skull and the first vertebra. In front of the fulcrum, on a relatively short shoulder, the force of gravity of the head acts R ; behind - traction force F muscles and ligaments attached to the occipital bone.

V. Testing knowledge and skills.

Option-1.

1. The lever is in equilibrium when the forces acting on it are directly proportional to the arms of these forces.

2. A stationary block gives a 2-fold gain in strength.

3. Wedge - a simple mechanism.

4. The moving block converts the force modulo.

5. Units of measurement of moment of force - N * m.

Option-2

1. The lever is in equilibrium when the forces acting on it are inversely proportional to the arms of these forces.

2. A stationary block gives a 4-fold increase in strength.

3. The inclined plane is a simple mechanism.

4. To lift a load weighing 100 N using a moving block, 40 N will be required

5. The equilibrium condition of the lever M clockwise = M counterclockwise.

Option-3.

1. A stationary block does not provide a gain in strength.

2.Simple mechanisms convert force only modulo.

3. To lift a load weighing 60 N using a moving block, 30 N will be required

4.Leverage of force - the distance from the axis of rotation to the point of application of force.

5. The compass is a simple mechanism.

Option-4.

1. The movable block gives a 2-fold increase in strength.

2.Simple mechanisms transform force only in direction.

3. The screw is not a simple mechanism.

4. To lift a load weighing 100 N using a moving block weighing 10 N

50 N will be required.

5.Leverage of force - the shortest distance from the axis of rotation to the line of action of the force.

Option - 5.

1. Moment of force - the product of force and shoulder.

2. Using a moving block, applying a force of 200 N, you can lift a load of -400 N.

3.The leverage of force is measured in Newtons.

4. The gate is a simple mechanism.

5.The fixed block converts the force in direction

VI. Summing up and homework.

In different reference systems, the movement of the same body looks different, and the simplicity or complexity of the description of the movement largely depends on the choice of the reference system. Usually used in physics inertial system reference, the existence of which was established by Newton by summarizing experimental data.

Newton's first law

There is a reference system relative to which a body (material point) moves uniformly and rectilinearly or maintains a state of rest if other bodies do not act on it. Such a system is called inertial.

If a body is stationary or moves uniformly and rectilinearly, then its acceleration is zero. Therefore, in an inertial reference frame, the speed of a body changes only under the influence of other bodies. For example, a soccer ball rolling across a field stops after a while. In this case, the change in its speed is due to influences from the field surface and air.

Inertial reference systems exist countless, because any reference system that moves uniformly rectilinearly relative to an inertial frame is also inertial.

In many cases inertial can be considered a frame of reference associated with the Earth.

4.2. Weight. Strength. Newton's second law. Addition of forces

In an inertial reference frame, the cause of a change in the speed of a body is the influence of other bodies. Therefore, when two bodies interact the speeds of both change.

Experience shows that when two material points interact, their accelerations have the following property.

The ratio of the acceleration values ​​of two interacting bodies is a constant value that does not depend on the conditions of interaction.

For example, when two bodies collide, the ratio of the acceleration values ​​does not depend either on the speeds of the bodies or on the angle at which the collision occurs.

That body which, in the process of interaction, acquires smaller acceleration is called more inert.

Inertia - the property of a body to resist changes in the speed of its movement (both in magnitude and direction).

Inertia is an inherent property of matter. A quantitative measure of inertia is a special physical quantity - mass.

Weight - a quantitative measure of body inertia.

In everyday life, we measure mass by weighing. However, this method is not universal. For example, it is impossible to weigh

The work done by a force can be either positive or negative. Its sign is determined by the magnitude of the angle a. If this angle ostry(the force is directed towards the movement of the body), then the work poloresident At stupid coal A Job negative.

If, when a point moves, the angle A= 90° (the force is directed perpendicular to the velocity vector), then the work is zero.

4.5. Dynamics of motion of a material point along a circle. Centripetal and tangential forces. Leverage and moment of force. Moment of inertia. Equations of rotational motion of a point

In this case, a material point can be considered a body whose dimensions are small compared to the radius of the circle.

