« Physics - 10th grade"

Remember what friction is.
What factors is it due to?
Why does the speed of movement of the block on the table change after a push?

Another type of force dealt with in mechanics is frictional forces. These forces act along the surfaces of bodies when they are in direct contact.

Friction forces in all cases prevent the relative motion of contacting bodies. Under certain conditions, friction forces make this movement impossible. However, they not only slow down the movement of bodies. In a number of practically important cases, the movement of a body could not occur without the action of friction forces.

Friction that occurs during relative movement of the contacting surfaces of solid bodies is called dry friction.

There are three types of dry friction: static friction, sliding friction and rolling friction.

Rest friction.

Try moving a thick book lying on the table with your finger. You apply some force to it, directed along the surface of the table, and the book remains at rest. Consequently, a force arises between the book and the surface of the table, directed opposite to the force with which you act on the book, and exactly equal to it in magnitude. This is the friction force tr. You push the book with more force, but it still stays in place. This means that the friction force tr increases by the same amount.

The frictional force acting between two bodies stationary relative to each other is called force static friction.

If a body is acted upon by a force parallel to the surface on which it is located, and the body remains motionless, this means that it is acted upon by a static friction force tr, equal in magnitude and directed in the opposite direction to the force (Fig. 3.22). Consequently, the force of static friction is determined by the force acting on it:

If the force acting on a body at rest even slightly exceeds the maximum force of static friction, then the body will begin to slide.

The greatest value of the friction force, at which sliding does not yet occur, is called maximum static friction force.

To determine the maximum static friction force, there is a very simple, but not very accurate quantitative law. Let there be a block on the table with a dynamometer attached to it. Let's conduct the first experiment. Let's pull the dynamometer ring and determine the maximum static friction force. The block is acted upon by gravity m, force normal reaction supports 1, tension force 1, dynamometer springs and maximum static friction force tr1 (Fig. 3.23).

Let's place another similar block on the block. The force of pressure of the bars on the table will increase by 2 times. According to Newton's third law, the normal reaction force of support 2 will also increase by 2 times. If we measure the maximum static friction force again, we will see that it has increased as many times as the force 2 has increased, i.e. 2 times.

Continuing to increase the number of bars and measuring each time the maximum force of static friction, we will be convinced that

>the maximum value of the modulus of the static friction force is proportional to the modulus of the normal reaction force of the support.

If we denote the module of the maximum static friction force by F tr. max, then we can write:

F tr. max = μN (3.11)

where μ is a proportionality coefficient called the friction coefficient. The friction coefficient characterizes both rubbing surfaces and depends not only on the material of these surfaces, but also on the quality of their processing. The friction coefficient is determined experimentally.

This dependence was first established by the French physicist C. Coulomb.

If you place the block on the smaller face, then F tr. max will not change.

The maximum static friction force does not depend on the area of ​​contact between the bodies.

The static friction force varies from zero to a maximum value equal to μN. What can cause a change in the friction force?

The point here is this. When a certain force is applied to a body, it shifts slightly (imperceptibly to the eye), and this displacement continues until the microscopic roughness of the surfaces are positioned relative to each other in such a way that, hooking on one another, they will lead to the appearance of a force that balances the force. As the force increases, the body will again move slightly so that the smallest surface irregularities will cling to each other differently, and the friction force will increase.

And only at > F tr. max under no circumstances relative position surface roughness, the friction force is not able to balance the force, and sliding will begin.

Dependence of the sliding friction force modulus on the modulus acting force shown in Figure 3.24.

When walking and running, the soles of the feet are subject to static friction unless the feet slip. The same force acts on the drive wheels of the car. The driven wheels are also acted upon by a static friction force, but this time braking the movement, and this force is significantly less than the force acting on the drive wheels (otherwise the car would not be able to move).

For a long time, it was doubted that a steam locomotive could run on smooth rails. They thought that the friction braking the driven wheels would be equal to the friction force acting on the driving wheels. It was even proposed to make the drive wheels geared and lay special geared rails for them.

Sliding friction.

When sliding, the friction force depends not only on the state of the rubbing surfaces, but also on the relative speed of the bodies, and this dependence on speed is quite complex. Experience shows that often (though not always) at the very beginning of sliding, when the relative speed is still low, the friction force becomes somewhat less than the maximum static friction force. Only then, as the speed increases, does it grow and begin to exceed F tr. max.

