2.2.4. Friction force

The force of friction acts not only on a moving body, but also on a body at rest, if there are forces that tend to break this rest. A body rolling on a support is also subjected to frictional force.

static friction force numerically equal to the component of the force directed along the surface on which the given body is located, and tending to move it from its place (Fig. 2.7):

F tr.pok \u003d F x.

Rice. 2.7

When the specified component reaches a certain critical value (F x = F crit), the body begins to move. The critical value of the force, which corresponds to the beginning of the movement, is determined by the formula

F x \u003d F crit \u003d µ until N,

where µ so - coefficient of static friction; N is the modulus of the force of the normal reaction of the support (this force is numerically equal to the weight of the body).

At the moment of the beginning of the movement, the static friction force reaches its maximum value:

F tr. until max = μ until N .

sliding friction force is constant and is determined by the product:

F tr.sk = µ sk N ,

where µ sk - coefficient of sliding friction; N is the modulus of the force of the normal reaction of the support.

When solving problems, it is considered that the coefficients of static friction µ so and sliding µ sk are equal to each other:

µ until = µ sk = µ.

On fig. 2.8 shows a graph of the dependence of the magnitude of the friction force F tr on the projection of the force F x tending to move the body onto an axis directed along the surface of the intended movement.

Rice. 2.8

In order to determine whether this body will be in at rest or moving under the action of an applied force of a certain magnitude and direction, it is necessary:

F crit = µN,

where µ is the coefficient of friction; N is the modulus of the force of the normal reaction of the support;

3) compare the values ​​of F crit and F x:

• if F x > F crit, then the body moves under the action of the applied force; in this case, the sliding friction force is calculated as

F tr.sk = µN ;

• if F x< F крит, то тело покоится под действием приложенной силы; в этом случае сила трения покоя рассчитывается как

F tr.pok \u003d F x.

Module rolling friction force F roll is proportional to the rolling friction coefficient µ roll, the modulus of the force of the normal reaction of the support N and is inversely proportional to the radius R of the rolling body:

F tr. qual = μ qual N R .

Example 13. A force of 25 N directed along the surface is applied to a body with a mass of 6.0 kg lying on a horizontal surface. Find the friction force if the coefficient of friction is 0.5.

Solution. Let us estimate the magnitude of the force capable of causing the motion of the body, according to the formula

F cr = µN,

where µ is the coefficient of friction; N is the modulus of the normal reaction force of the support, numerically equal to the weight of the body (P = mg).

The magnitude of the critical force sufficient to start the motion of the body is

F cr \u003d μ m g \u003d 0.5 ⋅ 6.0 ⋅ 10 \u003d 30 N.

The projection of the force applied to the body in the horizontal direction on the axis of the proposed movement Ox (see figure) is equal to

F x \u003d F \u003d 25 N.

Fx< F кр,

those. the magnitude of the force applied to the body is less than the magnitude of the force capable of causing its movement. Therefore, the body is at rest.

The desired friction force - the rest friction force - is equal to the external horizontal force tending to break this peace:

F tr.pok \u003d F x \u003d 25 N.

Example 14. The body is on an inclined plane with an angle at the base of 30°. Calculate the friction force if the coefficient of friction is 0.5 3 . Body weight is 3.0 kg.

Solution. The arrow in the figure shows the direction of the proposed movement.

Let us find out whether the body will remain at rest or whether it will start to move. To do this, we calculate the value of the critical force that can cause movement, i.e.

F cr = µN,

where µ is the coefficient of friction; N = mg  cos α is the magnitude of the normal reaction force of the inclined plane.

The calculation gives the value of the specified force:

F cr \u003d μ m g cos 30 ° \u003d 0.5 3 ⋅ 3.0 ⋅ 10 ⋅ 3 2 \u003d 22.5 N.

From a state of rest, the body seeks to bring the projection of gravity onto the axis Ox, the value of which is

F x = mg  sin 30° = 15 N.

Thus, there is an inequality

Fx< F кр,

those. the projection of the force seeking to cause the body to move is less than the magnitude of the force capable of doing so. Therefore, the body remains in a state of rest.

The desired force - the static friction force - is equal to

F tr \u003d F x \u003d 15 N.

Example 15. The puck is located on the inner surface of the hemisphere at a height of 10 cm from the bottom point. The radius of the hemisphere is 50 cm. Calculate the coefficient of friction between the washer and the sphere if the indicated height is known to be the maximum possible.

Solution. Let us illustrate the condition of the problem with a figure.

