Motion on an inclined plane of a body: speed, friction, time. Movement of a body up an inclined plane Body along an inclined plane

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

Basic dynamics formula

Before proceeding to the study of the physics of motion of a body 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 movement of macroscopic surrounding bodies, deduced three laws that currently bear his surname. All classical mechanics is based on these laws. We are only interested in the second law in this article. Its mathematical form is given below:

The formula says that the action of an external force F¯ will give an acceleration a¯ to a body of mass m. This simple expression will be further used to solve the problems of motion of a body along an inclined plane.

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

Inclined plane and forces acting on a body located on it

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

The figure below shows that the body (car) is at rest on a plane inclined at an angle to the horizon. What forces are acting on it?

The list below lists these powers:

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

Gravity

First of all, it is gravity (F g). 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 acceleration of the body and, in fact, is the only driving factor for the body in question. The second component causes the occurrence of the reaction force of the support.

Support reaction

The second force acting on the body is the support reaction (N). The reason for its appearance is connected with Newton's third law. The value of N shows the force with which the plane acts on the body. It is directed upward perpendicular to the inclined plane. If the body were on a horizontal surface, then N would be equal to its weight. In the case under consideration, N is only equal to the second component obtained by expanding the force of gravity (see the paragraph above).

The reaction of the support does not directly affect the nature of the movement of the body, since it is perpendicular to the plane of inclination. Nevertheless, it causes the appearance of friction between the body and the surface of the plane.

Friction force

The third force that should be taken into account when studying the motion of a body on an inclined plane is friction (F f). The physical nature of friction is not easy. Its appearance is associated with microscopic interactions of contacting bodies with inhomogeneous contact surfaces. There are three types of this force:

• rest;
• slip;
• rolling.

The static and sliding friction are described by the same formula:

where µ is a dimensionless coefficient, the value of which is determined by the materials of the rubbing bodies. So, when sliding friction of a tree on a tree µ = 0.4, and ice on ice - 0.03. The coefficient for static friction is always greater than that for sliding.

Rolling friction is described by a formula different from the previous one. It looks like:

Here r is the radius of the wheel, f is a coefficient having the dimension of the reciprocal length. This friction force is usually much less than the previous ones. Note that its value is affected by the radius of the wheel.

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

Thread tension

When solving problems of motion of a body along an inclined plane, this force is not always present. Its appearance is determined by the fact that a body located on an inclined plane is connected with another body by means of an inextensible thread. Often a second body hangs on a thread through a block outside the plane.

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

The appearance of this force in the problem significantly complicates the solution process, since it is necessary to consider simultaneously the motion of two bodies (on the plane and hanging down).

The task of determining the critical angle

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

Assume that a timber beam has a mass of 2 kg. It is on a wooden plane. It should be determined at what critical angle of inclination of the plane the beam will begin to slide along it.

The sliding of the beam will occur only when the total force acting downward along the plane on it is greater than zero. Thus, to solve this problem, it is enough to determine the resulting force and find the angle at which it becomes greater than zero. According to the condition of the problem, only two forces will act on the beam along the plane:

• component of gravity F g1 ;
• static friction F f .

For the sliding of the body to begin, the following condition must be met:

Note that if the gravity component exceeds the static friction, then it will also be greater than the sliding friction force, that is, the movement that has begun will continue with constant acceleration.

The figure below shows the directions of all acting forces.

Let us denote the critical angle by the symbol θ. It is easy to show that the forces F g1 and F f will be equal:

F g1 = m × g × sin(θ);

F f = µ × m × g × cos(θ).

Here m × g is the weight of the body, µ is the coefficient of the static friction force for a pair of wood-wood materials. From the corresponding table of coefficients, you can find that it is equal to 0.7.

We substitute the found values ​​into the inequality, we get:

m × g × sin(θ) ≥ µ × m × g × cos(θ).

Transforming this equality, we arrive at the condition of body motion:

tg(θ) ≥ µ =>

θ ≥ arctan(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 an inclined plane, but is uniquely determined by the coefficient of static friction µ. Substituting its value into the inequality, we obtain the value of the critical angle:

θ ≥ arctan(0.7) ≈ 35o.

The task of determining the acceleration when moving along an inclined plane of the body

Now let's solve a slightly different problem. Let there be a bar made of wood on a glass inclined plane. The plane is inclined to the horizon at an angle of 45 o . It is necessary to determine with what acceleration the body will move if its mass is 1 kg.

Let us write the main equation of dynamics for this case. Since the force F g1 will be directed along the movement, and F f against it, the equation will take the form:

F g1 - F f = m × a.

We substitute the formulas obtained in the previous problem for the forces F g1 and F f , we have:

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

Where do we get the formula for acceleration:

a = g × (sin(θ) - µ × cos(θ)).