In subsection (3.6) it was shown that the acceleration of a body moving in a circle consists of two components (see Fig. 3.20): centripetal acceleration - and I tangential acceleration a x, directed along the radius and tangent

respectively. These accelerations are created by projections of the resultant force onto the radius of the circle and the tangent to it, which are called centripetal force (F) and tangential force (F) accordingly (Fig. 4.5).

Centripetal force is called the projection of the resultant force onto the radius of the circle on which the body is currently located.

Tangential force is the projection of the resultant force onto the tangent to the circle drawn at the point where the body is currently located.

The role of these forces is different. Tangential force provides change quantities speed, and centripetal force causes a change directions movements. Therefore, to describe rotational motion, Newton’s second law is written for centripetal force:

Here T is the mass of the material point, and the magnitude of the centripetal acceleration is determined by formula (4.9).

In some cases, it is more convenient to use a non-centripetal force to describe circular motion { F.J., A moment of force acting on the body. Let us explain the meaning of this new physical quantity.

Let the body rotate around the axis (O) under the influence of a force that lies in the plane of the circle.

The shortest distance from the axis of rotation to the line of action of the force (lying in the plane of rotation) is called shoulder of strength (h).

In symmetrical homogeneous bodies, the CM is always located at the center of symmetry or lies on the axis of symmetry if the figure does not have a center of symmetry. The center of mass can be located both inside the body (disk, triangle, square) and outside it (ring, square, square with a cutout in the center). For a person, the position of the COM depends on the posture adopted. In Fig. 5.3. the position of the CM of the body of a water jumper at various stages of the jump is shown. Depending on the position of the parts of the body relative to each other, its CM is located at different points.

Example 1. Determine the support reactions of the beam (Fig. 1, a ), the ends of which are hinged. The beam is loaded by a couple of forces with a moment of kNm.

Fig.1

Solution. First of all, it is necessary to outline the direction of the support reactions (Fig. 1, b). Since a pair of forces is applied to the beam, it can only be balanced by a pair of forces. Consequently, the reactions of the supports are equal in magnitude, parallel, but opposite in direction. Let's replace the action of the supports with their reactions. Right support A- plane, therefore, the direction of the support reactionR Aperpendicular to this plane, and the support reactionR Bparallel to it and in the opposite direction. The beam is in equilibrium, so the sum of the moments of the pairs of forces applied to it is equal to zero:

where

KN.

Answer: kN.

Example 2. timber AB with a left articulated movable support and a right hinged fixed one, loaded with three pairs (Fig. 1), the moments of which kNm, kNm, kNm . Determine the reactions of the supports.

Fig.1

Solution. 1. Pairs of forces act on the beam, therefore, they can only be balanced by a pair, i.e. at points A And IN from the side of the supports, reactions of the supports must act on the beam, forming a pair of forces. At the point A the beam has a hinged and movable support, which means that the reaction is directed perpendicular to the supporting surface, i.e., in this case, perpendicular to the beam. Let's denote this reactionR Aand point it up. Then at the point IN from the side of the hinged-fixed support, a vertical force also actsR B, but down.

2. Based on the chosen direction of the pair forces (R A, R B) its moment (or ).

3. Let's create an equilibrium equation for pairs of forces:

Substituting the moment values ​​into this equation, we get

From here R A= 5 kN. Since the strengthR A And R Bform a pair, thenR B =R A= 5 kN.

Answer: kN.

Example3 . Load weighing G= 500 N suspended from a rope wound on a drum of radiusr= 10 cm. The drum is held by a pair of forces applied to the ends of a handle lengthl= 1.25 m, fastened to the drum and lying in the same plane with the rope. Determine Axle Reaction ABOUT drum and couple powerF, F", if they are perpendicular to the handle (Fig. 1, a).

Fig.1

Solution. Let's consider the balance of forces applied to the drum: vertical weight force G, a pair made up of forces F And F", and reactionsR o cylindrical hinge ABOUT, the magnitude and line of action of which are unknown. Since a pair of forces can only be balanced by a pair of forces lying in the same plane, then the forces G And R O must constitute a pair of forces, balanced by a pairF, F". Line of action of force G known, reactionR ohinge ABOUT direct parallel to the force G in the opposite direction (Fig. 1, b). The force modules must be equal, i.e.