You've probably noticed that a heavy object, such as a box, is difficult to move, but then moving it becomes easier. This is precisely explained by the decrease in friction force when sliding occurs at low speed (see Fig. 3.24).

At not too high relative speeds of movement, the sliding friction force differs little from the maximum static friction force. Therefore, it can be approximately considered constant and equal to the maximum static friction force:

F tr ≈ F tr. max = μN.

The force of sliding friction can be reduced many times by using a lubricant - most often a thin layer of liquid (usually some type of mineral oil) - between the rubbing surfaces.

Not a single modern machine, such as a car or tractor engine, can operate without lubrication. A special lubrication system is provided for in the design of all machines.

The friction between layers of liquid adjacent to solid surfaces is much less than between dry surfaces.

Rolling friction.

The rolling friction force is significantly less than the sliding friction force, so it is much easier to roll a heavy object than to move it.

The friction force depends on the relative speed of the bodies. This is its main difference from the forces of gravity and elasticity, which depend only on distances.

Resistance forces during the movement of solid bodies in liquids and gases.

When a solid body moves in a liquid or gas, it is acted upon by the drag force of the medium. This force is directed against the speed of the body relative to the medium and slows down the movement.

The main feature of the resistance force is that it appears only if there is relative motion body and environment.
The force of static friction in liquids and gases is completely absent.

This leads to the fact that with the effort of your hands you can move a heavy body, for example, a floating boat, while moving, say, a train with your hands is simply impossible.

The modulus of the resistance force F c depends on the size, shape and state of the surface of the body, the properties of the medium (liquid or gas) in which the body moves, and, finally, on the relative speed of movement of the body and the medium.

The approximate nature of the dependence of the modulus of the resistance force on the modulus of the relative velocity of the body is shown in Figure 3.25. At a relative speed equal to zero, the drag force does not act on the body (F c = 0). As the relative speed increases, the drag force grows slowly at first, and then faster and faster. At low speeds of movement, the resistance force can be considered directly proportional to the speed of movement of the body relative to the medium:

F c = k 1 υ, (3.12)

where k 1 is the resistance coefficient, depending on the shape, size, state of the surface of the body and the properties of the medium - its viscosity. It is not possible to calculate the coefficient k 1 theoretically for bodies of any complex shape; it is determined experimentally.

At high speeds of relative motion, the drag force is proportional to the square of the speed:

F c = k 2 υ 2 , υ, (3.13)

where k 2 is the resistance coefficient different from k 1 .

Which of the formulas - (3 12) or (3.13) - can be used in a particular case is determined experimentally. For example, for a passenger car, it is advisable to use the first formula at approximately 60-80 km/h; at higher speeds, the second formula should be used.

This force must be overcome in order to set two contacting bodies in motion relative to each other. Occurs during micromovements (for example, during deformation) of contacting bodies. It acts in the direction opposite to the direction of possible relative motion.

The maximum static friction force in the simplest approximation: , where k 0 is the static friction coefficient, N is the normal support reaction force.

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See what “Friction at rest” is in other dictionaries:

static friction- Friction of two bodies with microdisplacements without macrodisplacement. [GOST 27674 88] Topics: friction, wear and lubrication EN static friction ... Technical Translator's Guide

static friction- 3.3 static friction: Friction of two bodies with microdisplacements without macrodisplacement. Source: ST TsKBA 057 2008: Pipeline fittings. Friction coefficients in reinforcement units... Dictionary-reference book of terms of normative and technical documentation

static friction- rimties trintis statusas T sritis fizika atitikmenys: engl. friction of repose; friction of rest vok. Haftreibung, f; Ruhereibung, f rus. static friction, n pranc. frottement de repos, m … Fizikos terminų žodynas

static friction- static friction Friction at relative rest of two contacting bodies. IFToMM code: 3.5.47 Section: DYNAMICS OF MECHANISMS... Theory of mechanisms and machines

This term has other meanings, see Radiation friction. Friction is the process of interaction between bodies during their relative motion (displacement) or when a body moves in a gaseous or liquid medium. Otherwise called friction... ... Wikipedia

Mechanical impact between solids, which occurs at the places of their contact and prevents the relative movement of bodies in the direction lying in the plane of their contact. Distinguish: static friction between mutually motionless... ... Construction dictionary

