The load moves along an inclined plane. Friction. Basic formula of dynamics

21.05.2020

This article talks about how to solve problems about moving along inclined plane. Reviewed detailed solution problems on the motion of connected bodies on an inclined plane from the Unified State Examination in Physics.

Solving the problem of motion on an inclined plane

Before moving directly to solving the problem, as a tutor in mathematics and physics, I recommend carefully analyzing its condition. You need to start by depicting the forces that act on connected bodies:

Here and are the thread tension forces acting on the left and right bodies, respectively, are the support reaction force acting on the left body, and are the gravity forces acting on the left and right bodies, respectively. Everything is clear about the direction of these forces. The tension force is directed along the thread, the gravity force is vertically downward, and the support reaction force is perpendicular to the inclined plane.

But the direction of the friction force will have to be dealt with separately. Therefore, in the figure it is shown as a dotted line and signed with a question mark. It is intuitively clear that if the right load “outweighs” the left one, then the friction force will be directed opposite to the vector. On the contrary, if the left load “outweighs” the right one, then the friction force will be co-directed with the vector.

The right weight is pulled down by force N. Here we took the acceleration of gravity m/s 2. The left load is also pulled down by gravity, but not all of it, but only a “part” of it, since the load lies on an inclined plane. This “part” is equal to the projection of gravity onto an inclined plane, that is, a leg in a right triangle shown in the figure, that is, equal to N.

That is, the right load still “outweighs”. Consequently, the friction force is directed as shown in the figure (we drew it from the center of mass of the body, which is possible in the case when the body can be modeled by a material point):

The second important question that needs to be addressed is whether this coupled system will move at all? What if it turns out that the friction force between the left load and the inclined plane will be so great that it will not allow it to move?

This situation will be possible in the case when the maximum friction force, the modulus of which is determined by the formula (here - the coefficient of friction between the load and the inclined plane - the support reaction force acting on the load from the inclined plane), turns out to be greater than the force that is trying to bring the system into motion. That is, that very “outweighing” force that is equal to N.

The modulus of the support reaction force is equal to the length of the leg in the triangle according to Newton’s 3rd law (with the same magnitude of force the load presses on the inclined plane, with the same magnitude of force the inclined plane acts on the load). That is, the support reaction force is equal to N. Then the maximum value of the friction force is N, which is less than the value of the “overweighing force”.

Consequently, the system will move, and move with acceleration. Let us depict in the figure these accelerations and coordinate axes, which we will need later when solving the problem:

Now, after a thorough analysis of the problem conditions, we are ready to begin solving it.

Let's write down Newton's 2nd law for the left body:

And in the projection onto the axes of the coordinate system we get:

Here, projections are taken with a minus, the vectors of which are directed opposite the direction of the corresponding coordinate axis. Projections whose vectors are aligned with the corresponding coordinate axis are taken with a plus.

Once again we will explain in detail how to find projections and . To do this, consider the right triangle shown in the figure. In this triangle And . It is also known that in this right triangle . Then and.

The acceleration vector lies entirely on the axis, and therefore . As we already mentioned above, by definition, the modulus of the friction force is equal to the product of the friction coefficient and the modulus of the support reaction force. Hence, . Then the original system of equations takes the form:

Let us now write down Newton’s 2nd law for the right body:

In projection onto the axis we get.

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

Dynamics is one of the important branches of physics, which studies the reasons for the movement of bodies in space. In this article, we will consider from a theoretical point of view one of the typical problems of dynamics - the movement of a body along an inclined plane, and also give examples of solutions to some practical problems.

Basic formula of dynamics

Before moving on to studying the physics of body motion along an inclined plane, we present the necessary theoretical information for solving this problem.

In the 17th century, Isaac Newton, thanks to practical observations of the motion of macroscopic surrounding bodies, derived three laws that currently bear his name. Everything is based on these laws classical mechanics. We are interested in this article only in the second law. His mathematical form is given below:

Inclined plane and forces acting on the body located on it

The key point on which the success of solving problems of body motion along an inclined plane depends is the determination of the forces acting on the body. The definition of forces is understood as knowledge of their modules and directions of action.

Below is a drawing that shows that a body (car) is at rest on a plane inclined at an angle to the horizontal. What forces are acting on it?

The list below lists these forces:

heaviness;
support reactions;
friction;
thread tension (if present).

Gravity

First of all, this is the force of gravity (Fg). It is directed vertically downwards. Since the body has the ability to move only along the surface of the plane, when solving problems, the force of gravity is decomposed into two mutually perpendicular components. One of the components is directed along the plane, the other is perpendicular to it. Only the first of them leads to the appearance of acceleration in the body and, in fact, is the only driving factor for the body in question. The second component determines the occurrence of the support reaction force.

In our case F n = m g, because the surface is horizontal. But the normal force does not always coincide in magnitude with the force of gravity.

Normal force is the force of interaction between the surfaces of contacting bodies; the greater it is, the stronger the friction.

The normal force and the friction force are proportional to each other:

F tr = μF n

0 < μ < 1 - friction coefficient, which characterizes the roughness of surfaces.

