Study of the dependence of friction force on the surface area of ​​contact between bodies

Let's explore what the friction force depends on. To do this, we will use a smooth wooden board, a wooden block and a dynamometer.

Figure 1.

First, let's check whether the friction force depends on the surface area of ​​contact between the bodies. Place the block on a horizontal board with the edge with the largest surface area. Having attached a dynamometer to the block, we will gradually increase the force directed along the surface of the board and notice the maximum value of the static friction force. Then we place the same block on another face with a smaller surface area and again measure the maximum value of the static friction force. Experience shows that the maximum value of the static friction force does not depend on the surface area of ​​contact between the bodies.

By repeating the same measurements with a uniform movement of the block over the surface of the board, we are convinced that the sliding friction force also does not depend on the surface area of ​​contact between the bodies.

Study of the dependence of friction force on pressure force

Let's place a second block of the same type on the first block.

Figure 2.

By this we will increase the force perpendicular to the contact surface of the body and the table (it is called the pressure force~$\overline(P)$). If we now measure the maximum static friction force again, we will see that it has doubled. Having placed a third on two bars, we find that the maximum static friction force has increased three times.

Based on such experiments, we can conclude that the maximum value of the modulus of the static friction force is directly proportional to the pressure force.

The interaction between the body and the support causes deformation of both the body and the support.

The elastic force $\overline(N)$ arising as a result of the deformation of the support and acting on the body is called the support reaction force. According to Newton's third law, the pressure force and the support reaction force are equal in magnitude and opposite in direction:

Figure 3.

Therefore, the previous conclusion can be formulated as follows: the modulus of the maximum static friction force is proportional to the support reaction force:

The Greek letter $\mu$ denotes the coefficient of proportionality, called the coefficient of friction (rest or sliding, respectively).

Experience shows that the modulus of the sliding friction force $F_(mp) $, as well as the modulus of the maximum static friction force, is proportional to the modulus of the support reaction force:

The maximum value of the static friction force is approximately equal to the sliding friction force, and the coefficients of static and sliding friction are also approximately equal.

The dimensionless proportionality coefficient $\mu$ depends on:

• from the nature of rubbing surfaces;
• on the condition of the rubbing surfaces, in particular on their roughness;
• in the case of sliding, the coefficient of friction is a function of speed.

Example 1

Determine the minimum braking distance of a car that starts braking on a horizontal section of the highway at a speed of $20$ m/s. The friction coefficient is 0.5.

Given: $v=20$ m/s, $\mu =0.5$.

Find: $S_(\min ) $-?

Solution: The braking distance of the car will have a minimum value at the maximum value of the friction force. The modulus of the maximum value of the friction force is equal to:

\[(F_(mp))_(\max ) =\mu mg\]

The force vector $F_(mp) $during braking is directed opposite to the velocity vectors $\overline(v)_(0) $and displacement $\overline(S)$.

In case of rectilinear uniformly accelerated motion, the projection of the displacement $S_(x) $ of the car onto the axis parallel to the velocity vector $\overline(v)_(0) $ of the car is equal to:

Moving on to the modules of quantities, we get:

The time value can be found from the condition:

\ \

Then for the displacement module we get:

$a=\frac((F_(mp))_(\max ) )(m) =\frac(\mu mg)(m) =\mu g$, then

$S_(\min ) =\frac(v_(0) ^(2) )(2\mu g) \approx 40$m.

Answer: $S_(\min ) =40$ m.

Example 2

What force must be applied in the horizontal direction to a diesel locomotive weighing $8$t to reduce its speed by $0.3$ m/s in $5$ seconds? The coefficient of friction is $0.05.$

Given: $m=8000$ kg, $\Delta v=0.3$ m/s, $\mu =0.05$.

Find: $F$-?

Figure 4.

Let us write down the equation of motion of the body:

Let's project forces and acceleration onto the x-axis:

Since $F_(mp) =\mu mg$, and $a=\frac(v-v_(0) )(t) =\frac(\Delta v)(t) $, we get:

$F=m(\frac(\Delta v)(t) -\mu g)=3440$Н

Purpose of the work: get acquainted with the phenomenon of rolling friction, determine the rolling friction coefficient of a four-wheeled cart..

Equipment: a trolley as a model of a carriage, a horizontal rail track with a set of photocells, a stopwatch, a set of weights.