Friction- – a process that occurs on the surface of contact of bodies, both at rest and in mutual movement. … … Encyclopedia of terms, definitions and explanations of building materials

Mechanical resistance that arises in the plane of contact of two bodies pressed against each other when they are relative. moving. The resistance force F, directed in the opposite direction, relates. movement of a given body, called the frictional force acting on this body. T … Physical encyclopedia

Mechanical the resistance that arises in the plane of contact of two contacting bodies when they are relative. moving. The resistance force F, directed in the opposite direction, relates. movement of bodies, called strength training. T.v. dissipative process... ... Physical encyclopedia

Friction force is the force of mechanical resistance that arises in the plane of contact of two bodies pressed against each other during their relative movement.

The resistance force acting on a body is directed opposite to the relative movement of a given body.

The force of friction arises for two reasons: 1) the first and main reason is that at the points of contact, the molecules of substances are attracted to each other, and work must be done to overcome their attraction. The contacting surfaces touch each other only in very small areas. Their total area is 0.01 ÷ 0.001 0.01 \div 0.001 of the total (apparent) contact area. When sliding, the area of ​​actual contact does not remain unchanged. The friction (sliding) force will change during movement. If the body that is sliding is pressed more strongly against the body on which the sliding occurs, then due to the deformation of the bodies, theThe area of ​​the contact spots (and the frictional force) will increase in proportion to the pressing force.

$$F_\text(tr) \sim F_\text(prij)$$

2) the second reason for the occurrence of friction force isThis is the presence of roughness (irregularities) of surfaces, and their deformation when one body moves along the surface of another. The depth of penetration (engagement) of roughness depends onpressing force, and the magnitude of the deformations depends on this. The latter, in turn, determine the magnitude of the friction force: F tr ∼ F prj F_\mathrm(tr) \sim F_\mathrm(prj) .

With relative sliding, both causes take place, therefore the nature of the interaction has the form of a simple relationship:

F tr = μ N - \boxed(F_\mathrm(tr) =\mu N)\ - sliding friction force (Coulomb - Amonton formula), where

μ - \mu\ - sliding friction coefficient,

N - N\ - support reaction force equal to the pressing force.

The magnitude of the friction coefficient is different for different combinations of rubbing substances, even with the same treatment (attractive forces and elastic properties depend on the type of substance).

If there is a lubricant between the rubbing surfaces, the force of attraction will change noticeably (other molecules will be attracted, and the sliding friction force will be partially replaced by the force of viscous friction, which we will consider below).

If a body lying on a horizontal surface is acted upon by a horizontal force F → \vec F , then the movement will be caused by this force only when it becomes greater than a certain value (μ N) (\mu N) . Before the movement begins, the external the force is compensated by the force of static friction.












Rice. 13

The static friction force is always equal to the external force parallel to the surface, and arises due to the attraction between molecules in the areas of contact spots and deformation of roughness.

The force of static friction is different in different parts of the surface along which movement will occur. If the body lies on the surface for a long time, then due to vibrations (they are always present on the surface of the Earth), the area of ​​the contact spots will increase slightly. Therefore, to start moving, you will have to overcome a slightly greater frictional force than the sliding friction force. This phenomenon is called the stagnation phenomenon. We encounter this phenomenon, for example, when moving furniture in a room. (In Figure 13, the superiority of static friction over sliding friction is greatly exaggerated).

We use the force of static friction to move on skis or simply when walking.

The considered types of friction force relate to dry friction or external friction. But there is another type of friction force - viscous friction.

When a body moves in a liquid or gas, quite complex processes of exchange of molecules occur between the layers of the flowing liquid or gas. These processes are called transfer processes.

At low speeds of movement of a body relative to a gas or liquid, the resistance force will be determined by the expression:

F tr = 6 π η r v - \boxed(F_\mathrm(tr) = 6\pi \eta r v)\ - Stokes law for the ball, where

η - \eta\ - viscosity of the substance in which the body moves;

r - r\ - average transverse size (radius) of the body;

v - v\ - relative speed of the body;

6 π - 6\pi\ - coefficient corresponding to the spherical shape of the body.

A conclusion about the magnitude of the speed (whether it is large or small) can be made by determining a dimensionless coefficient called the Reynolds number:

R e = ρ r v η - \boxed(Re = \frac(\rho r v)(\eta))\ - Reynolds number, where

ρ - \rho\ is the density of the substance in which the body moves.

