Body movement inclined plane- This is a classic example of the movement of a body under the influence of several non-coordinated forces. Standard method solving problems of this kind of motion consists in decomposing the vectors of all forces into components directed along the coordinate axes. Such components are linearly independent. This allows us to write Newton's second law for components along each axis separately. Thus, Newton's second law, which is a vector equation, turns into a system of two (three for the three-dimensional case) algebraic equations.
The forces acting on the block are
case of accelerated downward movement
Consider a body that is sliding down an inclined plane. In this case, the following forces act on it:
- Gravity m g , directed vertically downwards;
- Ground reaction force N , directed perpendicular to the plane;
- Sliding friction force F tr, directed opposite to the speed (up along the inclined plane when the body slides)
When solving problems in which an inclined plane appears, it is often convenient to introduce an inclined coordinate system, the OX axis of which is directed downward along the plane. This is convenient, because in this case you will have to decompose only one vector into components - the gravity vector m g
, and the friction force vector F
tr and ground reaction forces N
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 is mg cos( α
) = N balances the ground reaction force, since there is no body movement along the OY axis.
Sliding friction force F tr = µN proportional to the ground reaction force. This allows us to obtain the following expression for the friction force: F tr = µmg cos( α
). This force is opposite to the "pulling" component of gravity. Therefore for body sliding down
, we obtain expressions for the total resultant force and acceleration:
F x = mg(sin( α
) – µ
cos( α
));
a x = g(sin( α
) – µ
cos( α
)).
It's not hard to see what if µ < tg(α ), then the expression has positive sign and we are dealing with uniformly accelerated motion down an inclined plane. If µ >tg( α ), then the acceleration will have negative sign and the movement will be equally slow. Such movement is possible only if the body is given an initial speed down the slope. In this case, the body will gradually stop. If provided µ >tg( α ) the object is initially at rest, it will not begin to slide down. Here the static friction force will completely compensate for the “pulling” component of gravity.
When the friction coefficient is exactly equal to the tangent of the angle of inclination of the plane: µ = tg( α ), we are dealing with mutual compensation of all three forces. In this case, according to Newton's first law, the body can either be at rest or move with constant speed(Wherein uniform motion only possible downwards).
The forces acting on the block are
sliding on an inclined plane:
case of slow motion upward
However, the body can also drive up an inclined plane. An example of such motion is the movement of a hockey puck up an ice slide. When a body moves upward, both the frictional force and the “pulling” component of gravity are directed downward along the inclined plane. In this case, we are always dealing with uniformly slow motion, since the total force is directed in the direction opposite to the speed. The expression for acceleration for this situation is obtained in a similar way and differs only in sign. So for body sliding up an inclined plane , we have.
This article talks about how to solve problems about moving along an inclined plane. A detailed solution to the problem of the motion of coupled bodies on an inclined plane from the Unified State Examination in Physics is considered.
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 with an image of the forces that act on connected bodies:
Here and are the thread tension forces acting on the left and right body, respectively, is the support reaction force acting on 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 the inclined plane, that is, the leg in 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):
Second important question, which needs to be dealt with, will this connected system 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 side of the inclined plane), turns out to be more than that force that tries 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 “overwhelming 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.
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 higher coefficient 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!
Dynamics and kinematics are two important branches of physics that study the laws of movement of objects in space. The first considers the forces acting on the body, while the second deals directly with the characteristics of the dynamic process, without delving into the reasons for what caused it. Knowledge of these branches of physics must be used to successfully solve problems involving motion on an inclined plane. Let's look at this issue in the article.
Basic formula of dynamics
Of course, we are talking about the second law, which was postulated by Isaac Newton in the 17th century while studying the mechanical motion of solid bodies. Let's write it in mathematical form:
Action external force F¯ causes the appearance of linear acceleration a¯ in a body with mass m. Both vector quantities (F¯ and a¯) are directed in the same direction. The force in the formula is the result of the action on the body of all the forces that are present in the system.