THEORETICAL INTRODUCTION

Rolling friction force is a motion resistance force tangent to the contact surface that occurs when cylindrical bodies roll.

When a wheel rolls on a rail, deformation occurs in both the wheel and the rail. Due to the non-ideal elasticity of the material, processes of plastic deformation of the microtubercles, surface layers of the wheel and rail occur in the contact zone. Due to residual deformation, the level of the rail behind the wheel turns out to be lower than in front of the wheel and the wheel constantly rolls onto the bump when moving. In the outer part of the contact zone, partial slipping of the wheel along the rail occurs. In all these processes, work is done by the rolling friction force. The work of this force leads to the dissipation of mechanical energy, its transformation into heat, therefore the rolling friction force is a dissipative force.

In the central part of the contact zone, another tangential force arises - this is the force of static friction or adhesion force wheel and rail material. For the driving wheel of a locomotive, the adhesion force is the traction force, and when braking with a shoe brake, it is the braking force. Since there is no movement of the wheel relative to the rail in the center of the contact zone, no work is done by the adhesion force.

The distribution of pressure on the wheel from the rail side turns out to be asymmetrical. There is more pressure in the front and less in the back (Fig. 1). Therefore, the point of application of the resultant force on the wheel is shifted forward by some small distance b relative to the axis . Let us imagine the force of the rail on the wheel in the form of two components. One is directed tangentially to the contact zone, it is the adhesion force F clutch. Another component Q directed normal to the contact surface and passes through the wheel axis.

Let us, in turn, expand the normal pressure force Q into two components: strength N, which is perpendicular to the rail and compensates for gravity and force F quality, which is directed along the rail against the movement. This force prevents the wheel from moving and is the rolling friction force. Pressure force Q does not create any torque. Therefore, the moments of its component forces relative to the wheel axis must compensate each other: . Where . Rolling friction force proportional to force N, acting on the wheel perpendicular to the rail:

. (1)

Here – rolling friction coefficient. It depends on the elasticity of the rail and wheel material, the condition of the surface, and the size of the wheel. As you can see, the larger the wheel, the less the rolling friction force. If the shape of the rail was restored behind the wheel, then the pressure diagram would be symmetrical and there would be no rolling friction. When a steel wheel rolls on a steel rail, the rolling friction coefficient is quite small: 0.003–0.005, hundreds of times less than the sliding friction coefficient. Therefore, rolling is easier than dragging.

The experimental determination of the rolling friction coefficient is carried out on a laboratory installation. Let a trolley, which is a model of a carriage, roll along horizontal rails. It is subjected to horizontal rolling friction and adhesion forces from the rails (Fig. 2). Let us write the equation of Newton's second law for the slow motion of a cart with a mass m in projection onto the direction of acceleration:

. (2)

Since the mass of the wheels makes up a significant part of the mass of the trolley, it is impossible not to take into account the rotational movement of the wheels. Let us imagine the rolling of wheels as the sum of two movements: translational movement together with the cart and rotational movement relative to the axes of the wheel pairs. We combine the forward motion of the wheels with the forward motion of the cart with their total mass m in equation (1) . The rotational movement of the wheels occurs under the influence of only the traction torque F sc R. Basic equation law of rotational dynamics(the product of the moment of inertia of all wheels and the angular acceleration is equal to the moment of force) has the form

. (3)

If there is no slipping of the wheel relative to the rail, the speed of the contact point is zero. This means that the speeds of translational and rotational movements are equal and opposite: . If we differentiate this equality, we obtain the relationship between the translational acceleration of the cart and the angular acceleration of the wheel: . Then equation (3) will take the form . Let's add this equation to equation (2) to eliminate the unknown adhesion force. As a result we get

. (4)

The resulting equation coincides with the equation of Newton’s second law for the translational motion of a cart with an effective mass: , which already takes into account the contribution of the inertia of wheel rotation to the inertia of the trolley. In the technical literature, the equation of rotational motion of wheels (3) is not used, but the rotation of the wheels is taken into account by introducing an effective mass. For example, for a loaded car the inertia coefficient γ is equal to 1.05, and for an empty car the influence of wheel inertia is greater: γ = 1,10.