If R e< 1700 Re движение газа (жидкости) вокруг тела ламинарное (слоистое), и скорости можно считать малыми.

If R e > 1700 Re > 1700 , then the movement of gas (liquid) around the body is turbulent(with turbulence), and the speeds can be considered high.

In the latter case, most of the kinetic energy of the body is spent on the formation of vortices, which means that the friction force becomes greater and the dependence ceases to be linear.

F tr = k v 2 ρ S - \boxed(F_\mathrm(tr) = kv^2\rho S)\ - viscous friction force at high speeds, where

S - S\ - cross-sectional area of ​​the body,

k - k\ - constant, depending on the transverse dimensions of the body.

Often the latter formula can be seen as:

The Reynolds number chosen to be 1700 1700 is actually determined by the specific problem (conditions) and can take other values ​​of the same order. This is explained by the fact that the dependence of the viscous friction force on speed is complex: at a certain speed linear dependence begins to break down, and at a certain speedthis dependence becomes quadratic.

Rice. 14

In the interval from v 1 v_1 to v 2 v_2 degree takes fractional values(Fig. 14) . The Reynolds number characterizes the state of a dynamic system in which the movement of layers remains laminar, and strongly depends on external conditions. For example: a steel ball, moving in water far from the boundaries of the liquid (in the ocean, lake), maintains laminar movement of the layers at R e = 1700 Re = 1700 , and the same ball moving in a vertical pipe of slightly larger radius than the ball, filled with water, already at R e = 2 Re = 2will cause the water to swirl around the ball. (Note that the Reynolds number is not the only one used to describe such a movement. For example, they also useFroude and Mach numbers.)

Let a small body be on an inclined plane with an angle of inclination a (Fig. 14.3, A). Let's find out: 1) what is the friction force if a body slides along an inclined plane; 2) what is the friction force if the body lies motionless; 3) at what minimum value of the inclination angle a does the body begin to slide off the inclined plane.

A) b)

The friction force will be hinder movement, therefore, it will be directed upward along the inclined plane (Fig. 14.3, b). In addition to the frictional force, the force of gravity and the normal reaction force also act on the body. Let us introduce the coordinate system HOU, as shown in the figure, and find the projections of all these forces onto the coordinate axes:

X: F tr X = –F tr, N X = 0, mg X = mg sina;

Y:F tr Y = 0, NY=N, mg Y = –mg cosa.

Since a body can accelerate only along an inclined plane, that is, along the axis X, then it is obvious that the projection of the acceleration vector onto the axis Y will always be zero: and Y= 0, which means the sum of the projections of all forces onto the axis Y must also be zero:

F tr Y + N Y + mg Y= 0 Þ 0 + N–mg cosa = 0 Þ

N = mg cosa. (14.4)

Then the sliding friction force according to formula (14.3) is equal to:

F tr.sk = m N= m mg cosa. (14.5)

If the body rests, then the sum of the projections of all forces acting on the body onto the axis X should be equal to zero:

F tr X + N X + mg X= 0 Þ – F tr + 0 +mg sina = 0 Þ

F tr.p = mg sina. (14.6)

If we gradually increase the angle of inclination, then the value mg sina will gradually increase, which means that the static friction force will also increase, which always “automatically adjusts” to external influences and compensates for it.

But, as we know, the “possibilities” of the static friction force are not unlimited. At some angle a 0, the entire “resource” of the static friction force will be exhausted: it will reach its maximum value, equal strength sliding friction. Then the equality will be true:

F tr.sk = mg sina 0 .

Substituting into this equality the value F tr.sk from formula (14.5), we obtain: m mg cosa 0 = mg sina 0 .

Dividing both sides of the last equality by mg cosa 0 , we get:

Þ a 0 = arctgm.

So, the angle a at which the body begins to slide along an inclined plane is given by the formula:

a 0 = arctgm. (14.7)

Note that if a = a 0, then the body can either lie motionless (if you don’t touch it), or slide at a constant speed down the inclined plane (if you push it a little). If a< a 0 , то тело «стабильно» неподвижно, и легкий толчок не произведет на него никакого «впечатления». А если a >a 0, then the body will slide off the inclined plane with acceleration and without any shocks.