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

Given that the coefficient µ for rubbing wood-glass materials is 0.2, we substitute all the parameters into equality, we get the answer:

Thus, the technique for solving problems with an inclined plane consists in determining the resulting force acting on the body, and in the subsequent application of Newton's second law.

Physics: motion of a body on an inclined plane. Examples of solutions and tasks - all interesting facts and achievements of science and education on the site

In our case F n \u003d m g, because the surface is horizontal. But, the normal force in magnitude does not always coincide with the force of gravity.

Normal force - the force of interaction between the surfaces of contacting bodies, the larger it is, the stronger the friction.

Normal force and friction force are proportional to each other:

F tr \u003d μF n

0 < μ < 1 - coefficient of friction, 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 force of friction does not depend on the area of ​​contact between two surfaces (if their masses do not change).

Please note: the equation F tr \u003d μF n is not a relation 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. Varieties of friction

Friction is of two types: static and kinetic.

Static friction (static friction) acts between contacting bodies that are at rest relative to each other. Static friction manifests itself 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 it is much easier to keep the cabinet moving. This is explained by the fact that when the surfaces of bodies move, they "do not have time" to switch to contact at the microscopic level.

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

We calculate the component of gravity. First we need to know 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 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 inc = F g cos(90° - α) = mgsinα

Required force to lift the ball:

F = F g inc + F friction = mgsinα + F friction

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

F friction = μF norm

Calculate the normal force F norms, 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 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 on a horizontal plane, rolling down an inclined plane with a length 10 meters with sliding friction coefficient μ = 0.05

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

Component of gravity along an inclined plane:

F g cos(90° - α) = mgsinα

Normal strength:

F n \u003d mgsin (90 ° - α) \u003d mgcos (90 ° - α)

Sliding friction force:

F friction = 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 \u003d 2as; V = 2as = 2 4.5 10 = 9.5 m/s

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

Total Strength:

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

The minus sign means that the force is in the opposite direction of the motion. Determine the acceleration deceleration of the ball:

a \u003d F / m \u003d -0.49 / 1 \u003d -0.49 m / s 2

Ball stopping distance:

V 1 2 - V 0 2 \u003d 2as; s \u003d (V 1 2 - V 0 2) / 2a

Since we are determining the path of the ball to a complete stop, then V1=0:

s \u003d (-V 0 2) / 2a \u003d (-9.5 2) / 2 (-0.49) \u003d 92 m

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

Bukina Marina, 9 V

Movement of a body on an inclined plane

with the transition to horizontal

As the body under study, I took a coin with a denomination of 10 rubles (edges are ribbed).

Specifications:

Coin diameter - 27.0 mm;

Coin weight - 8.7 g;

Thickness - 4 mm;

The coin is made of brass-cupronickel alloy.

For an inclined plane, I decided to take a book 27 cm long. It will be an inclined plane. The horizontal plane is unlimited, since the cylindrical body, and in the future the coin, rolling down 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.

The following were taken as additional equipment for measurements: a stopwatch, an ordinary ruler, a long thread, a protractor, a calculator.

On Fig.1. schematic representation of a coin on an inclined plane.

Let's launch a coin.

The results obtained will be entered in table 1

Table 1

The trajectory of the coin in all experiments was different, but some parts of the trajectory were similar. On an inclined plane, the coin moved in a straight line, and when moving on a horizontal plane, it moved curvilinearly.

Figure 2 shows the forces acting on a coin as it moves down an inclined plane:

With the help of Newton's II Law, we derive a formula for finding the acceleration of a coin (according to Fig. 2.):

First, let's write the formula II of Newton's Law in vector form.

Where is the acceleration with which the body moves, is the resultant force (forces acting on the body), https://pandia.ru/text/78/519/images/image008_3.gif" width="164" height="53" >, three forces act on our body during movement: gravity (Ftyazh), friction force (Ftr) and support reaction force (N);

Get rid of the vectors by projecting onto the X and Y axes:

Where is the coefficient of friction

Since we do not have data on the numerical value of the coefficient of friction of the coin on our plane, we will use another formula:

Where S is the path traveled by the body, V0 is the initial speed of the body, a is the acceleration with which the body moved, t is the time interval of the body's movement.

because ,

in the course of mathematical transformations, we obtain the following formula:

When projecting these forces onto the X axis (Fig. 2.), it is clear that the directions of the path and acceleration vectors coincide, we write the resulting form, getting rid of the vectors:

For S and t we take the average values ​​from the table, we find the acceleration and speed (the body moved along an inclined plane in a straight line with uniform acceleration).

https://pandia.ru/text/78/519/images/image021_1.gif" align="left" width="144" height="21">

Similarly, we find the acceleration of the body on a horizontal plane (on the horizontal plane the body moved rectilinearly with uniform slowness)

R=1.35 cm, where R is the radius of the coin

where - angular velocity, - centripetal acceleration, - frequency of rotation of the body in a circle

The movement of a body along an inclined plane with a transition to a horizontal one is rectilinear, uniformly accelerated, complex, which can be divided into rotational and translational movements.

