Equilibrium of a rigid body in the presence of friction. Equilibrium in the presence of friction forces Equilibrium in the presence of sliding friction

The study of the equilibrium of bodies taking into account sliding friction can be reduced to the consideration of the limit equilibrium, which occurs when the friction force is equal to

In an analytical solution, the reaction of a rough bond is represented by its two components N and

Then they compose the usual equilibrium equations and add to them the equality From the system of equations obtained in this way and determine the required quantities.

If the problem requires determining the equilibrium conditions for all values ​​that the friction force can have, i.e., for , then it can also be solved by considering the limit equilibrium and then reducing the friction coefficient in the resulting result to zero

Let us also note that if in the problem it is necessary to determine the value of the friction force F when equilibrium is not limiting and then, as already noted in § 23, this force F should be considered an unknown quantity and found from the corresponding equations (see the second part of problem 29, and also problems 151, 152, § 130).

In a geometric solution, it is more convenient to represent the reaction of a rough bond by a single force R, which in the limiting equilibrium position is deviated from the normal to the surface by an angle

Problem 29. A weight lies on a horizontal plane (Fig. 77). Determine what force Q, directed at an angle to this plane, must be applied to the load in order to move it from its place if the static coefficient of friction of the load on the plane

Solution. Let us consider the limiting equilibrium of the load. Then forces act on it. Composing the equilibrium conditions in projections on the axis, we obtain:

From the last equation. Then

Substituting this value into the first equation and solving it, we finally find

If a smaller force is applied to the load, for example force N, then the shear force will be ; the maximum friction force that can develop in this case will be . Therefore, the load will remain at rest. In this case, the friction force F holding it in equilibrium will be determined from the equilibrium equation in projection onto the axis and will be equal to the shear force, and not to the force

We draw attention to the fact that in all calculations it should be determined by the formula, finding N from the equilibrium conditions. A mistake that is often made in problems similar to the one solved is that when calculating, the force of pressure on the plane is not equal to the weight of the Load R.

Problem 30. Determine at what values ​​of the angle of inclination a a load lying on an inclined plane remains in equilibrium if its coefficient of friction on the plane is equal to

Solution. Let us first find the limiting equilibrium position at which the angle a is equal to . In this position (Fig. 78), the load is acted upon by the force of gravity V, the normal reaction N and the limiting force of friction. By constructing a closed triangle from the listed forces, we find from it that But on the other hand, Consequently,

If you reduce the resulting equality, the value will also decrease. From here we conclude that equilibrium is also possible at Finally, all values ​​of the angle a at which the load will be in equilibrium will be determined by the inequality

The result obtained in the problem, expressed by equality (a), can be used to experimentally determine the friction coefficient by finding the angle from experiment.

Let us also note that since where is the friction angle, then, therefore, i.e., the largest angle a at which the load lying on the inclined plane remains in equilibrium is equal to the friction angle.

Problem 31. A beam bent at a right angle rests its vertical part on protrusions A and B, the distance between which (vertically) is h (Fig. 79, a). Neglecting the weight of the beam, find at what width d the beam with the load lying on its horizontal part will be in equilibrium at any position of the load. The friction coefficient of the beam on the guides is equal to

Solution. Let us denote the weight of the load by P, and its distance from the vertical part of the beam by I. Let us consider the limiting equilibrium of the beam at which its width Then the forces P, N, F, N, F act on the beam, where are the limiting friction forces. Compiling the equilibrium conditions (29) and taking the moments about the center A, we obtain:

where From the first two equations we find:

Substituting these values ​​into the third equation and reducing by N, we get

If we decrease zero in this equality, then its right side will increase to infinity. Therefore, equilibrium is possible for any value of . In turn has highest value, when So the beam will be in equilibrium at any position of the load (at 10), if the inequality is satisfied. The less friction, the greater d should be. In the absence of friction, equilibrium is impossible, since in this case it turns out

Let us also present a geometric solution to the problem.

With this solution, instead of normal reactions and friction forces, we depict complete reactions at points A and B, which in the limiting position are deviated from the normals by the friction angle (Fig. 79, b). Then three forces RA, RB, P will act on the beam. In equilibrium, the lines of action of these forces must intersect at one point, i.e. at point K, where the forces RA and RB intersect. From here we obtain the obvious (see Fig. 79, b) equality or since . As a result, we find for the same value as in the analytical solution.

The problem gives an example of a self-braking device, often used in practice.

