purpose of work- experimental determination of the work of the friction force when the load slides along an inclined plane.
1. Theoretical part
Fig. 1. Block on an inclined plane
On a block of mass m located on an inclined plane, several forces act (Fig. 1) - gravity 
, the force of the normal reaction of the support 
and friction force 
... Under the influence of these forces, the bar can move or be at rest.
Let us first consider the state of rest, when the resultant of all forces is equal to zero:
(1)
where 
- the force of friction at rest. Let us introduce the coordinate axes as shown in Fig. 1. Since 
then the projection of equation (1) onto the axis 
gives
That. at rest, the static friction force balances the rolling force 
If you increase the angle of inclination 
then at some of its limiting values 
this balance will be disturbed and the bar will begin to slide off the inclined plane. At the moment of the beginning of sliding, the static friction force 
takes on a maximum value equal to the sliding friction force 
.
According to the Amonton - Coulomb law, the modulus of sliding friction force is
,
where - coefficient of friction.
The sliding of a bar along an inclined plane is described by the equation of dynamics
(2)
The projection of equation (2) on the axis y gives
.
.
Figure 2 shows the dependence of the forces of static friction and sliding friction on the angle of inclination 
Each of these dependencies has its own domain of definition. For function 
it lies within 
... Function scope 
lies in the interval 
... Outside these areas, both functions have no physical meaning.
Fig. 2. Dependencies 
and 
as a function of angle 
As seen from Fig. 2, with increasing angle 
the static friction force changes according to the sinusoidal law, and the sliding friction force changes according to the cosine law. The intersection of these two functions occurs at an angle 
, upon reaching which the bar will begin to slide down the inclined plane. Meaning 
is found from the equality
from where you can find the coefficient of friction
(3)
By measuring the length of the path l a bar along an inclined plane and its angle of inclination 
, it is possible to determine the work of the friction force by the limiting angle 
and the corresponding coefficient of friction
Now let's force the block of mass m 1 slide not down, but up an inclined plane. To do this (see Fig. 3) we tie the end of the thread thrown over the block to the bar; at the other end of the thread we will tie a weight of mass m 2, when lowered, the thread will pull the bar up an inclined plane with acceleration a.
Rice. 3. Diagram of the system inclined plane - bar-load.
On the length of the path l along the inclined plane (coordinate 
) a bar with a mass m 1, when moving from point 1 - the state of rest to point 2, it acquires a certain speed 
and, accordingly, the kinetic energy 
Kinetic energy can be calculated as the total work of all forces applied to the bar:
... - the work of the rolling force,
because 
-working of the thread tension force.
Further, we will assume that the thread and the block are weightless, so the tension of the thread on both sides of the block is the same: T 1 = T 2 = T... Equation of motion (Newton's second law) of cargo m 2 projected onto the axis at gives
whence we have the value T
The lowering height of the load according to the laws of kinematics is equal to:
Therefore, the acceleration of the load can be expressed in terms of the measured values - the height h and the time of the cargo descent m 2 -
All bodies of the system under consideration are connected by an inextensible thread and, therefore, move with the same speed and acceleration. Therefore, the speed of the mass bar m 1 at the end of a path segment of length l(position 2) equals
.
Taking into account the measured and calculated values, equation (5) will be rewritten as
,
Consider that the length 
section 1-2 lifting of a bar along an inclined plane is equal to the height 
lowering the load ( 
), then from (5) we obtain expression for determining the work of frictional force
by kinematic parameters (tilt angle
,length
and time
)moving the bar along an inclined plane
. (7)
Appliances and accessories:
1. Laboratory installation.
This article describes how to solve problems about movement on an inclined plane. A detailed solution to the problem of the motion of connected bodies on an inclined plane from the exam in physics is considered.