The motion of a body on an inclined plane is rectilinear and uniformly accelerated.

According to Newton's II Law, it can be seen that acceleration depends only on the resultant force (R), and it remains constant throughout the entire path along the inclined plane, because in the final formula, after projecting Newton's II Law, the quantities involved in the formula are constant https://pandia.ru/text/78/519/images/image029_1.gif" width="15" height="17">rotation from some initial position.

Translational is such a motion of an absolutely rigid body, in which any straight line, rigidly connected with the body, moves, remaining parallel to itself. All points of a body moving forward at each moment of time have the same speeds and accelerations, and their trajectories are completely combined with parallel transfer.

Factors affecting the time of movement of the body

on an inclined plane

with the transition to horizontal

Dependence of time on coins of different denominations (i.e., having different d (diameter)).

table 2

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

Dependence of time on the angle of inclination

Table 3

The body that slides down an inclined plane. In this case, the following forces act on it:

Gravity mg directed vertically downwards;

Support reaction force N, directed perpendicular to the plane;

The sliding friction force Ftr is directed opposite to the speed (up along the inclined plane when the body slides).

Let us introduce an inclined coordinate system, the OX axis of which is directed downward along the plane. This is convenient, because in this case it will be necessary to decompose into components only one vector - the vector of gravity mg, and the vectors of the friction force Ftr and the reaction force of the support N are already directed along the axes. With this expansion, the x-component of gravity is equal to mg sin(α) and corresponds to the “pulling force” responsible for the accelerated downward movement, and the y-component - mg cos(α) = N balances the support reaction force, since the movement of the body along the OY axis missing.

The force of sliding friction Ftr = µN is proportional to the reaction force of the support. This allows one to obtain the following expression for the friction force: Ffr = µmg cos(α). This force is opposite to the "pulling" component of gravity. Therefore, for a body sliding down, we obtain the expressions for the total resultant force and acceleration:

Fx = mg(sin(α) – µcos(α));

ax = g(sin(α) – μ cos(α)).

acceleration:

speed is

v=ax*t=t*g(sin(α) – µ cos(α))

after t=0.2 s

speed is

v=0.2*9.8(sin(45)-0.4*cos(45))=0.83 m/s

The force with which a body is attracted to the Earth under the influence of the Earth's gravitational field is called gravity. According to the law of universal gravitation, on the surface of the Earth (or near this surface), a body of mass m is affected by the force of gravity

Fт=GMm/R2 (2.28)

where M is the mass of the Earth; R is the radius of the Earth.

If only gravity acts on the body, and all other forces are mutually balanced, the body is in free fall. According to Newton's second law and formula (2.28), the free fall acceleration modulus g is found by the formula

g=Ft/m=GM/R2. (2.29)

From formula (2.29) it follows that the free fall acceleration does not depend on the mass m of the falling body, i.e. for all bodies in a given place on the Earth it is the same. From formula (2.29) it follows that Fт = mg. In vector form

In § 5 it was noted that since the Earth is not a sphere, but an ellipsoid of revolution, its polar radius is less than the equatorial one. From formula (2.28) it can be seen that for this reason the force of gravity and the acceleration of free fall caused by it at the pole is greater than at the equator.

The force of gravity acts on all bodies in the Earth's gravitational field, but not all bodies fall to the Earth. This is explained by the fact that the movement of many bodies is hindered by other bodies, such as supports, suspension threads, etc. Bodies that restrict the movement of other bodies are called bonds. Under the action of gravity, the bonds are deformed and the reaction force of the deformed bond, according to Newton's third law, balances the force of gravity.

In § 5 it was also noted that the acceleration of free fall is affected by the rotation of the Earth. This influence is explained as follows. The frames of reference associated with the surface of the Earth (except for the two associated with the poles of the Earth) are not, strictly speaking, inertial frames of reference - the Earth rotates around its axis, and with it move along circles with centripetal acceleration and such frames of reference. This non-inertiality of reference systems is manifested, in particular, in the fact that the value of the acceleration of free fall turns out to be different in different places on the Earth and depends on the geographical latitude of the place where the reference frame associated with the Earth is located, relative to which the acceleration of gravity is determined.

Measurements carried out at different latitudes showed that the numerical values ​​of the gravitational acceleration differ little from each other. Therefore, with not very accurate calculations, one can neglect the non-inertiality of reference systems associated with the Earth's surface, as well as the difference in the shape of the Earth from a spherical one, and assume that the acceleration of free fall in any place on the Earth is the same and equal to 9.8 m/s2.