R o =G= 500 H.

The algebraic sum of the moments of two pairs of forces applied to the drum must be equal to zero:

Where l- couple's shoulder F, F";

r - couple's shoulder G, R o .

Finding force modules F:

N.

Answer: N; N.

Example 4. Beam length AB= 10 m has a hinged-fixed support A and articulated movable support IN with an inclined reference plane making an angle = 30° with the horizon. The beam is acted upon by three pairs of forces lying in the same plane, the absolute values ​​of the moments of which are:

kNm ; kNm ; kNm.

Determine the reactions of the supports (Fig. 1, a).

Fig.1

Solution. Let us consider the equilibrium of forces applied to the beam AB: three pairs of forces, ground reactionR B, directed perpendicular to the reference plane, and the reaction of the supportR A, the line of action of which is unknown (Fig. 1, b). Since the load consists only of pairs of forces lying in the same plane, the reaction of the supports R A And R Bmust form a pair of forces lying in the same plane and balancing the given pairs of forces.

Let's direct the reactionR Aparallel to the reactionR Bso that strength R A And R Bformed a pair of forces directed in the direction opposite to the clockwise rotation (Fig. 1, b).

For four pairs of forces applied to the beam, we use the equilibrium condition for pairs of forces lying in the same plane:

Where

From here

kN.

The plus sign in the answer indicates that the accepted direction of the support reactionsR A And R B matches with true:

kN.

Answer: kN.

Example 5. Two disks with diametersD 1 = 200 mm and D 2 = 100 mm fixed to the shaft (Fig. 1). The shaft axis is perpendicular to their plane. The disks rotate at a constant angular velocity. PowersF 1 and F 2 located in the plane of the disks and directed tangentially to them. Define strengthF 2 if F 1 = 500 N.

Fig.1

Solution.The shaft with disks, according to the conditions of the problem, rotates at a constant angular velocity, therefore, the torques must be balanced, i.e. Since the shaft axis is perpendicular to the plane of action of the forces, then

.

(The minus sign indicates the direction of the moment counterclockwise when viewed along the axis from its positive direction.)

from here

N.

When calculating the strength of shafts, it is necessary to determine the moments of internal forces in sections perpendicular to the shaft axis. The resulting moment of internal forces relative to the longitudinal axis of the shaft is usually called torque and is designated differently from the moments of external forces, which are usually called torques.

Answer: N.

Example6 . To a rectangular parallelepiped, the length of whose edges is A=100 cm,b= 120 cm, With= 160 cm, three mutually balanced pairs of forces are appliedF 1 , F" 1 , F 2 , F" 2 and F 3 , F" 3. The forces of the first pair have a modulusF 1 = F" 1 = 4 N. Determine the modules of the remaining forces (Fig. 1).

Fig.1

Solution. When three pairs of forces that do not lie in the same plane are in equilibrium, the geometric sum of the moments of these pairs must be equal to zero, i.e., the triangle of their moments must be closed:

We build at a point ABOUT the moment of each pair of forces, directing it perpendicular to the plane of action of the pair so that, looking towards it, we see the corresponding pair of forces tending to rotate this plane in the direction opposite to the clockwise rotation:

Moment modules:

Ncm;

We construct a closed triangle of moments of pairs of forces.

From DEOC

From the triangle of moments

Ncm;

Ncm.

Modules of the forces that make up the pairs:

N;

N.

Answer: N; N.

Example 7. The ends of the beam are hinged at points A And IN(Fig. 1, a). Pairs of forces are applied to the beam, the moments of which are equal to kNm; kNm. Beam axis AB coincides with the plane of action of the pair of forces. Distance between supportsl= 3 m. Determine the support reactions of the beam, not taking into account the gravity of the beam.