In the case of rotational motion, Newton's second law is written as:
Here M and I are inertia, respectively, α is angular acceleration.
Kinematics formulas
Solving problems involving motion on an inclined plane requires knowledge of not only the main formula of dynamics, but also the corresponding expressions of kinematics. They connect acceleration, speed and distance traveled into equalities. For uniformly accelerated (uniformly decelerated) rectilinear motion the following formulas apply:
S = v 0 *t ± a*t 2 /2
Here v 0 is the value of the initial velocity of the body, S is the path traveled along a straight path during time t. A "+" sign should be added if the speed of the body increases over time. Otherwise (uniformly slow motion), the “-” sign should be used in the formulas. This is an important point.
If the movement is carried out along a circular path (rotation around an axis), then the following formulas should be used:
ω = ω 0 ± α*t;
θ = ω 0 *t ± α*t 2 /2
Here α and ω are the speed, respectively, θ is the angle of rotation of the rotating body during time t.
Linear and angular characteristics are related to each other by the formulas:
Here r is the radius of rotation.
Movement on an inclined plane: forces
This movement is understood as the movement of an object along a flat surface that is inclined at a certain angle to the horizon. Examples include a block sliding across a board or a cylinder rolling on an inclined sheet of metal.
To determine the characteristics of the type of movement under consideration, it is necessary first of all to find all the forces that act on the body (bar, cylinder). They may be different. IN general case these could be the following forces:
- heaviness;
- support reactions;
- and/or slipping;
- thread tension;
- external traction force.
The first three of them are always present. The existence of the last two depends on the specific system of physical bodies.
To solve problems involving movement along an inclined plane, it is necessary to know not only the magnitudes of forces, but also their directions of action. If a body rolls down a plane, the friction force is unknown. However, it is determined from the corresponding system of equations of motion.
Solution method
Solving problems of this type begins with determining the forces and their directions of action. To do this, the force of gravity is first considered. It should be decomposed into two component vectors. One of them should be directed along the surface of the inclined plane, and the second should be perpendicular to it. The first component of gravity, in the case of a body moving downwards, provides its linear acceleration. This happens anyway. The second is equal to All these indicators can have different parameters.
The friction force when moving along an inclined plane is always directed against the movement of the body. When it comes to sliding, the calculations are quite simple. To do this, use the formula:
Where N is the support reaction, µ is the friction coefficient, which has no dimension.
If only these three forces are present in the system, then their resultant along the inclined plane will be equal to:
F = m*g*sin(φ) - µ*m*g*cos(φ) = m*g*(sin(φ) - µ*cos(φ)) = m*a
Here φ is the angle of inclination of the plane to the horizon.
Knowing the force F, we can use Newton's law to determine the linear acceleration a. The latter, in turn, is used to determine the speed of movement along an inclined plane after a known period of time and the distance traveled by the body. If you look into it, you can understand that everything is not so complicated.
In the case when a body rolls down an inclined plane without slipping, the total force F will be equal to:
F = m*g*sin(φ) - F r = m*a
Where F r - It is unknown. When a body rolls, the force of gravity does not create a moment, since it is applied to the axis of rotation. In turn, F r creates the following moment:
Considering that we have two equations and two unknowns (α and a are related to each other), we can easily solve this system, and therefore the problem.
Now let's look at how to use the described technique to solve specific problems.
Problem involving the movement of a block on an inclined plane
The wooden block is at the top of the inclined plane. It is known that it has a length of 1 meter and is located at an angle of 45 o. It is necessary to calculate how long it will take the block to descend along this plane as a result of sliding. Take the friction coefficient equal to 0.4.
We write down Newton's law for this physical system and calculate the linear acceleration value:
m*g*(sin(φ) - µ*cos(φ)) = m*a =>
a = g*(sin(φ) - µ*cos(φ)) ≈ 4.162 m/s 2
Since we know the distance that the block must travel, we can write the following formula for the path during uniformly accelerated motion without an initial speed:
Where should the time be expressed, and substitute known values:
t = √(2*S/a) = √(2*1/4.162) ≈ 0.7 s
Thus, the time it takes to move along the inclined plane of the block will be less than a second. Note that the result obtained does not depend on body weight.
