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 these laws are based on classical mechanics. We are only interested in the second law in this article. Its mathematical form is given below:
The formula says that the action 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
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 movement 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:
You will be interested:
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 (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 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.
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, is the support reaction force acting on left body, and are the gravity forces acting on the left and right body, 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 weight is pulled down by the force N. Here we have taken the acceleration free fall 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 right triangle shown in the figure, that is, equal to N.
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):
Second important question, with which you need to figure out whether this coupled 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. Hence, . 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.
Let a small body be on an inclined plane with an inclination angle a (Fig. 14.3, A). Let us find out: 1) what is the force of friction if the body slides along an inclined plane; 2) what is the force of friction if the body lies motionless; 3) at what minimum value of the angle of inclination a the body begins to slide off the inclined plane.
A) b) 
The friction force will hinder movement, therefore, it will be directed upward along the inclined plane (Fig. 14.3, b). In addition to the friction force, the body is also affected by the force of gravity and the normal reaction force. We introduce the coordinate system HOW, as shown in the figure, and find the projections of all these forces on the coordinate axes:
X: F tr X = –F tr, N X = 0, mg X = mg sina;
Y:F tr Y = 0, N Y = N, mg Y = –mg cosa.
Since the body can accelerate only along an inclined plane, that is, along the axis X, it is obvious that the projection of the acceleration vector onto the axis Y will always be zero: and Y= 0, which means that the sum of the projections of all forces on the axis Y should also be zero:
F tr Y + NY + mgY= 0 z 0 + N-mg cosa = 0
N=mg cosa. (14.4)
Then the force of sliding friction 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 zero:
F tr X + N X + mg X= 0 Þ – F tr + 0 + mg sina = 0
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 the external influence 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 to the force of sliding friction. Then the equality will be true:
F tr.sk = mg sina 0 .
Substituting into this equality the value F tr.ck from formula (14.5), we get: m mg cosa 0 = mg sina 0 .
Dividing both sides of the last equality by mg cosa 0 , we get:
Þ 
a 0 = arctanm.
So, the angle a, at which the body begins to slide along the inclined plane, is given by the formula:
a 0 = arctanm. (14.7)
Note that if a = a 0 , then the body can either lie motionless (if it is not touched), or slide down the inclined plane at a constant speed (if it is slightly pushed). 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 sledges connected to each other (Fig. 14.4, A) by applying force F at an angle a to the horizontal. The masses of the sleigh are the same and equal T. The coefficient of friction of skids on snow m. Find the acceleration of the sleigh and the force of tension T ropes between the sledges, as well as force F 1, with which a person must pull the rope so that the sled moves evenly.
| F a m m |
 A)
|  b) Rice. 14.4 |
| A = ? T = ? F 1 = ? |
Solution. We write Newton's second law for each sleigh 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 , 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 thread, as shown in Fig. 14.5, A. How fast is the body moving M, if the coefficient of friction on the surface of the table is m. What is the thread tension T? What is the force of pressure on the axis of the block?
| T M m | Solution. We write Newton's second law in projections on the axis X 1 and X 2 (Fig. 14.5, b), given 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 are not moving, 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 by a thread thrown over a block (Fig. 14.6). Body T 1 is on an inclined plane with an inclination angle a. Coefficient of friction on the plane m. body mass T 2 hangs on a thread. Find the acceleration of the bodies, the force of the thread tension and the pressure force of the block on the axis, provided that T 2 < T 1 . Read tga > m.
Rice. 14.7 
We 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 \u003d m 2 a.
Solving this system of equations, we find:
, .
Because A>0, then
So, if inequality (1) holds, then the load T 2 goes up, and if inequality (2) is satisfied, then it goes down. Therefore, if none of these conditions is met, i.e.
,
the system is immobile.
It remains to find the force of pressure on the axis of the block (Fig. 14.9). The force of pressure on the axis of the block R in this case can be found as the diagonal of the rhombus ABCD. Because
Ð ADC\u003d 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 there is a horizontal force F(Fig. 14.10, A). Coefficient of friction between the load T and the trolley 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 trolley 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 magnitude force - (Fig. 14.10, b). Slippage begins at the moment when it has already reached its maximum value , but the system is still moving as one body with mass T+M with acceleration. Then according to Newton's second law
A body of mass 2 kg under the action of a force F moves up the inclined plane at a distance the distance of the body from the surface of the Earth while increasing by
Force vector F directed parallel to the inclined plane, the modulus of force F equal to 30 N. What work did gravity do during this movement? (Give your answer in joules.) Take the free fall acceleration equal to the coefficient of friction
Solution.
The work of a force is defined as the scalar product of the force vector and the displacement vector of the body. Therefore, the force of gravity when lifting the body up the inclined plane did the work ( - angle at the base of the inclined plane)
Answer: -60.
Alternative solution.
Gravity refers to a type of force called potential. These forces have the property that their work on any closed path is always zero (this can be considered a definition). As other examples of potential forces, one can mention the elastic force, which obeys Hooke's law, the Coulomb force of the interaction of charges, the force of universal gravitation (as a generalization of the simple force of gravity).
As it is easy to see, for all forces that are called potential here, the value of potential energy is determined: - for the force of gravity, - for the force of elasticity, - for the forces of the Coulomb interaction, and, finally, for the force of universal gravitation. It turns out that it is the remarkable property of potential forces, which formed the basis of their definition, that allows us to introduce the concepts of the corresponding potential energies for them. In general, this is done as follows. Let the potential force do the work when transferring the body from point 1 to point 2 Then, by definition, they say that the difference in the values of the corresponding potential energy at points 2 and 1 is Since this definition always contains only the difference in potential energies at two points, potential energy is always determined up to a constant. This should be a fact well known to you. Let's apply this to this problem.
We need to find the work of gravity, for gravity we know what potential energy is. According to the formula written earlier, we get. That the desired work is equal to the change in the potential energy of the body, taken with a minus sign. The height of the body above the surface of the Earth has increased by, therefore, its energy has increased by
So the work done by gravity is
As a consolidation of the material, I propose to consider the following problem. A rocket of mass starts from the surface of the Earth Determine what work the force of attraction from the Earth will do by the moment when the rocket is at a distance of two earth radii from the center of the Earth.
Solution.
It will not be possible to use the formula "" directly, since the force of attraction decreases with distance from the Earth, the only chance to apply this formula is to start integrating. We will leave it and try to apply our knowledge again. The force of attraction to the Earth is potential. For it, we know the value of potential energy. Determine how much the potential energy of the rocket will change.
Therefore, the gravitational force has done work
As expected, this work is negative.
An example for self-parsing:
A spring with a stiffness of 10 N/m is stretched 5 cm, what work will be done by the elastic force when it is stretched another 5 cm?