Fig.1

Solution. Since 2 pairs of forces are applied to the beam, they can only be balanced by a pair of forces. This means that the reactions of the supports are equal in magnitude, parallel, but opposite in direction. We replace the actions of the supports with their reactions (Fig. 1 , b). The beam is in equilibrium, so the sum of the moments of pairs of forces opposite to it is equal to zero:

kN.

Answer: kN.

Example8 . The shaft, on which three gears are mounted, rotates around a fixed axis. PowersF 1 , F 2 and F 3 located in planes perpendicular to the axis of rotation and directed tangentially to the circles of the gears, as shown schematically in Fig. 1. PowersF 2 = 400 H, F 3 = 200 H . Gear diameters = 100 mm, = 200 mm,= 400 mm. Calculate the magnitude of the moments of forces F 1 , F 2 and F 3 relative to the axis of rotation and force modulus F 1 attached to a disk with a diameterD 1 .

Fig.1

Solution. Since the shaft axis is perpendicular to the plane of action of the forces, then:

Nm;

Nm.

(The minus sign for a moment indicates the clockwise direction of the moment when viewed along the axis from its positive direction.)

The torques must be balanced:

Then

Nm;

N.

Answer: Nm, Nm, N × m, N.

Example 9.CargoGcreates pressing force using a leverFper detail A(Fig. 1, a ). Lever arms A= 300 mm,b= 900 mm. Determine the force of gravity of the load if the clamping force is 400 N.

Fig.1

Solution. On the design diagram of the lever (Fig. 1, b) to the point A load weight appliedG, to the point IN– joint reaction force, to the point WITH a reaction force is applied equal in modulus to the clamping forceF(Newton's 3rd law).

Let's create an equilibrium equation for the lever relative to the point IN :

in this case the moment of force relative to the point IN equals 0.

Answer: N.

Example 10. Determine clamping forceFper detail A(Fig. 1, a ), created using a lever and a weightG= 300 H . Lever Ratiob / a = 3.

Fig.1

Solution.Let us consider the equilibrium of the lever. To do this, we replace the action of the supports with their reactions (Fig. 1, b).

Clamping forceFper detail A modulo equal to the reaction force (this follows from Newton’s 3rd law).

Let us write down the equilibrium condition of the lever relative to the point IN :

Answer: N.

Example 11.Three disks are rigidly fixed to the shaft (Fig. 1, a). Drive disk 1 transmits torque Nm. Moment applied to driven disk 2, Nm. Disc diametersD 1 = 0.2 m, D 2 = 0.4 m, D 3 = 0.6 m. Determine the magnitude and direction of the moment on disk 3, provided that the shaft rotates uniformly. Also calculate the circumferential forcesF 1 , F 2 and F 3 , attached to the corresponding discs. These forces are directed tangentially to the circumference of the disk and are located in planes perpendicular to the shaft axis.

Fig.1

Solution. The shaft with disks, according to the conditions of the problem, rotates uniformly, therefore, the torques must be balanced (Fig. 1, b):

, Nm.

Let's determine the circumferential forcesF 1 , F 2 , F 3 :

, , N, kN;

, , N, kN;

, , N, N.

Answer: N × m, N, N, N.

Example 12. To a rod supported at points A And IN (Fig. 1, a), two pairs of forces are applied, the moments of which To Nm and to Nm. Distance A= 0.4 m. Determine the reactions of the stops A And IN, without taking into account the gravity of the rod. The plane of action of the force pairs coincides with the axis of the rod.

Fig.1

Solution. Since only pairs of forces are applied to the rod, they can only be balanced by a pair of forces. This means that the reactions of the supports are equal in magnitude, but opposite in direction (Fig. 1, b).

The rod is in equilibrium, so

, ,

kN,

The minus sign indicates the direction of the moment of force pairs and .

Answer: kN, kN.

Example 13. On the lever at the point WITH force actsF= 250 H (Fig. 1, a ). Determine the force applied to the brake discs at the point A, if the length of the leverC.B.= 900 mm, distanceCD= 600 mm.

Fig.1

Solution.Let us replace the actions of the supports with lever by their reactions (Fig. 1, b). Lever equilibrium equation:

;

N.

The force applied to the brake discs at the point A, is equal in modulus (according to Newton’s third law).