Substituting the rolling friction force into equation (4), we obtain the calculation formula for the rolling friction coefficient

. (5)

To determine the rolling friction coefficient using formula (5), the acceleration of the trolley should be experimentally measured. To do this, push the cart with some speed V 0 on horizontal rails. The equation of kinematics of uniformly slow motion has the form .

Path S and driving time t can be measured, but the initial speed of movement is unknown V 0 . However, the installation (Fig. 3) has seven stopwatches that measure the time of movement from the starting photocell to the next seven photocells. This allows you to either create a system of seven equations and exclude the initial velocity from them, or solve these equations graphically. For a graphical solution, we rewrite the equation of uniformly slow motion, dividing it by time: .

The average speed of movement to each photocell depends linearly on the time of movement to the photocells. Therefore, the dependence graph<V>(t) is a straight line with an angular coefficient equal to half the acceleration (Fig. 4)

. (6)

The moment of inertia of the four wheels of a trolley, which are shaped like cylinders of radius R with their total mass m count, can be determined by the formula . Then the correction for the inertia of wheel rotation will take the form .

GETTING THE WORK DONE

1. Determine by weighing the mass of the cart along with some cargo. Measure the radius of the wheels along the rolling surface. Record the measurement results in the table. 1.

Table 1 Table 2

S, m	t, With	, m/s
0,070
0,140
0,210
0,280
0,350
0,420
0,490

2. Check the horizontality of the rails. Place the cart at the beginning of the rails so that the rod of the cart is in front of the holes of the starting photocell. Connect the power supply to a 220 V network.

3. Push the cart along the rails so that it reaches the trap and falls into it. Each stopwatch will show the time the cart moves from the starting photocell to its photocell. Repeat the experiment several times. Record the readings of seven stopwatches in one of the experiments in the table. 2.

4. Make calculations. Determine the average speed of the cart on the path from the start to each photocell

5. Plot the dependence of the average speed of movement to each photocell on the time of movement. The size of the chart is at least half a page. Specify a uniform scale on the coordinate axes. Draw a straight line near the points.

6. Determine the average acceleration value. To do this, construct a right triangle on the experimental line as on the hypotenuse. Using formula (6), find the average acceleration value.

7. Calculate the correction for the inertia of rotation of the wheels, considering them homogeneous disks . Determine the average value of the rolling friction coefficient using formula (5)<μ>.

8. Estimate the measurement error graphically

. (7)

Record the result μ = <μ>± δμ, Р = 90%.

Draw conclusions.

TEST QUESTIONS

1. Explain the cause of the rolling friction force. What factors influence the magnitude of rolling friction force?

2. Write down the law for the rolling friction force. What does the coefficient of rolling friction depend on?

3. Write down the equations for the dynamics of the translational motion of the cart on horizontal rails and the rotational motion of the wheels. Derive the equation of motion for a cart with effective mass.

4. Derive a formula for determining the rolling friction coefficient.

5. Explain the essence of the graphical method for determining the acceleration of a trolley when rolling on rails. Derive the acceleration formula.

6. Explain the effect of wheel rotation on the inertia of the cart.

Work 17-b

Related information.

Friction force is the force that occurs when two bodies come into contact and prevents their relative movement. It is applied to the bodies along the contact surface. The friction that occurs between the surfaces of different bodies is called external friction. If friction occurs between parts of the same body, then it is called internal friction.

The friction between the surfaces of two contacting solids in the absence of a liquid or gaseous layer between them is called dry friction.

The friction between the surface of a solid body and the surrounding liquid or gaseous medium in which the body moves is called viscous friction.

There are static friction, sliding friction and rolling friction.

The force of static friction arises between stationary solid bodies when there are forces acting in the direction of the possible movement of the body.

The static friction force is always equal in magnitude and directed opposite to the force parallel to the surface of contact and tending to cause this body to move. An increase in this external force applied to the body leads to an increase in the static friction force. The static friction force is directed in the direction opposite to the possible movement of the body (Fig. 1 a, b). . The maximum static friction force is proportional to the modulus of the normal pressure force produced by the body on the support:

Since according to Newton's third law. Here is the coefficient of static friction, depending on the material and state of the rubbing surfaces. The force of static friction prevents the start of movement. But there are cases when the force of static friction causes the movement of a body. For example, a person walking. When walking, the static friction force acting on the sole gives us acceleration. The sole does not slide back, and therefore the friction between it and the road is static friction.