Problem 14.1. A man is carrying two sleds connected to each other (Fig. 14.4, A), applying force F at an angle a to the horizontal. The masses of the sleds are the same and equal T. Coefficient of friction of runners on snow m. Find the acceleration of the sled and the tension force T ropes between the sleds, as well as force F 1, with which a person must pull the rope in order for the sled to move evenly.

F a m m	A)	b) Rice. 14.4
A = ? T = ? F 1 = ?

Solution. Let's write down Newton's second law for each sled in projections on the axis X And at(Fig. 14.4, b):

I at: N 1 + F sina – mg = 0, (1)

x: F cosa - T–m N 1 = ma; (2)

II at: N 2 – mg = 0, (3)

x: T–m N 2 = ma. (4)

From (1) we find N 1 = mg–F sina, from (3) and (4) we find T = m mg+ + ma. Substituting these values N 1 and T in (2), we get

.

Substituting A in (4), we get

T= m N 2 + ma= m mg + that =

M mg + T .

To find F 1, let us equate the expression for A to zero:

Answer: ; ;

.

STOP! Decide for yourself: B1, B6, C3.

Problem 14.2. Two bodies with masses T And M tied with a thread, as shown in Fig. 14.5, A. With what acceleration is the body moving? M, if the coefficient of friction on the table surface is m. What is the thread tension T? What is the force of pressure on the block axis?

T M m	Solution. Let's write Newton's second law in projections on the axis X 1 and X 2 (Fig. 14.5, b), considering that: X 1: T - m Mg = Ma, (1) X 2: mg – T = ma. (2) Solving the system of equations (1) and (2), we find:
A = ? T = ? R = ?

If the loads do not move, then .

Answer: 1) if T < mM, That A = 0, T = mg, ; 2) if T³m M, That , , .

STOP! Decide for yourself: B9–B11, C5.

Problem 15.3. Two bodies with masses T 1 and T 2 are connected with a thread thrown over a block (Fig. 14.6). Body T 1 is on an inclined plane with an angle of inclination a. Coefficient of friction about the plane m. Body mass T 2 hanging on a thread. Find the acceleration of the bodies, the tension force of the thread and the pressure force of the block on the axis provided that T 2 < T 1 . Consider tga > m.

Rice. 14.7

Let's write Newton's second law in projections on the axis X 1 and X 2, given that and:

X 1: T 1 g sina – T - m m 1 g cosa = m 1 a,

X 2: T–m 2 g = m 2 a.

, .

Because A>0, then

If inequality (1) is not satisfied, then the load T 2 is definitely not moving up! Then two more options are possible: 1) the system is motionless; 2) cargo T 2 moves down (and the load T 1, respectively, up).

Let's assume that the load T 2 moves down (Fig. 14.8).

Rice. 14.8

Then the equations of Newton's second law on the axis X 1 and X 2 will look like:

X 1: T – t 1 g sina – m m 1 g cosa = m 1 a,

X 2: m 2 g – T = m 2 a.

Solving this system of equations, we find:

, .

Because A>0, then

So, if inequality (1) is satisfied, then the load T 2 goes up, and if inequality (2) is satisfied, then down. Therefore, if none of these conditions are met, i.e.

,

the system is motionless.

It remains to find the pressure force on the block axis (Fig. 14.9). Pressure force on the block axis R in this case can be found as the diagonal of a rhombus ABCD. Because

Ð ADC= 180° – 2,

where b = 90°– a, then by the cosine theorem

R 2 = .

From here .

Answer:

1) if , That , ;

2) if , That , ;

3) if , That A = 0; T = T 2 g.

In all cases .

STOP! Decide for yourself: B13, B15.

Problem 14.4. On a trolley weighing M horizontal force acts F(Fig. 14.10, A). Friction coefficient between load T and cart is equal to m. Determine the acceleration of the loads. What should be the minimum force F 0 to load T started to slide on the cart?

M, T F m	A)	b) Rice. 14.10
A 1 = ? A 2 = ? F 0 = ?

Solution. First, note that the force driving the load T in motion is the static friction force with which the cart acts on the load. The maximum possible value of this force is m mg.

According to Newton's third law, the load acts on the cart with the same force - (Fig. 14.10, b). Slip begins at the moment when it has already reached its maximum value, but the system is still moving as one body of mass T+M with acceleration. Then according to Newton's second law

Thanks to this force, cars slow down at traffic lights, a boat stops in the water, and a wheel slips in a hole. As you already understand, in this article we will figure out how to solve problems on friction force.