At μ=0 there is no friction (idealized case)

When μ=1 the maximum friction force is equal to the normal force.

The friction force does not depend on the area of contact of two surfaces (if their masses do not change).

Please note: Eq. F tr = μF n is not a relationship between the vectors, since they are directed in different directions: the normal force is perpendicular to the surface, and the friction force is parallel.

1. Types of friction

There are two types of friction: static And kinetic.

Static friction (static friction) acts between bodies in contact that are at rest relative to each other. Static friction occurs at the microscopic level.

Kinetic friction (sliding friction) acts between bodies in contact and moving relative to each other. Kinetic friction manifests itself at the macroscopic level.

Static friction is greater than kinetic friction for the same bodies, or the coefficient of static friction is greater than the coefficient of sliding friction.

Surely you know this from personal experience: The cabinet is very difficult to move, but keeping the cabinet moving is much easier. This is explained by the fact that when moving, the surfaces of bodies “do not have time” to contact each other at the microscopic level.

Task #1: what force is required to lift a ball weighing 1 kg along an inclined plane located at an angle α = 30° to the horizontal. Friction coefficient μ = 0.1

We calculate the component of gravity. First, we need to find out the angle between the inclined plane and the gravity vector. We have already done a similar procedure when considering gravity. But repetition is the mother of learning :)

The force of gravity is directed vertically downwards. The sum of the angles of any triangle is 180°. Consider a triangle formed by three forces: the gravity vector; inclined plane; the base of the plane (in the figure it is highlighted in red).

The angle between the gravity vector and the base of the plane is 90°.
The angle between the inclined plane and its base is α

Therefore, the remaining angle is the angle between the inclined plane and the gravity vector:

180° - 90° - α = 90° - α

Components of gravity along an inclined plane:

F g slope = F g cos(90° - α) = mgsinα

Required force to lift the ball:

F = F g incl + F friction = mgsinα + F friction

It is necessary to determine the friction force F tr. Taking into account the static friction coefficient:

Friction F = μF norm

Calculate normal force F normal, which is equal to the component of gravity perpendicular to the inclined plane. We already know that the angle between the gravity vector and the inclined plane is 90° - α.

F norm = mgsin(90° - α) = mgcosα
F = mgsinα + μmgcosα

F = 1 9.8 sin30° + 0.1 1 9.8 cos30° = 4.9 + 0.85 = 5.75 N

We will need to apply a force of 5.75 N to the ball in order to roll it to the top of the inclined plane.

Task #2: determine how far a ball of mass will roll m = 1 kg along a horizontal plane, rolling down an inclined plane of length 10 meters at sliding friction coefficient μ = 0.05

The forces acting on a rolling ball are shown in the figure.

Gravity component along an inclined plane:

F g cos(90° - α) = mgsinα

Normal strength:

F n = mgsin(90° - α) = mgcos(90° - α)

Sliding friction force:

Friction F = μF n = μmgsin(90° - α) = μmgcosα

Resultant force:

F = F g - F friction = mgsinα - μmgcosα

F = 1 9.8 sin30° - 0.05 1 9.8 0.87 = 4.5 N

F = ma; a = F/m = 4.5/1 = 4.5 m/s 2

Determine the speed of the ball at the end of the inclined plane:

V 2 = 2as; V = 2as = 2 4.5 10 = 9.5 m/s

The ball finishes moving along an inclined plane and begins moving along a horizontal straight line at a speed of 9.5 m/s. Now, in the horizontal direction, only the friction force acts on the ball, and the component of gravity is zero.

Total force:

F = μF n = μF g = μmg = 0.05 1 9.8 = -0.49 N

The minus sign means that the force is directed in the opposite direction from the movement. We determine the acceleration of the deceleration of the ball:

a = F/m = -0.49/1 = -0.49 m/s 2

Ball braking distance:

V 1 2 - V 0 2 = 2as; s = (V 1 2 - V 0 2)/2a

Since we determine the path of the ball until it comes to a complete stop, then V 1 =0:

s = (-V 0 2)/2a = (-9.5 2)/2·(-0.49) = 92 m

Our ball rolled in a straight line for as much as 92 meters!

Projection of forces. Movement on an inclined plane

Dynamics problems.

Newton's I and II laws.

Input and direction of axes.

Non-collinear forces.

Projection of forces on the axes.

Solving systems of equations.

The most typical tasks by dynamics

Let's start with Newton's I and II laws.

Let's open a physics textbook and read it. Newton's first law: there are such inertial frames of reference in which... Let's close this tutorial, I don't understand either. Okay, I’m joking, I understand, but I’ll explain it more simply.

Newton's first law: if a body stands still or moves uniformly (without acceleration), the sum of the forces acting on it is zero.

Conclusion: If a body moves at a constant speed or stands still, the vector sum of forces will be zero.

Newton's II law: if a body moves uniformly accelerated or uniformly decelerated (with acceleration), the sum of the forces acting on it is equal to the product of mass and acceleration.

Conclusion: If a body moves with varying speed, then the vector sum of the forces that somehow influence this body (traction force, friction force, air resistance force) is equal to the mass of this body times the acceleration.