The washer, according to the condition of the problem, is at the maximum possible height. Consequently, the static friction force acting on the washer has a maximum value coinciding with the projection of gravity on the axis Ox:

F tr. until max = F x ,

where F x = mg  cos α is the modulus of the projection of gravity onto the axis Ox ; m is the mass of the washer; g - free fall acceleration modulus; α is the angle shown in the figure.

The maximum force of static friction coincides with the force of sliding friction:

F tr. until max = F tr. ck,

where F tr.sk \u003d µN - the modulus of the sliding friction force; N = mg  sin α is the magnitude of the force of the normal reaction of the surface of the hemisphere; µ is the coefficient of friction.

We determine the friction coefficient by writing the indicated equality in explicit form:

mg  cos α = µmg  sin α.

It follows that the desired friction coefficient is determined by the tangent of the angle α:

The indicated angle is determined from the additional construction:

tg α = R − h 2 h R − h 2 ,

where h is the maximum height at which the puck can be; R is the radius of the hemisphere.

The calculation gives the value of the tangent:

tan α = 0.5 − 0.1 2 ⋅ 0.1 ⋅ 0.5 − (0.1) 2 = 4 3

and allows you to calculate the desired coefficient of friction.

Chapter 15

15.3. Theorem on the change in the energy of a kinetic point and solid body during forward movement.

15.3.1. What work is done by the forces acting on a material point if its kinetic energy decreases from 50 to 25 J? (Answer -25)

15.3.2. Free fall material point mass m starts from rest. Neglecting air resistance, determine the path traveled by the point at the time when it has a speed of 3 m/s. (Answer 0.459)

15.3.3. A material point of mass m = 0.5 kg is thrown from the Earth's surface with an initial velocity v o \u003d 20 m / s and in position M has a speed v= 12 m/s. Determine the work of gravity when moving a point from position M o to position M (Answer -64)

15.3.4. A material point of mass m is thrown from the Earth's surface at an angle α = 60° to the horizon with initial speed v 0 = 30 m/s. Determine the maximum height h of the point. (Answer 34.4)

15.3.5. A body of mass m = 2 kg from a push rises along an inclined plane with an initial speed v o = 2 m/s. Determine the work done by gravity on the path traveled by the body to a stop. (Answer -4)

15.3.6. A material point M of mass m, suspended on a thread of length OM = 0.4 m to a fixed point O, is retracted at an angle α = 90° from the equilibrium position and released without initial velocity. Determine the speed of this point during its passage through the equilibrium position. (Answer 2.80)

15.3.7. The swing cabin is suspended on two rods with a length l= 0.5 m. Determine the speed of the cabin when it passes the lower position, if at the initial moment the rods were deflected by an angle φ = 60° and released without initial velocity. (Answer 2.21)

15.3.8. A material point M with mass m moves under the action of gravity along the inner surface of a half-cylinder of radius r = 0.2 m. Determine the speed of the material point at point B on the surface if its speed at point A is zero. (Answer 1.98)

15.3.9. On the wire ABC, located in a vertical plane and bent in the form of arcs of circles of radii r 1, = 1 m, r 2 = 2 m, a ring D of mass m can slide without friction. Determine the speed of the ring at point C if its speed at point A is zero. (Answer 9.90)

15.3.10. A body of mass m = 2 kg moves along a horizontal plane, to which an initial speed was given v 0 = 4 m/s. Before stopping, the body traveled a distance equal to 16 m. Determine the modulus of the sliding friction force between the body and the plane. (Answer 1)

15.3.11. A body of mass m = 100 kg begins to move from a state of rest along a horizontal rough plane under the action of a constant force F. Having traveled a distance of 5 m, the speed of the body becomes equal to 5 m/s. Determine the modulus of force F if the sliding friction force F tr \u003d 20 N. (Answer 270)

15.3.12. A hockey player, being at a distance of 10 m from the goal, with a stick informs the puck lying on the ice, the speed is 8 m/s. The puck, sliding on the surface of the ice, flies into the gate at a speed of 7.7 m/s. Determine the coefficient of sliding friction between the puck and the ice surface.
(Answer 2.40 10 -2)

15.3.13. A body of mass m = 1kg descends down an inclined plane without initial velocity. Determine the kinetic energy of the body at the moment when it has traveled a distance of 3 m, if the coefficient of sliding friction between the body and the inclined plane f= 0.2. (Answer 9.62)

15.3.14. A load of mass m descends down an inclined plane without initial velocity. What speed v will the load have when it has traveled a distance of 4 m from the start of movement if the coefficient of sliding friction between the load and the inclined plane is 0.15? (Answer 5.39)

15.3.15. A spring 2 is attached to the slider 1 with a mass m = 1 kg. The spring is compressed from a free state by 0.1 m, after which the load is released without initial speed. Determine the stiffness of the spring if the load, having traveled a distance of 0.1 m, acquires a speed of 1 m/s.
(Answer 100)

The force of friction () is the force arising from the relative motion of bodies. It has been empirically established that the force of sliding friction depends on the force of mutual pressure of bodies (reaction of support) (N), materials of surfaces of rubbing bodies, and speeds of relative motion.