Answer: N.

Example 14. The shoe brake holds the shaft at rest, to which a pair of forces with a torque of Nm is applied. Brake disc diameterD= 400 mm (Fig. 1 , A). Determine with what force the pads must be pressed against the brake disc so that the shaft remains at rest. The coefficient of static friction between the brake disc and pads is assumed to bef = 0,15.

Fig.1

Solution. In order for the shaft to remain at rest, the moments must be equal M and (Fig. 1, b):

where is the moment created by a pair of friction forces.

Let us determine the friction force by knowing the friction coefficientfrest between the brake disc and pads:

Then

N.

Answer: kN.

Example 15. Two disks with diameters ofD 1 = 220 mm and D 2 = 340 mm (Fig. 1, a). To the first disc force applied F 1 = 500 N. The line of action of the force is located in a plane perpendicular to the shaft axis. Determine the magnitude and direction of the force that must be applied to the second disk so that the shaft rotates evenly. Calculate the torques on each disk.

Fig.1

Solution. Disc torques:

(The minus sign for a moment indicates the direction of the moment counterclockwise when viewed along the axis from its positive direction.)

Since the shaft rotates uniformly, the torques must be balanced (Fig. 1, b):

N × m,N × m,

, , N.

The direction of the force is opposite to the direction of the force

Answer: N × m,N × m, N.

Example 16.A load kN, raised using a cable wound on a drum with a diameter of m, is held at rest by a ratchet mechanism consisting of a gear wheel with a design diameter of m and a thrust lever (Fig. 1, a). Neglect the weight of the mechanism parts, as well as friction. Determine the force loading the thrust lever.

Fig.1

Solution.We will consider the equilibrium of the block. An external connection is applied to it - a persistent lever. Let's replace it with a reaction. In this problem there is one unknown, which, according to Newton’s third law, is equal to the reaction (Fig. 1, b).

,

where we have:

, kN.

kN.

Answer: kN.

Example 17.The force applied by a person to the end of the handle of a manual lever press is equal toF= 120 H. Having accepted AC= 220 mm and AB= 40 mm, determine the pressure force of the piston on the pressed material (Fig. 1, a). Fastening at points A And IN articulated. Neglect the weight of the mechanism parts, as well as friction.

Fig.1

Solution. The piston pressure force is equal to the reaction force acting from the piston on the handle (Fig. 1, b). Let's create an equation for the moments of force for the handle:

. N.

Answer: N.

Example 18.In the tape transport mechanism of the device, the tape is kept taut using a double-armed lever ABC(Fig. 1, a) . There is a pressure roller at one end of the lever, the other end is pulled back by a spring band with an elastic force of 4 N. Determine the force of pressure of the roller on the tape, assuming that the common normal at the point of contact is vertical. Accept AB= 50 mm and Sun= 10 mm. Neglect the weight of the mechanism parts, as well as friction.

Fig.1

Solution. On the lever ABC external connections are imposed. Let's get rid of them by replacing their action with reaction forces (Fig. 1, b). In this problem, one unknown is the pressure force of the roller on the tape, which is equal to the reaction force

Let's create an equation for the moments of force:

Where do we get:

N.

Answer: N.

Example 19.A load weighing 950 N is lifted uniformly using a gate consisting of a drum with a diameter of 0.14 m and a handle with a shoulder of 0.4 m (Fig. 1). For a given position of the mechanism, determine the forceF, applied by the worker, considering it to be directed vertically. Neglect the weight of the mechanism parts, as well as friction.

Fig.1

Solution. In this problem, there is one unknown – force (Fig. 1, b). To find it, we write the equation of moments of forces:

, , .

N.

Answer: N.

Example 20.To transfer a homogeneous column AB from a horizontal to a vertical position, one end of it was hooked with a crane cable, and a stop was attached to the other end (Fig. 1, a). Determine the tension force of the cable at the moment the column begins to rise, if its weight is 3 kN and length is 4 m.

Fig.1

Solution. To find the tension force of the cable, we create an equation for the moments of force (Fig. 1, b):

;

KN.

Answer: kN.