The friction force is electromagnetic in nature. This means that this force is manifested as a result of the interaction of the particles that make up the substance.

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What you need to know about friction force to solve problems

Friction is one of the types of interaction between bodies that occurs when they come into contact.

The friction force is always directed in the direction opposite to the movement and tangential to the contacting surfaces. Dry friction occurs between solid bodies, and when bodies move in liquids or gases they speak of viscous friction.

We have already established the nature of this force. In addition, you need to know that there are different types friction forces:

• static friction;
• sliding friction;
• rolling friction (when bodies roll over each other);
• resistance of the medium (for movement in a liquid).

Here is an example of the types of friction force: the block lies on the table and no one touches it. In this case, only gravity and the normal ground reaction force act. If we start pushing the block, but so hard as to move it, it will be acted upon by a static friction force, which, according to Newton’s third law, is equal to the external force applied to the block. The static friction force has a limiting value. If the external force is greater than this value, the block will begin to slide along the table. In this case, they talk about the sliding friction force. And here is the simplest formula for the friction force:

“Mu” is the coefficient of sliding friction. This is a dimensionless quantity that depends on the materials of the interacting bodies and the quality of their surfaces. The friction coefficient does not exceed unity.

When solving simple physical problems, the sliding friction force is often taken to be equal to the maximum static friction force.

Questions on the topic “Friction Force”

Question 1. What does friction force depend on?

Answer. Let's take a look at the formula above and the answer will come to you. The friction force depends on the properties of the contacting bodies, the force of the normal reaction of the support, and the speed of the relative movement of the bodies.

Question 2. Does the force of friction depend on the area of ​​contacting surfaces?

Answer. No, area does not affect the force of friction.

Question 3. In what ways can you reduce or increase the force of friction?

Answer. You can reduce the coefficient of friction by making dry friction viscous. To increase the friction force, it is necessary to increase the pressure on them.

Question 4. The body is at rest on a plane. Does friction force act on it?

Answer. If there is no effect on the body external forces, then the static friction force, according to Newton’s third law, is equal to zero.

Question 5. Which of these forces is the largest in magnitude: static friction force, rolling friction force or sliding friction force?

Answer. The sliding friction force is of greatest importance.

Question 6. What are some examples? useful action frictional forces?

Answer. Among the useful uses of friction force, we can highlight the operation of brakes Vehicle, the production of fire by primitive people.

Friction problems with solutions

By the way! There is a discount for our readers 10% on any type of work.

Task No. 1. Finding the friction force

Condition

A block of mass 5 kilograms slides along a horizontal surface. The sliding friction force is 20 N. Find the friction force if the mass of the block is halved and the friction coefficient remains unchanged.

Solution

Let's apply the formulas:

Answer: 10 N.

Task No. 2. Finding the coefficient of friction

Condition

A body slides along a horizontal plane. Find the coefficient of friction if the friction force is 5 N and the pressure force of the body on the plane is 20 N.

Solution

The force of body pressure on the plane is equal to the force of the normal support reaction.

Answer: 0,25

Task No. 3. Finding the friction force and friction coefficient

Condition

A skier weighing 60 kg, having a speed of 10 m/s at the end of the descent, stops 40 s after the end of the descent. Determine the friction force and friction coefficient.

Solution

First, let's find the acceleration with which the skier is moving. Then, using Newton’s second law, we find the force that acts on it:

Answer: 15 N; 0.025.

Task No. 4. Finding the friction force

Condition

A block with a mass of 20 kg moves uniformly along a horizontal surface under the action of a constant force directed at an angle of 30° to the surface and equal to 75 N. What is the coefficient of friction between the block and the plane?

Solution

First, let's use Newton's second law, given that the acceleration is zero. Then we find the projections of force on the vertical and horizontal axes:

Answer: 0,4

Task No. 5. Finding the static friction force

Condition

A box with a mass of 10 kg stands on a horizontal floor. The coefficient of friction between the floor and the box is 0.25. A force of 16 N is applied to the box in the horizontal direction. Will it move? What is the force of friction between the box and the floor?

Solution

Let's calculate the maximum static friction force:

Since the applied force is by condition less than the maximum static friction force, the box will remain in place. The friction force between the floor and the box, according to Newton's third law, is equal to the applied force.

Answer: 16 N.

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