Consider a block lying on a cart (Fig. 2). A force acts on it, trying to move it from its place. In the opposite direction, the static friction force acts on the block from the side of the cart. A force of the same magnitude and opposite in direction acts on the cart from the side of the block, leading to the movement of the cart to the right. The force of static friction plays a fundamental role in the movement of cars. The tires of the driving wheels of cars seem to push away from the road, and in the absence of slipping, the force pushing the car is the static friction force.

The sliding friction force occurs when bodies moving relative to each other come into contact and complicates their movement. The sliding friction force is directed along the contact surface in the direction opposite to the speed of movement. The sliding friction force is directly proportional to the normal pressure force:

where is the coefficient of sliding friction, which depends on the quality of surface treatment and their material.

for these tel.

(slightly more) - moving a body is more difficult than continuing its sliding).

The friction force does not depend on the area of ​​the contacting surfaces of the bodies and their position relative to each other, as well as on the velocity module at low speeds, but depends on the direction of the speed: when the direction of the speed changes, the direction also changes (Fig. 3). The action of sliding friction forces is accompanied by the transformation of mechanical energy into internal energy.

The existence of friction forces is explained by the manifestation of electromagnetic interaction forces. Static friction forces are caused mainly by elastic deformations of microprotrusions on the surface of rubbing bodies; sliding friction forces arise as a result of plastic deformations of microprotrusions and their partial destruction, as well as intermolecular interaction forces in the contact area.

Scientific and practical conference

Friction coefficient and m methods his calculation

Penza 2010

Chapter I Theoretical part

1. Types of friction, friction coefficient

Chapter II. Practical part

Calculation of static, sliding, and rolling friction

Calculation of the static friction coefficient

References

Chapter I Theoretical part

1. Types of friction, friction coefficient

We encounter friction at every step. It would be more accurate to say that without friction we cannot take a single step. But despite the large role that friction plays in our lives, a sufficiently complete picture of the occurrence of friction has not yet been created. This is not even due to the fact that friction has a complex nature, but rather to the fact that experiments with friction are very sensitive to surface treatment and are therefore difficult to reproduce.

Exists external And internal friction (otherwise calledviscosity ). External This type of friction is called in which forces arise at the points of contact of solid bodies that impede the mutual movement of the bodies and are directed tangentially to their surfaces.

Internal friction (viscosity) is a type of friction that occurs during mutual movement. layers of liquid or gas, tangential forces arise between them, preventing such movement.

External friction is divided intostatic friction (static friction ) And kinematic friction . Static friction occurs between fixed solid bodies when they try to move one of them. Kinematic friction exists between mutually touching moving solid bodies. Kinematic friction, in turn, is divided intosliding friction And rolling friction .

Friction forces play an important role in human life. In some cases he uses them, and in others he fights them. Friction forces are electromagnetic in nature.

If a body slides on any surface, its movement is impededsliding friction force.

Where N - ground reaction force, aμ - coefficient of sliding friction. Coefficientμ depends on the material and quality of processing of the contacting surfaces and does not depend on body weight. The friction coefficient is determined experimentally.

The sliding friction force is always directed opposite to the movement of the body. When the direction of speed changes, the direction of the friction force also changes.

The force of friction begins to act on the body when they try to move it. If an external forceF less productμN, then the body will not move - the beginning of movement, as they say, is prevented by the force of static friction. The body will begin to move only when the external forceF will exceed the maximum value that the static friction force can have

Static friction – frictional force that prevents the movement of one body on the surface of another.

Chapter II. Practical part

1. Calculation of static, sliding and rolling friction

Based on the above, I empirically found the force of static, sliding and rolling friction. To do this, I used several pairs of bodies, as a result of the interaction of which a friction force would arise, and a device for measuring force - a dynamometer.

Here are the following pairs of bodies:

a wooden block in the form of a rectangular parallelepiped of a certain mass and a varnished wooden table.

a wooden block in the form of a rectangular parallelepiped with less mass than the first and a varnished wooden table.

a wooden block in the form of a cylinder of a certain mass and a varnished wooden table.

a wooden block in the form of a cylinder with less mass than the first and a varnished wooden table.

After the experiments were carried out, the following conclusion could be drawn:

The force of static, sliding and rolling friction is determined experimentally.