In this case, the same body most often moves differently (uniformly or with acceleration) in different axes. Let's consider just such an example.

Task 1. Determine the coefficient of friction of the tires of a car weighing 600 kg if the engine traction force of 4500 N causes an acceleration of 5 m/s².

In such problems, it is necessary to make a drawing and show the forces that act on the machine:

On X Axis: movement with acceleration

On the Y Axis: no movement (here the coordinate, as it was zero, will remain the same, the machine does not go up the mountains or down)

Those forces whose direction coincides with the direction of the axes will be plus, in the opposite case - minus.

Along the X axis: the traction force is directed to the right, just like the X axis, the acceleration is also directed to the right.

Ftr = μN, where N is the support reaction force. On the Y axis: N = mg, then in this problem Ftr = μmg.

We get that:

The friction coefficient is a dimensionless quantity. Therefore, there are no units of measurement.

Answer: 0.25

Problem 2. A load weighing 5 kg, tied to a weightless inextensible thread, is lifted upward with an acceleration of 3 m/s². Determine the tension of the thread.

Let's make a drawing and show the forces that act on the load

T - thread tension force

On the X axis: no power

Let's figure out the direction of forces on the Y axis:

Let's express T (tension force) and substitute the numerical values:

Answer: 65 N

The most important thing is not to get confused with the direction of forces (along the axis or against), everything elsemake a calculator or everyone’s favorite column.

Not always all forces acting on a body are directed along the axes.

A simple example: a boy pulling a sled

If we also construct the X and Y axes, then the tension (traction) force will not lie on any of the axes.

To project the traction force onto the axes, recall a right triangle.

The ratio of the opposite side to the hypotenuse is the sine.

The ratio of the adjacent leg to the hypotenuse is the cosine.

Traction force on the Y axis - segment (vector) BC.

The traction force on the X axis is a segment (vector) AC.

If this is not clear, look at problem #4.

The longer the rope and, accordingly, the smaller the angle α, the easier it will be to pull the sled. Ideal when the rope is parallel to the ground, because the force that acts on the X axis is Fнcosα. At what angle is the cosine maximum? The larger this leg is, the stronger the horizontal force.

Task 3. The block is suspended by two threads. The tension force of the first is 34 N, the second- 21Н, θ1 = 45°, θ2 = 60°. Find the mass of the block.

Let's introduce the axes and project the forces:

We get two right triangle. Hypotenuses AB and KL are tension forces. LM and BC - projections on the X axis, AC and KM - on the Y axis.

Answer: 4.22 kg

Task 4. A block with a mass of 5 kg (mass is not needed in this problem, but so that everything is known in the equations, let’s take a specific value) slides off a plane that is inclined at an angle of 45°, with a coefficient of friction μ = 0.1. Find the acceleration of the block?

When there is an inclined plane, it is best to direct the axes (X and Y) in the direction of movement of the body. Some forces in this case (here it is mg) will not lie on any of the axes. This force must be projected so that it has the same direction as the taken axes.
ΔABC is always similar to ΔKOM in such problems (by right angle and the angle of inclination of the plane).

Let's take a closer look at ΔKOM:

We obtain that KO lies on the Y axis, and the projection of mg onto the Y axis will be with a cosine. And the vector MK is collinear (parallel) to the X axis, the projection mg onto the X axis will be with a sine, and the vector MK is directed against the X axis (that is, it will be with a minus).

Do not forget that if the directions of the axis and the force do not coincide, it must be taken with a minus!

From the Y axis we express N and substitute it into the equation of the X axis, we find the acceleration:

Answer: 6.36 m/s²

As you can see, the mass in the numerator can be taken out of brackets and reduced with the denominator. Then it is not necessary to know it; it is possible to get an answer without it.
Yes Yes, under ideal conditions (when there is no air resistance, etc.), both the feather and the weight will roll (fall) at the same time.

Task 5. A bus slides down a hill at a slope of 60° with an acceleration of 8 m/s² and a traction force of 8 kN. The coefficient of friction between tires and asphalt is 0.4. Find the mass of the bus.

Let's make a drawing with forces:

Let's introduce the X and Y axes. Project mg onto the axes:

Let's write Newton's second law for X and Y:

Answer: 6000 kg

Task 6. A train moves along a curve of radius 800 m at a speed of 72 km/h. Determine how much the outer rail should be higher than the inner one. The distance between the rails is 1.5 m.

The most difficult thing is to understand which forces act where, and how the angle affects them.

Remember, when you drive in a circle in a car or on a bus, where does it push you? This is why the tilt is needed so that the train does not fall on its side!

Corner α specifies the ratio of the difference in the height of the rails to the distance between them (if the rails were horizontal)

Let's write down what forces act on the axis:

The acceleration in this problem is centripetal!

Let's divide one equation by another:

Tangent is the ratio of the opposite side to the adjacent side:

Answer: 7.5 cm

As we found out, the solution similar tasks comes down to arranging the directions of forces, projecting them onto axes and solving systems of equations, almost a mere trifle.

To reinforce the material, solve several similar problems with hints and answers.