DEFINITION

Physical quantity, which characterizes friction surfaces, is called coefficient of friction. Most often, the coefficient of friction is denoted by the letters k or.

In the general case, the coefficient of friction depends on the speed of the bodies relative to each other. It should be noted that the dependence is usually not taken into account and the coefficient of sliding friction is considered constant. In most cases, the force of friction

The coefficient of sliding friction is a dimensionless quantity. The coefficient of friction depends on: the quality of surface treatment, rubbing bodies, the presence of dirt on them, the speed of movement of bodies relative to each other, etc. The coefficient of friction is determined empirically (experimentally).

The coefficient of friction, which corresponds to the maximum force of static friction, is in most cases greater than the coefficient of sliding friction.

For more pairs of materials, the value of the friction coefficient is not more than unity and lies within

The value of the coefficient of friction of any pair of bodies, between which the friction force is considered, is influenced by pressure, the degree of contamination, the surface area of ​​the bodies, and other things that are usually not taken into account. Therefore, those values ​​of the coefficients of friction forces, which are indicated in the reference tables, fully coincide with reality only under the conditions in which they were obtained. Consequently, the values ​​of the coefficients of the friction forces cannot be considered unchanged for the same de pair of rubbing bodies. So, there are coefficients of thorns for dry surfaces and surfaces with lubrication. For example, the coefficient of sliding friction for a body made of bronze and a body made of cast iron, if the surfaces of the materials are dry, is For the same pair of materials, the friction coefficient of sliding in the presence of

Examples of problem solving

EXAMPLE 1

Exercise	A thin metal chain lies on a horizontal table (Fig. 1). Its length is , mass . The end of the chain hangs over the edge of the table. If the length of the hanging part of the chain is a fraction of the length of the entire chain, it begins to slide down the table. What is the friction coefficient of the chain on the table, if the chain is considered uniform in length?
Solution	The chain moves under the influence of gravity. Let the force of gravity acting on a unit length of the chain be . In this case, at the moment of the beginning of sliding, the force of gravity, which acts on the overhanging part, will be: Before sliding, this force is balanced by the friction force that acts on the part of the chain that lies on the table: Since the forces are balanced, we can write ():
Answer

EXAMPLE 2

Exercise	What is the coefficient of friction of the body on an inclined plane, if the angle of inclination of the plane is and its length is . The body moves along the plane with constant acceleration during the time t.
Solution	In accordance with Newton's second law, the resultant of forces applied to a body moving with acceleration is: In projections on the X and Y axes of equation (2.1), we obtain:

If the bar is pulled with a dynamometer at a constant speed, then the dynamometer shows the modulus of sliding friction force (F tr). Here, the elastic force of the dynamometer spring balances the force of sliding friction.

On the other hand, the force of sliding friction depends on the force of the normal reaction of the support (N), which arises as a result of the action of the body weight. The greater the weight, the greater the force of the normal reaction. And the greater the normal reaction force, the greater the friction force. There is a direct proportional relationship between these forces, which can be expressed by the formula:

Here μ is coefficient of friction. It shows exactly how the force of sliding friction depends on the force of the normal reaction (or, one might say, on the weight of the body), what proportion of it it is. The coefficient of friction is a dimensionless quantity. For different pairs of surfaces, μ has a different value.

So, for example, wooden objects rub against each other with a coefficient of 0.2 to 0.5 (depending on the type of wooden surfaces). This means that if the force of the normal reaction of the support is 1 N, then during movement the sliding friction force can be in the range from 0.2 N to 0.5 N.

From the formula F tr \u003d μN, it follows that knowing the forces of friction and normal reaction, it is possible to determine the coefficient of friction for any surfaces:

The strength of the normal support reaction depends on the weight of the body. It is equal to it in modulus, but opposite in direction. Body weight (P) can be calculated by knowing the mass of the body. Thus, if we do not take into account the vectorial nature of quantities, we can write that N = P = mg. Then the coefficient of friction is found by the formula:

μ = F tr / (mg)

For example, if it is known that the friction force of a body with a mass of 5 kg moving along the surface is 12 N, then you can find the coefficient of friction: μ = 12 N / (5 kg ∙ 9.8 N/kg) = 12 N / 49 N ≈ 0.245.