Problem 32. Neglecting the weight of the ladder AB (Fig. 80), find at what values ​​of the angle a, a person can climb the ladder to its end B, if the angle of friction of the ladder with the floor and with the wall is equal

Solution. Let us consider the limiting equilibrium position of the ladder and apply the geometric method to solve it. In the limiting position, the reactions of the floor and steps act on the staircase, deviated from the normals to these planes by the angle of triplicity. The lines of action of the reactions intersect at point K. Consequently, in equilibrium, the third force P acting on the staircase (numerically equal to the weight of a person) must also pass through point K? Therefore, in the position shown in the drawing, a person cannot rise above point D. In order for a person to rise to point B, the lines of action of the forces RA and RB must intersect somewhere on the straight line BO, which is possible only when the force RA is directed along AB, i.e. when the angle

Consequently, a person can climb the ladder to its end, then it forms an angle with the wall that does not exceed the angle of friction of the ladder with the floor. Friction against the wall does not play a role in this case, i.e. the wall can be smooth.

Consider a cylinder (roller) resting on a horizontal plane when a horizontal active force acts on it; In addition to it, gravity forces act, as well as normal reaction and friction force. As experience shows, with a sufficiently small force the cylinder remains at rest. But this fact cannot be explained if we are satisfied with the introduction of the forces shown in Fig. According to this scheme, equilibrium is impossible, since the main moment of all forces acting on the cylinder is nonzero and one of the equilibrium conditions is not satisfied.

The reason for the discrepancy that has emerged is that in our reasoning we continue to use the idea of ​​an absolutely rigid body and assume that the cylinder touches the surface along a generatrix. To eliminate the noted discrepancy between theory and experience, it is necessary to abandon the hypothesis of an absolutely rigid body and take into account that in reality a cylinder and a plane near the point WITH are deformed and there is a certain contact area of ​​finite width. As a result, in its right part the cylinder is pressed harder than in the left, and the full reaction applied to the right of the point WITH(dot ).

The now obtained diagram of the acting forces is statically satisfactory, since the moment of the couple can be balanced by the moment of the couple. Assuming the deformation to be small, let us replace this system of forces with the system shown in Fig. Unlike the first scheme, a pair of forces with a moment is applied to the cylinder

. (6.11) This moment is called rolling friction moment .

Let's create the equilibrium equations for the cylinder:

The first two equations give , , and from the third equation we can find . Then from (6.11) we determine the distance between the points WITH And :

. (6.13) As can be seen, as the modulus of the active force increases, the distance increases. But this distance is related to the contact surface area and, therefore, cannot increase indefinitely. This means that a state will arise where an increase in strength will lead to an imbalance. Let us denote the maximum possible value by the letter . It has been experimentally established that the value is proportional to the radius of the cylinder and is different for different materials.

Therefore, if there is an equilibrium, then the condition is satisfied

The quantity is called rolling friction coefficient ; it has the dimension of length.

Condition (6.14) can also be written in the form

or, taking into account (6.12),

It is obvious that the maximum rolling friction moment is proportional to the normal pressure force.

The reference tables show the ratio of the rolling friction coefficient to the cylinder radius for various materials.

Problem 6.8. There is a cylinder on an inclined plane. Find at what angles of inclination of the plane to the horizon the cylinder will be in equilibrium if is the radius of the cylinder, is the sliding friction coefficient, and is the rolling friction coefficient. , then inequality (6.16) will be violated and the cylinder will begin to slide.

3.4.1 Equilibrium of a rigid body in the presence of sliding friction

Sliding friction is the resistance that occurs during the relative sliding of two bodies in contact.

The magnitude of the sliding friction force is proportional to the normal pressure of one of the contacting bodies on the other:

The reaction of a rough surface is deviated from the normal by a certain angle φ (Fig. 3.7). The largest angle that the total reaction of a rough bond makes with the normal to the surface is called the friction angle.

where is the static friction coefficient.

This coefficient is usually greater than the coefficient of friction during movement.

From Fig. 3.7 it is clear that the friction angle is equal to the value

. (3.26)

Equality (3.26) expresses the relationship between the friction angle and the friction coefficient.

The technique for solving statics problems in the presence of friction remains the same as in the case of the absence of friction, i.e., it comes down to compiling and solving equilibrium equations. In this case, the reaction of a rough surface should be represented by two components - the normal reaction and the friction force.

