Movement of the roller on the inclined plane. How do inclined planes work? Inclined plane and forces acting on the body on it

Similarly, the lever, the inclined planes reduce the effort required to lift the tel. For example, a concrete block weighing 45 kilograms to raise hands is quite difficult, but it is quite possible to drag it upstairs on the inclined plane. The weight of the body placed on the inclined plane is unfolded into two components, one of which is parallel, and the other is perpendicular to its surface. To move the block upwards, a person should overcome only the parallel component, the value of which grows with an increase in the angle of inclination of the plane.

Inclined planes are very diverse in constructive execution. For example, the screw consists of an inclined plane (thread), which spoils its cylindrical part. When screwing the screw into the part, its carving penetrates the body of the part, forming a very strong connection due to the large friction between the detail and the coils of the thread. Vice convert the action of the lever and rotary traffic Screws in linear squeezing force. By the same principle, the jack is also used to raise heavy loads.

Power on the inclined plane

The body located on the inclined plane, the strength of gravity acts in parallel and perpendicular to its surface. To move the body up the inclined plane requires a force equal to the magnitude of the gravity component, the parallel surface of the plane.

Inclined planes and screws

The kinship of the screw with the inclined plane is easy to trace if wrap the cylinder cut diagonally to a sheet of paper. The resulting spiral is identical by the arrangement of the screw thread.

Forces acting on the screw

When turning the screw, its carving creates a very greater force applied to the material of the part in which he will be screwed. This force drags the screw forward, if it turns clockwise, and back, if it turns counterclockwise.

Screw for lifting weights

Rotating jack screws develop great strength, allowing them to raise such heavy bodies like passenger or trucks. When turning the central screw lever, two end of the jack is tightened together, producing the necessary rise.

Inclined planes for splitting

Wedge consists of two inclined planes connected by its bases. When clogging a wedge into a tree, inclined planes develop lateral forces sufficient to split the most durable lumber.

Strength and work

Despite the fact that the inclined plane can facilitate the task, it does not reduce the amount of work required for its execution. The lifting of the concrete block weighing 45 kg (w) by 9 meters vertically upwards (the far drawing on the right) requires the operation of 45x9 kilogram meters, which corresponds to the product of the block weight by the value of movement. When the block is on an inclined plane with an angle of inclination of 44.5 °, the force (F) required to accumulate the block decreases to 70 percent of its weight. Although it makes it easier to move the block, but now, to raise a block to a height of 9 meters, it must be dragged on a plane of 13 meters. In other words, the winnings in force is equal to the height of the lift (9 meters) divided by the length of movement along the inclined plane (13 meters).

The body movement on the inclined plane is a classic example of the body movement under the action of several uninforced forces. The standard method for solving problems of such a movement is to decompose the vectors of all forces by components directed along the coordinate axes. Such components are linearly independent. This allows you to record Newton's second law for a component along each axis separately. Thus, the second law of Newton, which is a vector equation, turns into a system of two (three for a three-dimensional case) of algebraic equations.

Forces acting on the bar
accelerated movement

Consider the body that slides down the inclined plane. In this case, the following strengths are valid for it:

• Gravity m. g. directed vertically down;
• Power reaction support N. directed perpendicularly plane;
• Slip friction force F. Tr, directed oppositely speed (upwards along the inclined plane when scaling the body)

When solving the tasks in which the inclined plane appears often conveniently enter the inclined coordinate system, the OX axis of which is directed along the plane down. This is convenient, because in this case you have to lay out only one vector - vector of gravity m. g. , and friction strength vector F. Tr and reaction force support N. Already directed along the axes. With such decomposition of the X-component of gravity is equal to mG. sin ( α ) and corresponds to the "pulling force" responsible for the accelerated movement down, and the Y component - mG. COS ( α ) = N. Babs balancing the power of the support, because there is no body movement along the Oy axis.
Slip friction force F. Tr \u003d. μN. Proportional to the strength of the support reaction. This allows you to get the following expression for friction force: F. Tr \u003d. μmg. COS ( α ). This force is contamited by the "pulling" component of gravity. Therefore for body scaling down , we obtain expressions of the total automatic strength and acceleration:

F. x \u003d. mG.(sin ( α ) – µ COS ( α ));
a. x \u003d. g.(sin ( α ) – µ COS ( α )).

It is not difficult to see that if µ < tg(α ), the expression has a positive sign and we are dealing with an equilibrium movement down the inclined plane. If µ \u003e TG ( α ), the acceleration will have a negative sign and movement will be equivalent. Such a movement is possible only if the body is granted initial speed towards down the slope. In this case, the body will gradually stop. If provided µ \u003e TG ( α ) The subject is initially resting, it will not start scoring down. Here the friction force of peace will fully compensate for the "pulling" component of gravity.