Solving the problem of motion on an inclined plane
Before proceeding directly to the solution of the problem, as a tutor in mathematics and physics, I recommend that you carefully analyze its condition. You need to start by depicting the forces that act on connected bodies:
Here and are the thread tension forces acting on the left and right bodies, respectively, are the support reaction force acting on the left body, and are the gravity forces acting on the left and right bodies, respectively. With the direction of these forces, everything is clear. The tension force is directed along the thread, gravity is directed vertically downward, and the reaction force of the support 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 with 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 in the opposite direction 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 took the gravitational acceleration m / s 2. The left weight is also pulled down by the force of gravity, but not entirely, but only its "part", since the weight lies on an inclined plane. This "part" is equal to the projection of gravity on an inclined plane, that is, the leg in the right-angled triangle shown in the figure, that is, it is equal to N.
That is, the right weight still "outweighs". Therefore, the friction force is directed as shown in the figure (we drew it from the center of mass of the body, which is possible in the case when the body can be modeled by a material point):
The second important question that needs to be dealt with is whether this connected system will move at all? What if it turns out that the friction force between the left weight and the inclined plane will be so great that it will not allow it to budge?
Such a situation will be possible in the case when the maximum friction force, the modulus of which is determined by the formula (here is the coefficient of friction between the load and the inclined plane, is the reaction force of the support acting on the load from the side of the inclined plane), is greater than the force that is trying bring the system into motion. That is, the same "outweighing" force, which is equal to N.
The modulus of the reaction force of the support is equal to the length of the leg in the triangle according to Newton's 3-muzacon (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 equal to 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's depict in the figure these accelerations and coordinate axes, which we will need further when solving the problem:
Now, after a thorough analysis of the problem conditions, we are ready to start solving it.
Let's write Newton's 2nd law for the left body:
And in the projection on the axis of the coordinate system, we get:
Here, with a minus, the projections are taken, the vectors of which are directed against the direction of the corresponding coordinate axis. The projections are taken with a plus, the vectors of which are co-directional with the corresponding coordinate axis.
Once again, we will explain in detail how to find the projections and. To do this, consider the right-angled triangle shown in the figure. In this triangle 
and 
... Also known to be in this right-angled triangle. Then and.
The acceleration vector lies entirely on the axis, therefore and. As we already recalled above, by definition, the modulus of the friction force is equal to the product of the coefficient of friction by the modulus of the reaction force of the support. Hence, . Then the original system of equations takes the form:
Let us now write down Newton's second law for the right body:
In projection onto the axis we get.
The movement of a body on an inclined plane is a classic example of the movement of a body under the influence of several non-directional forces. The standard method for solving problems of this kind of motion is to expand 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 the 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 bar
downward acceleration case
Consider a body that slides down an inclined plane. In this case, the following forces act on him:
- Gravity m g directed vertically downward;
- Support 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 only one vector will have to be decomposed into components - the gravity vector m g
, and the vector of the friction force F
tr and reaction forces of support N
already directed along the axes. With this expansion, the x-component of the gravity force is 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 reaction force of the support, since there is no body movement along the OY axis.
Sliding friction force F tr = µN proportional to the reaction force of the support. 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 the gravity force. Therefore for body sliding down
, we obtain the 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 that if µ < tg(α ), then the expression has a positive sign and we are dealing with uniformly accelerated motion down an inclined plane. If µ > tg ( α ), then the acceleration will have a negative sign and the motion will be equally slow. Such a movement is possible only if the body is given an initial velocity down the slope. In this case, the body will gradually stop. If provided µ > tg ( α ) the object is initially at rest, then it will not begin to slide down. Here, the static friction force will fully compensate for the "pulling" component of the gravity force.
When the coefficient of friction 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 at a constant speed (In this case, uniform motion is possible only downward).
The forces acting on the bar
sliding on an inclined plane:
case of slow motion upward
However, the body can also drive up the inclined plane. An example of such a movement is the movement of a hockey puck up an ice slide. When the body moves upward, both the friction force and the "pulling" component of the gravity force are directed downward along the inclined plane. In this case, we are always dealing with an equally slow motion, since the total force is directed in the direction opposite to the speed. Expression for acceleration for this situation is obtained in a similar way and differs only in sign. So for a body sliding up an inclined plane , we have.