It should be borne in mind that in such problems the calculation is usually carried out for the maximum value of the friction force, which is determined by formula (3.25).

Example 3.6:

Weight A weight Q lies on a rough plane inclined to

horizontal at an angle α, and is held by a thread wound on a block step of radius R. At what weight R load B, the system will be in equilibrium if the coefficient of sliding friction of the load on the plane is equal to f, and the radius of the smaller block step (Fig. 3.8).

Let us consider the equilibrium of the load B, which is acted upon by the force of gravity and the reaction of the thread, and numerically (Fig. 3.8, a). The force of gravity, the reaction of the thread, the normal reaction of the inclined plane and the force of friction act on the load A. Since the radius r the smaller stage of the block is half the size of the larger stage, then in the equilibrium position, or

Let us consider the case in which there is equilibrium of load A, but in such a way that the increase in gravity P load B will cause load A to move upward (Fig. 3.8, b). In this case, the friction force is directed down the inclined plane, and . Let us select the x and y axes indicated in the figure and draw up two equilibrium equations for a system of converging forces on the plane:

(3.27)

We get that , then the friction force .

Let us substitute the values ​​and into equality (3.27), and find the value R:

Now consider the case when there is equilibrium of load A, but in such a way that the decrease in gravity R load B will cause load A to move downward (Fig. 3.8, c). Then the friction force will be directed upward along the inclined plane. Since the value N does not change, then it is enough to create one equation in projection onto the x-axis:

. (3.29)

Substituting the values ​​and into equality (3.29), we obtain that

Thus, the equilibrium of this system will be possible under the condition

3.4.2. Equilibrium of a rigid body in the presence of rolling friction

Rolling friction is the resistance that occurs when one body rolls over the surface of another.

An idea of ​​the nature of rolling friction can be obtained by going beyond the statics of a rigid body. Consider a cylindrical roller of radius R and weight R resting on a horizontal plane. Let us apply a force to the roller axis that is less than the friction force (Fig. 3.9, a). Then the friction force, numerically equal to , prevents the cylinder from sliding along the plane. If a normal reaction is applied at point A, then it will balance the force, and the forces form a pair that causes the cylinder to roll even at a low force value S.

In fact, due to the deformations of the bodies, their contact occurs along a certain area AB (Fig. 3.9, b). When a force is applied, the pressure intensity at point A decreases, and at point B increases. As a result, the normal reaction shifts towards the force by an amount k, which is called the rolling friction coefficient. This coefficient is measured in units of length.

In the ideal equilibrium position of the roller, two mutually balanced pairs will be applied to it: one pair of forces with a moment and the second pair of forces keeping the roller in balance. The moment of the couple, called the rolling friction moment, is determined by the formula

From this equality it follows that in order for pure rolling to take place (without sliding), it is necessary that the rolling friction force was less than the maximum sliding friction force: , where f- coefficient of sliding friction. Thus, clean rolling is possible under the condition.

It is necessary to distinguish the direction of displacement of the point of application of the normal reaction of the driving and driven wheels. For the drive wheel, the deformation roller, which causes a displacement of the point of application of the normal reaction of the plane, is located to the left of its center C if the wheel moves to the right. Therefore, for this wheel, the direction of the friction force coincides with the direction of its movement (Fig. 3.10, a). In the driven wheel, the deformation roller is shifted relative to center C in the direction of movement. Consequently, the friction force in this case is directed in the direction opposite to the direction of movement of the wheel center.

Example 3.7:

Weight cylinder R=10 N and radius R= 0.1 m is located on a rough plane inclined at an angle α = 30˚ to the horizontal. A thread is tied to the axis of the cylinder, thrown over a block and carrying a load B at the other end. At what weight Q the load will not roll into the cylinder if the rolling friction coefficient is equal to k= 0.01 m (Fig. 3.11, a)?

Let us consider the equilibrium of the cylinder in two cases. If the magnitude of the force Q has the smallest value, then the cylinder can move down the inclined plane (Fig. 3.11, b). The weight of the cylinder and the tension of the thread are applied to the cylinder. In this case, the normal reaction of the inclined plane will be shifted by a distance k to the left of a perpendicular dropped from the center of the cylinder onto an inclined plane. The friction force is directed along the inclined plane opposite to the possible movement of the center of the cylinder.