When friction coefficient exactly equal to Tangent The angle of inclination of the plane: µ \u003d TG ( α ), We are dealing with mutual compensation for all three forces. In this case, according to the first law of Newton, the body can either rest, or move at a constant speed (at the same time uniform traffic It is possible only down).

Forces acting on the bar
Sliding on the inclined plane:
Slow motion case

However, the body can and drive up the inclined plane. An example of such a movement is the movement of a hockey washer up the ice hill. When the body moves up, the friction force and the "pulling" component of gravity are directed down along the inclined plane. In this case, we are always dealing with an equilibrium movement, since the total force is directed in the opposite speed of the side. An expression for acceleration for this situation is obtained similarly and differs only by the sign. So for body sliding up the inclined plane We have.

Bukina Marina, 9 in

Body movement on the inclined plane

with the transition to horizontal

As a studied body, I took the coin the dignity of 10 rubles (the edge of the ribbed).

Specifications:

The diameter of the coin is 27.0 mm;

Mass of coins - 8.7 g;

Thickness - 4 mm;

The coin is made of brass-melchior alloy.

For an inclined plane, I decided to take a book with a length of 27 cm. It will be an inclined plane. The horizontal plane is unlimited, since the cylindrical body, and in the further coin, rolling from the book, will continue its movement on the floor (parquet board). The book is raised to a height of 12 cm from the floor; The angle between the vertical plane and the horizontal is 22 degrees.

As an additional measurement equipment, a stopwatch, a ruler, a long thread, a vehicle, a calculator were taken.

In Fig.1. Sketchy image of coins on an inclined plane.

Perform the start of the coin.

The results obtained will be in Table 1

Table 1

The trajectory of the movement of the coin in all experiments was different, but some parts of the trajectory were similar. On the inclined plane of the coin moved straightforwardly, and when moving on a horizontal plane - curved.

Figure 2 shows the forces acting on the coin during its movement along the inclined plane:

With the help of the II of the Newton law, we will bring the formula for finding the acceleration of the coin (in Fig. 2):

To begin with, I will write Newton's law formula II in vector formula.

Where - the acceleration with which the body moves is the resultant force (the forces acting on the body), https://pandia.ru/text/78/519/images/image008_3.gif "width \u003d" 164 "height \u003d" 53 " \u003e, on our body, during the movement there are three forces: the strength of gravity (FLAYED), the friction force (FTR) and the power of the support reaction (N);

Get rid of vectors using projects on the X and Y axis:

Where is the friction coefficient

T. K. We do not have data on the numerical value of the coin coefficient of the coin on our plane, we use another formula:

Where S is the path passing by the body, the V0- initial velocity of the body, and the acceleration with which the body was moving, T is the interval of the body movement time.

t. ,

in the course of mathematical transformations we get the following formula:

When projection of these forces on the X axis (Fig.2.) It can be seen that the direction of the path vectors and acceleration coincide, write down the resulting form, getting rid of the vectors:

For S and T, we will take the average values \u200b\u200bfrom the table, we will find acceleration and speed (according to the inclined plane, the body moved straightly equally).

https://pandia.ru/text/78/519/images/image021_1.gif "Align \u003d" Left "width \u003d" 144 "height \u003d" 21 "\u003e

Similarly, we find the acceleration of the body on the horizontal plane (the body has moved straightly equifiable) to the horizontal plane)

R \u003d 1, 35 cm, where R is a radius of coins

where - angular velocity, CENTROSTREMTREMENTAL ACCELERATION, - Body circulation frequency

The movement of the body on the inclined plane with the transition to the horizontal - straight equative, complex, which can be divided into rotational and translational motion.

The movement of the body on the inclined plane is straight-linean equal.

According to the II, Newton's law it can be seen that the acceleration depends only on the relay force (R), and it remains the magnitude of the entire path along the inclined plane, because of the inclined plane, since eltimate formula, after projection II of Newton's law, the values \u200b\u200binvolved in the formula are permanent https://pandia.ru/text/78/519/images/image029_1.gif "width \u003d" 15 "height \u003d" 17 "\u003e turn from some initial position .

Progressive is called such a movement absolutely solidIn which any straight, rigidly connected with the body, moves, remaining parallel to itself. All points of the body moving progressively, at each moment of time have the same speeds and acceleration, and their trajectories are fully combined with parallel transfer.