Static friction:

For 1) Fp=0.6 N, 2) Fp=0.4 N, 3) Fp=0.2 N, 4) Fp=0.15 N

Sliding friction:

For 1) Fс=0.52 N, 2) Fс=0.33 N, 3) Fс=0.15 N, 4) Fс=0.11 N

Rolling friction:

For 3) Fk=0.14 N, 4) Fk=0.08 N

Thus, I determined experimentally all three types of external friction and obtained that

Fп> Fс > Fк for the same body.

2. Calculation of the static friction coefficient

But what is more interesting is not the friction force, but the friction coefficient. How to calculate and determine it? And I found only two ways to determine the friction force.

The first method is very simple. Knowing the formula and determining empirically and N, the coefficient of static, sliding and rolling friction can be determined.

1) N  0.81 N, 2) N  0.56 N, 3) N  2.3 N, 4) N  1.75

Static friction coefficient:

= 0,74; 2)  = 0,71; 3)  = 0,087; 4)  = 0,084;

Sliding friction coefficient:

= 0,64; 2)  = 0,59; 3)  = 0,063; 4)  = 0,063

Rolling friction coefficient:

3)  = 0,06; 4)  = 0,055;

By checking the tabular data, I confirmed the correctness of my values.

But the second method of finding the friction coefficient is also very interesting.

But this method determines the coefficient of static friction well, but a number of difficulties arise when calculating the coefficient of sliding and rolling friction.

Description: A body is at rest with another body. Then the end of the second body on which the first body lies begins to be lifted until the first body moves from its place.

 = sin  /cos  =tg  =BC/AC

Based on the second method, I calculated a certain number of static friction coefficients.

Wood to wood:

AB = 23.5 cm; BC = 13.5 cm.

P = BC/AC = 13.5/23.5 = 0.57

2. Polystyrene foam on wood:

AB = 18.5 cm; BC = 21 cm.

P = BC/AC = 21/18.5 = 1.1

3. Glass on wood:

AB = 24.3 cm; BC = 11 cm.

P = BC/AC = 11/24.3 = 0.45

4. Aluminum on wood:

AB = 25.3 cm; BC = 10.5 cm.

P = BC/AC = 10.5/25.3 = 0.41

5. Steel on wood:

AB = 24.6 cm; BC = 11.3 cm.

P = BC/AC = 11.3/24.6 = 0.46

6. Org. Glass on wood:

AB = 25.1 cm; BC = 10.5 cm.

P = BC/AC = 10.5/25.1 = 0.42

7. Graphite on wood:

AB = 23 cm; BC = 14.4 cm.

P = BC/AC = 14.4/23 = 0.63

8. Aluminum on cardboard:

AB = 36.6 cm; BC = 17.5 cm.

P = BC/AC = 17.5/36.6 = 0.48

9. Iron on plastic:

AB = 27.1 cm; BC = 11.5 cm.

P = BC/AC = 11.5/27.1 = 0.43

10. Org. Glass on plastic:

AB = 26.4 cm; BC = 18.5 cm.

P = BC/AC = 18.5/26.4 = 0.7

Based on my calculations and experiments, I concluded that P >  C >  K , which undeniably corresponded to the theoretical basis taken from the literature. The results of my calculations did not go beyond the tabular data, but even supplemented them, as a result of which I expanded the tabulated values ​​of the friction coefficients of various materials.

Literature

1. Kragelsky I.V., Dobychin M.N., Kombalov V.S. Basics of calculations for friction and wear. M.: Mechanical Engineering, 1977. 526 p.

Frolov, K.V. (ed.):Modern tribology: Results and prospects. Publishing house LKI, 2008

Elkin V.I. “Unusual educational materials in physics.” “Physics at school” magazine library, No. 16, 2000.

Wisdom of millennia. Encyclopedia. Moscow, Olma - press, 2006.

COEFFICIENT OF FRICTION

COEFFICIENT OF FRICTION, a quantitative characteristic of the force required to slide or move one material along the surface of another. If we denote the weight of an object as N, and the coefficient of FRICTION as m, then the force (F) required to move an object on a flat surface without acceleration is F = mN. The coefficient of static friction determines the force required to initiate motion; The coefficient of kinetic friction (friction of motion) determines the (less) force required to maintain motion.

Scientific and technical encyclopedic dictionary.

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