An inclined plane is a flat surface located at one or another angle to the horizontal. It allows you to lift a load with less force than if this load was lifted vertically upwards. On an inclined plane, the load rises along this plane. At the same time, he overcomes a greater distance than if he rose vertically.
Note 1
Moreover, how many times there is a gain in strength, so many times the distance that the load will overcome will be greater.
Figure 1. Inclined plane
If the height to which the load must be lifted is $ h $, and the force $ F_h $ would be expended, and the length of the inclined plane is $ l $, and the force $ F_l $ is expended, then $ l $ is related to $ h $ as $ F_h $ refers to $ F_l $: $ l / h = F_h / F_l $ ... However, $ F_h $ is the weight of the load ($ P $). Therefore, it is usually written like this: $ l / h = P / F $, where $ F $ is the force that lifts the load.
The magnitude of the force $ F $, which must be applied to a load of weight $ P $ so that the body is in equilibrium on an inclined plane, is equal to $ F_1 = P_h / l = Psin (\ mathbf \ alpha) $, if the force $ P $ is applied parallel to the inclined plane (Fig. 2, a), and $ F_2 $ = $ P_h / l = Ptg (\ mathbf \ alpha) $, if the force $ P $ is applied parallel to the base of the inclined plane (Fig. 2, b).
Figure 2. The movement of cargo on an inclined plane
a) the force is parallel to the plane b) the force is parallel to the base
The inclined plane gives a gain in strength, with its help you can more easily raise the load to a height. The smaller the angle $ \ alpha $, the greater the strength gain. If the angle $ \ alpha $ is less than the angle of friction, then the load will not spontaneously move, and an effort is needed to pull it down.
If we take into account the forces of friction between the load and the inclined plane, then the following values are obtained for $ F_1 $ and $ F_2 $: $ F_1 = Psin ($$ (\ mathbf \ alpha) $$ \ pm $$ (\ mathbf \ varphi) $) / cos $ (\ mathbf \ varphi) $; $ F_2 = Рtg ($$ (\ mathbf \ alpha) $$ \ pm $$ (\ mathbf \ varphi) $)
The plus sign refers to moving up, the minus sign to lowering the load. Inclined plane efficiency $ (\ mathbf \ eta) $ 1 = sin $ (\ mathbf \ alpha) $ cos $ (\ mathbf \ alpha) $ / sin ($ (\ mathbf \ alpha) $ + $ (\ mathbf \ varphi ) $), if the force $ Р $ is directed parallel to the plane, and $ (\ mathbf \ eta) $ 2 = tg $ (\ mathbf \ alpha) $ / tg ($ (\ mathbf \ alpha) $ + $ (\ mathbf \ varphi ) $), if the force $ P $ is directed parallel to the base of the inclined plane.
The inclined plane obeys the "golden rule of mechanics". The smaller the angle between the surface and the inclined plane (that is, the flatter it is, not steeply rising up), the less forces must be applied to lift the load, but the greater distance will need to be overcome.
In the absence of friction forces, the gain in force is $ K = P / F = 1 / sin $$ \ alpha = l / h $. In real conditions, due to the action of the friction force, the efficiency of the inclined plane is less than 1, the gain in force is less than the $ l / h $ ratio.
Example 1
A load weighing 40 kg is lifted along an inclined plane to a height of 10 m while applying a force of 200 N (Fig. 3). How long is the ramp? Friction is neglected.
$ (\ mathbf \ eta) $ = 1
When a body moves along an inclined plane, the ratio of the applied force to the weight of the body is equal to the ratio of the length of the inclined plane to its height: $ \ frac (F) (P) = \ frac (l) (h) = \ frac (1) ((sin (\ mathbf \ alpha) \)) $. Therefore, $ l = \ frac (Fh) (mg) = \ \ frac (200 \ cdot 10) (40 \ cdot 9.8) = 5.1 \ m $.
Answer: The length of the inclined plane is 5.1 m
Example 2
Two bodies with masses $ m_1 $ = 10 g and $ m_2 $ = 15 g are connected by a thread thrown over a fixed block mounted on an inclined plane (Fig. 4). The plane makes an angle with the horizon $ \ alpha $ = 30 $ () ^ \ circ $. Find the acceleration with which these bodies will move.