Factors affecting the time of the body

by inclined plane

with the transition to horizontal

The dependence of time from coins of different dignity (i.e. having different D (diameter)).

table 2

The larger the diameter of the coin, the more time its movement.

The dependence of time from the angle of inclination

Table 3.

On the surface of the Earth gravity (gravitis) is constant and equal to the product of the mass of the incident body to accelerate the free fall: F G \u003d MG

It should be noted that the acceleration of the free incidence is permanent: G \u003d 9.8 m / s 2, and is directed towards the center of the Earth. Based on this, we can say that the bodies with different mass will fall on the ground equally quickly. How so? If you throw a piece of cotton wool and brick from the same height, the latter will do your way to the ground faster. Do not forget about air resistance! For cotton, it will be essential because its density is very small. In airless space, brick and wool will fall at the same time.

The ball moves along the inclined plane of 10 meters long, the angle of inclination of the plane is 30 °. What will be the speed of the ball at the end of the plane?

Only the strength of gravity F G acts on the ball, directional perpendicular to the base of the plane. Under the action of this force (component directed along the surface of the plane), the ball will move. What will be the component of gravity, acting along the inclined plane?

To determine the component, it is necessary to know the angle between the power vector F G and the inclined plane.

Determine the angle is quite simple:

• the sum of the angles of any triangle is 180 °;
• the angle between the power vector F G and the base of the inclined plane is 90 °;
• the angle between the inclined plane and its base is α

Based on the foregoing, the desired angle will be equal to: 180 ° - 90 ° - α \u003d 90 ° - α

From trigonometry:

F G None \u003d F G · COS (90 ° -α)

SINα \u003d COS (90 ° -α)

F G None \u003d F G · SINα

This is true:

• at α \u003d 90 ° (vertical plane) f g totted \u003d f g
• at α \u003d 0 ° (horizontal plane) F G None \u003d 0

We define the acceleration of the ball from the famous formula:

F g · sinα \u003d m · a

A \u003d f g · sinα / m

A \u003d m · g · sinα / m \u003d g · sinα

The acceleration of the ball along the inclined plane does not depend on the mass of the ball, but only on the angle of the plane.

Determine the ball speed at the end of the plane:

V 1 2 - V 0 2 \u003d 2 · A · s

(V 0 \u003d 0) - the ball starts moving from the spot

V 1 2 \u003d √2 · A · s

V \u003d 2 · G · sinα · s \u003d √2 · 9.8 · 0.5 · 10 \u003d √98 \u003d 10 m / s

Pay attention to the formula! The body velocity at the end of the inclined plane will depend only on the angle of the plane and its length.

In our case, the speed of 10 m / s at the end of the plane will have a billiard ball and a passenger car, and a dump truck, and a schoolboy on sledding. Of course, we do not consider friction.

The dynamics is one of the important sections of physics, which studies the causes of the movement of bodies in space. In this article, consider from the point of view of the theory one of typical tasks The speakers are the movement of the body on the inclined plane, as well as we give examples of solutions of some practical problems.

Basic dynamics formula

Before moving to the study of body motion physics on the plane by inclined, we present the necessary theoretical information to solve this problem.

In XVII, Isaac Newton, thanks to the practical observations of the movement of macroscopic surrounding bodies, brought three laws that are currently hosting his last name. On these laws are all classical mechanics. We are interested in this article only the second law. Its mathematical species is below:

The formula suggests that the effect of external power F¯ will give acceleration A¯ body mass M. This is a simple expression will be further used to solve the problems of the body movement on the plane inclined.

Note that strength and acceleration are the magnitude of the vector, directed in the same side. In addition, force is an additive characteristic, that is, in the above formula, F¯ can be considered as a resulting effect on the body.

Inclined plane and forces acting on the body on it

The key point that depends on the success of solving the problems of the body movement on the plane of the inclined, is to determine the forces acting on the body. Under the definition of forces understand the knowledge of their modules and directions of action.

Below is a drawing, where it is shown that the body (car) is located at rest on the plane inclined at the angle to the horizon. What are the forces on it?

The list below lists these forces:

• gravity;
• support reactions;
• friction;
• thread tension (if present).

Gravity

First of all it is the power of gravity (F G). It is directed vertically down. Since the body has the ability to move only along the surface of the plane, then when solving tasks, 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 and, in fact, is the only driving factor for the body under consideration. The second component determines the emergence of the reaction force of the support.