From the law of universal gravitation it follows that the force of gravity and the acceleration of free fall caused by it decrease with increasing distance from the Earth. At a height h from the Earth's surface, the gravitational acceleration module is determined by the formula

It has been established that at a height of 300 km above the Earth's surface, the free fall acceleration is less than at the Earth's surface by 1 m/s2.

Consequently, near the Earth (up to heights of several kilometers), the force of gravity practically does not change, and therefore the free fall of bodies near the Earth is a uniformly accelerated motion.

Body weight. Weightlessness and overload

The force in which, due to attraction to the Earth, the body acts on its support or suspension, is called the weight of the body. Unlike gravity, which is a gravitational force applied to a body, weight is an elastic force applied to a support or suspension (i.e., to a connection).

Observations show that the weight of the body P, determined on a spring balance, is equal to the force of gravity Ft acting on the body only if the balance with the body relative to the Earth is at rest or moving uniformly and rectilinearly; In this case

If the body is moving with acceleration, then its weight depends on the value of this acceleration and on its direction relative to the direction of free fall acceleration.

When a body is suspended on a spring scale, two forces act on it: the force of gravity Ft=mg and the force of elasticity Fyp of the spring. If at the same time the body moves vertically up or down relative to the direction of free fall acceleration, then the vector sum of the forces Ft and Fup gives the resultant, which causes the acceleration of the body, i.e.

Ft + Fup \u003d ma.

According to the above definition of the concept of "weight", we can write that Р=-Fyп. taking into account the fact that Ft=mg, it follows that mg-ma=-Fyp. Therefore, P \u003d m (g-a).

Forces Ft and Fup are directed along one vertical straight line. Therefore, if the acceleration of the body a is directed downward (i.e., it coincides in direction with the acceleration of free fall g), then modulo

If the acceleration of the body is directed upwards (i.e., opposite to the direction of free fall acceleration), then

P \u003d m \u003d m (g + a).

Consequently, the weight of a body whose acceleration coincides in direction with the acceleration of free fall is less than the weight of a body at rest, and the weight of a body whose acceleration is opposite to the direction of acceleration of free fall is greater than the weight of a body at rest. The increase in body weight caused by its accelerated movement is called overload.

In free fall a=g. it follows that in this case P=0, i.e., there is no weight. Therefore, if bodies move only under the influence of gravity (i.e., freely fall), they are in a state of weightlessness. A characteristic feature of this state is the absence of deformations and internal stresses in freely falling bodies, which are caused in resting bodies by gravity. The reason for the weightlessness of bodies is that the force of gravity imparts the same accelerations to a freely falling body and its support (or suspension).

This article talks about how to solve problems about moving along an inclined plane. A detailed solution of the problem of the motion of bound bodies along an inclined plane from the Unified State Examination in Physics is considered.

Solution of the problem of motion on an inclined plane

Before proceeding directly to solving the problem, as a tutor in mathematics and physics, I recommend that you carefully analyze its condition. You need to start with the image of the forces that act on the 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. With the direction of these forces, everything is clear. 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 weight "outweighs" the left one, then the friction force will be directed opposite to the vector. On the contrary, if the left weight "outweighs" the right one, then the friction force will be co-directed with the vector.

The right load is pulled down by the force N. Here we have taken the free fall acceleration m/s 2 . The left load is also pulled down by gravity, but not all of it, but only its "part", since the load lies on an inclined plane. This "part" is equal to the projection of gravity on an inclined plane, that is, the leg in a right triangleshown in the figure, that is, equal to H.

That is, it “outweighs” the right load. 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 when the body can be modeled by a material point):

The second important question to be dealt with is whether this bound system will move at all? Suddenly it turns out that the friction force between the left weight and the inclined plane will be so great that it will not let it move?

This situation will be possible in the case when the maximum friction force, the modulus of which is determined by the formula set the system in motion. That is, the very "outweighing" force, which is equal to N.

The module of the reaction force of the support is equal to the length of the leg in a triangle according to Newton's 3-mouse law (with what force the load presses on the inclined plane, with the same force the inclined plane acts on the load). That is, the reaction force of the support is N. Then the maximum value of the friction force is N, which is less than the value of the "outweighing force".

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

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

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

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

Here, projections are taken with a minus, the vectors of which are directed against the direction of the corresponding coordinate axis. With a plus, projections are taken, the vectors of which are co-directed with the corresponding coordinate axis.

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 recalled above, by definition, the friction force modulus is equal to the product of the friction coefficient and the support reaction force modulus. Consequently, . Then the original system of equations takes the form:

We now write Newton's 2nd law for the right body:

In the projection onto the axis, we get.