Rice. 3.11

To determine the value, it is enough to create an equilibrium equation relative to the point WITH. When calculating the moment of force about this point, we will decompose the force into components: the component is perpendicular to the inclined plane, and the component is parallel to this plane. The moment of force and relative to point C are equal to zero, since they are applied at this point:

Where

In the second case, when the force Q reaches its maximum value, it is possible to move the center of the cylinder up the inclined plane (Fig. 3.11, c). Then the forces will be directed similarly to the first case. The reaction of the inclined plane will be applied at a point and displaced by a distance k to the right along an inclined plane. The friction force is directed opposite to the possible movement of the center of the cylinder. Let's create an equation of moments about the point.

Friction is the resistance that occurs when one body tries to move along the surface of another.

Depending on the nature of the movement (whether the body slides or rolls), two types of friction are distinguished: sliding friction and rolling friction.

If two bodies I And II(Fig. 1.48) interact with each other, touching at point A, then there is always a reaction A, acting, for example, from the body II and attached to the body I, can be decomposed into two components: A, directed along the common normal to the surface of the contacting bodies at the point A, And A, lying in the tangent plane. Component A called normal reaction, force A called sliding friction force - it prevents the body from sliding I over the body II. In accordance with axiom 6 (I. Newton’s third law) on the body II from the body side I a reaction force of equal magnitude and opposite direction acts. Its component perpendicular to the tangent plane is called the normal pressure force. As noted earlier, the friction force A is equal to zero if the contacting surfaces are perfectly smooth. In real conditions, surfaces are rough, and in many cases the friction force cannot be neglected.

Numerous studies have shown that if a body is at rest, the friction force is determined only by the magnitude and direction of the active forces applied to this body. But the friction force cannot exceed a certain fixed value, which coincides with the maximum friction force. That is, if the body is in equilibrium, then

T≤T max (1.61)

Maximum friction force T max depends on the properties of the materials from which the bodies are made, their condition (for example, on the nature of surface treatment), as well as on normal pressure . Experience shows that the maximum friction force is approximately proportional to normal pressure:

T max = f N.(1.62)

This relationship is called the Amonton-Coulomb law.

Dimensionless coefficient f called friction coefficient slip. As follows from experience, its value within a wide range does not depend on the area of ​​the contacting surfaces, but depends on the material and the degree of roughness of the contacting surfaces. Friction coefficient values ​​are determined empirically and can be found in reference tables.

Thus, inequality (1.61) can be written as

T≤f·N. (1.63)

The case of strict equality in (1.63) corresponds to maximum value friction forces. This means that the friction force can be calculated using the formula

T=f·N. (1.64)

only in cases where it is known in advance that a critical incident is occurring. In all other cases, the friction force should be determined from the equilibrium equations.

Corner φ between limit reaction and the normal to the surface is called friction angle(Fig. 1.49,a).

It is easy to show that

tan φ = f.(1.65)

Therefore, instead of the friction coefficient, you can set the friction angle (both values ​​are given in the reference tables).

Depending on the action of active forces, the direction of the limiting reaction may change. Geometric locus of all possible directions of limiting reaction a cone of friction forms a conical surface (Fig. 1.49b). If the friction coefficient f is the same in all directions, then the cone of friction will be circular. In cases where the coefficient of friction depends on the direction possible movement body, the cone of friction will not be circular.

It is easy to show that if the resultant of the active forces is located inside the friction cone, then the body will be in equilibrium, and by increasing the modulus of the resultant in this case it is impossible to disturb the equilibrium of the body. In order for a body to start moving, it is necessary (and sufficient) that the resultant of the active forces F be outside the friction cone.

If the body under study does not slide, but rolls along a certain surface (Fig. 1.50), then it is convenient to represent the resistance to movement as a pair of forces with a moment:

M T =δN. (1.66)

This moment is called rolling friction moment. Magnitude δ called rolling friction coefficient, it has the dimension of length. It has been experimentally established that the value δ is proportional to the radius of the cylinder and varies for different materials.

The reference tables show the ratio of the rolling friction coefficient to the cylinder radius:

λ = δ/R,

for various materials.

If the active forces applied to the body are insufficient to make it roll, that is, equilibrium takes place, then the rolling friction moment will be determined by the expression:

M T ≤ δN.(1.67)

Size M T in this case should be determined from the equilibrium equations.

When solving problems on rolling friction, it is necessary to take into account that pure rolling is possible only in the absence of slipping between the surfaces of bodies. This happens if the friction force between the bodies is strictly less than the maximum friction force, that is:

T (1.68)