Reaction support

The second body acting on the body is the support reaction (N). The reason for its appearance is related to the third law of Newton. N value shows how power the plane affects the body. It is aimed up perpendicular to the plane inclined. If the body was on the horizontal surface, then n would be equal to its weight. In the case under consideration, n is equal to only the second component obtained in the decomposition of gravity (see paragraph above).

The support reaction does not have direct impact on the nature of the body's movement, since it is perpendicular to the tilt plane. Nevertheless, it causes the appearance of friction between the body and the surface of the plane.

Friction force

The third force, which should be taken into account when studying the movement of the body on the inclined plane, is friction (F f). Physical nature friction is difficult. Its appearance is associated with microscopic interactions of contacting bodies having inhomogeneous contact surfaces. Highlight three types of this force:

• rest;
• slip;
• rolling.

The friction of peace and slip are described by the same formula:

where μ is a dimensionless coefficient whose value is determined by the materials of the rubbing tel. So, with friction of the sliding of the tree about the tree μ \u003d 0.4, and ice about the ice is 0.03. The coefficient for friction rest is always more than such for slipping.

Friction of rolling is described by different from the previous formula. It has the form:

Here R is the radius of the wheel, F is a coefficient having a dimension of the back length. This friction force is usually much smaller than the previous ones. Note that the radius of the wheel affects its value.

The force F f, whatever type it, is always directed against the movement of the body, that is, F f seeks to stop the body.

Thread tension

When solving the problems of the body movement on the inclined plane, this force is not always present. Its appearance is determined by the fact that the body located on the inclined plane is associated with the help of a non-express thread with another body. Often the second body hangs on the thread through the block outside the plane.

On the object located on the plane, the force of the thread tension affects or accelerating it, or slowing down. It all depends on the modules of the forces acting in the physical system.

The appearance of this force in the task significantly complicates the decision-making process, since it is necessary to consider at the same time the movement of two bodies (on the plane and hanging).

The task of determining the critical angle

Now it's time to apply the described theory to solve real movement problems on the inclined plane of the body.

Suppose that the timber bar has a lot of 2 kg. It is on a wooden plane. It should be determined, with what critical angle of inclination of the plane, the timber will begin to slide it.

The slip of the bar will come only when the total power acting along the plane will be larger than zero. Thus, to solve this task, it is enough to determine the resulting force and find the angle at which it will become more zero. According to the condition of the problem of the bar, only two forces will be opened along the plane:

• gravity component F G1;
• friction of rest F f.

To begged body gliding, condition must be performed:

It should be noted that if the component of gravity will exceed the friction of peace, it will also be greater than the friction force of the slip, that is, the starting movement will continue with constant acceleration.

The figure below shows the directions of all existing forces.

Denote the critical angle symbol θ. It is easy to show that the forces f g1 and f f will be equal:

F g1 \u003d m × g × sin (θ);

F f \u003d μ × m × g × cos (θ).

Here M × G is the weight of the body, μ is the coefficient of the friction force for a pair of wood-tree materials. From the corresponding coefficient table, it can be found that it is 0.7.

We substitute the found values \u200b\u200binto inequality, we get:

m × G × sin (θ) ≥ μ × m × g × cos (θ).

Transforming this equality, come to the condition of the body:

tG (θ) ≥ μ \u003d\u003e

θ ≥ arctg (μ).

We got a very interesting result. It turns out that the value of the critical angle θ does not depend on the mass of the body on the inclined plane, and is uniquely determined by the coefficient of friction of rest μ. Substituting its meaning to inequality, we obtain the size of the critical angle:

θ ≥ arctg (0.7) ≈ 35 o.

The task of determining the acceleration when moving on the inclined plane of the body

Now we decide a somewhat different task. Suppose on a glass inclined plane there is a timber bar. The plane to the horizon is tilted at an angle of 45 o. It should be determined which an acceleration will move the body if its mass is 1 kg.

We write the main equation of dynamics for this case. Since the force F G1 is directed along the movement, and F f against it, the equation will take the form:

F g1 - f f \u003d m × a.

We substitute the formula obtained in the previous problem for the forces F G1 and F F, we have:

m × g × sin (θ) - μ × m × g × cos (θ) \u003d m × a.

Where do you get a formula for acceleration:

a \u003d g × (sin (θ) - μ × cos (θ)).

Again, we got a formula in which there is no body weight. This fact means that bars of any mass will be slipped in the same time on the inclined plane.

Given that the coefficient μ for rubbing materials of the tree-glass is 0.2, we will substitute all the parameters to the equality, we will receive the answer:

Thus, the method of solving problems with an inclined plane is to determine the resulting force acting on the body, and in the subsequent application of the Second Law of Newton.

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