Problem with a cylinder rolling down a plane
A cylinder with a radius of 20 cm and a mass of 1 kg is placed on a plane inclined at an angle of 30 o. Its maximum should be calculated linear speed, which he will gain when rolling down the plane if its length is 1.5 meters.
Let's write the corresponding equations:
m*g*sin(φ) - F r = m*a;
F r *r = I*α = I*a/r
The moment of inertia of cylinder I is calculated by the formula:
Let's substitute this value into the second formula, express the friction force F r from it and replace it with the resulting expression in the first equation, we have:
F r *r = 1/2*m*r 2 *a/r = >
m*g*sin(φ) - 1/2*m*a = m*a =>
a = 2/3*g*sin(φ)
We found that linear acceleration does not depend on the radius and mass of the body rolling off the plane.
Knowing that the length of the plane is 1.5 meters, we find the time of movement of the body:
Then the maximum speed of movement along the inclined plane of the cylinder will be equal to:
v = a*t = a*√(2*S/a) = √(2*S*a) = √(4/3*S*g*sin(φ))
We substitute all the quantities known from the problem conditions into the final formula, and we get the answer: v ≈ 3.132 m/s.
Bukina Marina, 9 V
Movement of a body along an inclined plane
with transition to horizontal
As a body to be studied, I took a coin of 10 rubles (ribbed edges).
Specifications:
Coin diameter – 27.0 mm;
Coin weight - 8.7 g;
Thickness - 4 mm;
The coin is made of brass-nickel silver alloy.
I decided to take a book 27 cm long as an inclined plane. It will be an inclined plane. The horizontal plane is unlimited, since it is a cylindrical body, and in the future the coin, rolling off 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 additional equipment for measurements was taken: a stopwatch, an ordinary ruler, a long thread, a protractor, and a calculator.
In Fig.1. schematic image of a coin on an inclined plane.
Let's launch the coin.
We will enter the results obtained in Table 1
plane view | ||||
inclined plane | ||||
horizontal plane | ||||
*0.27 m constant value ttotal=90.04 |
Table 1
The trajectory of the coin's movement was different in all experiments, but some parts of the trajectory were similar. On an inclined plane, the coin moved rectilinearly, and when moving on a horizontal plane, it moved curvilinearly.
Figure 2 shows the forces acting on a coin as it moves along an inclined plane:
Using Newton's II Law, we derive a formula for finding the acceleration of a coin (according to Fig. 2):
To begin with, let's write down 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 (Ft), friction force (Ftr) and ground reaction force (N);
Let's get rid of vectors by projecting onto the X and Y axes:
Where is the friction coefficient
Because we have no data about numerical value 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, and is the acceleration with which the body moved, t is the time period of movement of the body.
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; let’s write the resulting form, getting rid of the vectors:
Let's take the average values from the table for S and t, find the acceleration and speed (the body moved rectilinearly with uniform acceleration along the inclined plane).
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 a horizontal plane the body moved rectilinearly at equal speed)
R=1.35 cm, where R is the radius of the coin
where is the angular velocity, is the centripetal acceleration, is the frequency of rotation of the body in a circle
The movement of a body along an inclined plane with a transition to a horizontal plane 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 is clear that acceleration depends only on the resultant force (R), and it remains a constant value throughout the entire path along the inclined plane, since 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.
Such a movement is called progressive solid, in which any straight line rigidly connected to the body moves while remaining parallel to itself. All points of a body moving translationally at each moment of time have the same speeds and accelerations, and their trajectories are completely combined during parallel translation.
Factors affecting body movement time
on an inclined plane
with transition to horizontal
Dependence of time on coins of different denominations (i.e., having different d (diameter)).
Coin denomination | d coins, cm | tav, s |
||
table 2
The larger the diameter of the coin, the longer the time it takes to move.
Dependence of time on the angle of inclination
Tilt angle | tav, s |
||