$ (\ mathbf \ alpha) $ = 30 degrees
$ g $ = 9.8 $ m / c_2 $
Let us direct the OX axis along the inclined plane, and the OY axis perpendicular to it, and project on these axes the vectors $ \ (\ overrightarrow (P)) _ 1 \ and \ (\ overrightarrow (P)) _ 2 $. As can be seen from the figure, the resultant of the forces applied to each of the bodies is equal to the difference between the projections of the vectors $ \ (\ overrightarrow (P)) _ 1 \ and \ (\ overrightarrow (P)) _ 2 $ on the OX axis:
\ [\ left | \ overrightarrow (R) \ right | = \ left | P_ (2x) -P_ (1x) \ right | = \ left | m_2g (sin \ alpha \) -m_1g (sin \ alpha \) \ right | = g (sin \ alpha \ left | m_2-m_1 \ right | \) \] \ [\ left | \ overrightarrow (R) \ right | = 9.8 \ cdot (sin 30 () ^ \ circ \) \ cdot \ left | 0.015-0.01 \ right | = 0.0245 \ H \] \
Answer: Acceleration of bodies $ a_1 = 2.45 \ frac (m) (s ^ 2); \ \ \ \ \ \ a_2 = 1.63 \ m / s ^ 2 $
Let us recall that when one speaks of a smooth surface, one means that the friction between the body and this surface can be neglected.
A body of mass m, located on a smooth inclined plane, is acted upon by the force of gravity m and the force of normal reaction (Fig. 19.1). 
It is convenient to direct the x-axis along the inclined plane downward, and the y-axis perpendicular to the inclined plane upward (Fig. 19.1). The angle of inclination of the plane is denoted by α.
The equation of Newton's second law in vector form has the form 
1. Explain why the following equations are true:
2. What is the projection of the body's acceleration onto the x-axis?
3. What is the modulus of the normal reaction force?
4. At what angle of inclination is the body's acceleration on a smooth plane 2 times less than the gravitational acceleration?
5. At what angle of inclination of the plane is the force of normal reaction 2 times less than the force of gravity?
When performing the next task, it is useful to note that the acceleration of a body located on a smooth inclined plane does not depend on the direction of the initial velocity of the body.
6. The puck was pushed upward along a smooth ramp with an angle of inclination α. The initial speed of the washer is v 0.
a) How far will the puck go to stop?
b) How long will it take for the puck to return to the starting point?
c) How fast will the puck return to the starting point?
7. A bar of mass m is located on a smooth inclined plane with an angle of inclination α.
a) What is the modulus of the force holding the bar on an inclined plane if the force is directed along the inclined plane? Horizontally?
b) What is the force of normal reaction when the force is directed horizontally?
2. Condition of rest of the body on an inclined plane
We will now take into account the force of friction between the body and the inclined plane.
If the body is at rest on an inclined plane, it is acted upon by the force of gravity m, the force of normal reaction and the force of friction at rest, tr.poc (Fig. 19.2). 
The static friction force is directed upward along the inclined plane: it prevents the bar from slipping. Consequently, the projection of this force onto the x-axis directed downward along the inclined plane is negative:
F p.p. x = –F p.p.
8. Explain why the following equations are true:
9. A bar of mass m rests on an inclined plane with an angle of inclination α. The coefficient of friction between the bar and the plane is μ. What is the frictional force acting on the bar? Is there extra data in the condition?
10. Explain why the condition of rest of the body on an inclined plane is expressed by the inequality
Prompt. Take advantage of the fact that the static friction force satisfies the inequality F tr.poc ≤ μN.
The last inequality can be used to measure the coefficient of friction: the angle of inclination of the plane is gradually increased until the body begins to slide along it (see laboratory work 4).
11. The block lying on the board began to slide along the board when its angle of inclination to the horizon was 20º. What is the coefficient of friction between the bar and the board?
12. A brick weighing 2.5 kg lies on a 2 m long board. The coefficient of friction between the brick and the board is 0.4.
a) What is the maximum height that one end of the board can be raised to prevent the brick from moving?
b) What will the friction force acting on the brick be equal to?
The static frictional force acting on a body on an inclined plane is not necessarily directed upward along the plane. It can also be directed downward along the plane!
13. A bar of mass m is located on an inclined plane with an angle of inclination α. The coefficient of friction between the bar and the plane is μ, and μ< tg α. Какую силу надо приложить к бруску вдоль наклонной плоскости, чтобы сдвинуть его вдоль наклонной плоскости:
a) down? b) up?
3. The movement of the body along an inclined plane, taking into account friction
Now let the body slide down an inclined plane (Fig. 19.3). In this case, the force of sliding friction acts on it, directed opposite to the speed of the body, that is, along the inclined plane upward. 
? 15. Draw on a drawing in a notebook the forces acting on the body and explain why the following equations are true: 
16. What is the projection of the body's acceleration onto the x-axis?
17. The bar slides down an inclined plane. The coefficient of friction between the bar and the plane is 0.5. How does the speed of the bar change over time if the angle of inclination of the plane is:
a) 20º? b) 30º? c) 45º? d) 60º?
18. The bar begins to slide on the board when it is tilted at an angle of 20 degrees to the horizon. Why is the coefficient of friction between the bar and the board injured? With what magnitude and direction of acceleration will the bar slide down the board tilted at an angle of 30º? 15º?
Now let the initial velocity of the body be directed upward (Fig. 19.4). 
19. Draw on a drawing in a notebook the forces acting on the body and explain why the following equations are true:
20. What is the projection of the body's acceleration onto the x-axis?
21. The bar begins to slide on the board when it is tilted at an angle of 20 degrees to the horizon. The bar was pushed up the board. With what acceleration will it move if the board is tilted at an angle: a) 30º? b) 15º? In which of these cases will the bar stop at the top?
22.The washer was pushed up the inclined plane with an initial speed v 0. The angle of inclination of the plane α, the coefficient of friction between the washer and the plane μ. After some time, the puck returned to its original position.
a) How long did it take to move the puck up to stop?
b) How far did the puck go to stop?
c) How long after that did the puck return to its original position?
23. After the push, the bar moved for 2 s up the inclined plane and then for 3 s down until it returned to its initial position. The angle of inclination of the plane is 45º.
a) How many times is the acceleration modulus of the bar when moving upward than when moving down?
b) What is the coefficient of friction between the bar and the plane?
Additional questions and tasks
24. The bar slides off without initial speed from a smooth inclined plane of height h (Fig. 19.5). The angle of inclination of the plane is α. What is the speed of the bar at the end of the descent? Is there any extra data here? 
25. (Galileo's problem) A straight-line smooth groove is drilled in a vertical disk of radius R (Fig. 19.6). What is the time for the bar to slide along the entire gutter from rest? The angle of inclination of the gutter α, at the initial moment the bar is at rest. 
26. The trolley rolls along a smooth inclined plane with an angle of inclination α. A tripod is mounted on the trolley, on which a load is suspended on a thread. Make a drawing, depict the forces acting on the load. At what angle to the vertical is the thread located when the load is at rest relative to the cart?
27. The bar is at the top of an inclined plane 2 m long and 50 cm high. The coefficient of friction between the bar and the plane is 0.3.
a) With what modulus the acceleration will the bar move if it is pushed down along the plane?
b) What speed must be imparted to the bar in order for it to reach the base of the plane?
28. A body weighing 2 kg is on an inclined plane. The coefficient of friction between the body and the plane is 0.4.
a) At what angle of inclination of the plane is the greatest possible value of the friction force achieved?
b) What is the greatest value of the friction force?
c) Build an approximate graph of the dependence of the friction force on the angle of inclination of the plane.
Prompt. If tg α ≤ μ, the body is acted upon by the static friction force, and if tan α> μ, it is the sliding friction force.
