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What is an inclined plane used for? Methods of teaching solving problems on movement on an inclined plane

It allows you to lift a load upward, applying a force to it that is noticeably less than the force of gravity acting on this load.

Examples of inclined planes are ramps and stairways. The principle of an inclined plane can also be seen in such piercing and cutting tools as a chisel, axe, plow, wedge, screw.

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✪ Inclined plane - Physics in experiments and experiments

✪ Lesson 87. Movement on an inclined plane (part 1)

✪ Simple mechanisms. Inclined plane

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Story

Ramps and ladders were widely used in the construction of early stone structures, roads and aqueducts, and during the assault on military fortifications.

Thought and real experiments with inclined planes, which at the dawn of modern times were carried out by Leonardo da Vinci, Simon Stevin, Galileo Galilei and other physicists, led to the knowledge of the laws of nature related to gravity, mass, inertia.

The first proof of the equilibrium condition on an inclined plane without friction was given by Stevin; this proof was based on the postulate about the impossibility of perpetual motion.

Movement on an inclined plane

Here μ (\displaystyle \mu )- coefficient of friction of the body on the surface, α (\displaystyle \alpha )- plane inclination angle.

The limiting case is when the angle of inclination of the plane is 90°, α = g (\displaystyle \alpha =g), and the body falls along the wall. Another extreme case is when the plane's inclination angle is 0° and the plane is parallel to the ground. In this case, the body cannot move without the application of an external force.

The nature of the motion of a body lying on an inclined plane depends on the value of the critical angle. The body is at rest if the angle of inclination of the plane is less than the critical angle, is at rest or moves uniformly if the angle of inclination of the plane is equal to the critical angle, and moves uniformly accelerated if the angle of inclination of the plane is greater than the critical angle.

So, I will try to describe in detail the course of my reasoning on this issue. In the first lesson, I pose the question to the students: how can a body move along an inclined plane? Together we answer: roll down evenly, with acceleration; rest on an inclined plane; hold on to it; move down under the influence of traction force evenly, with acceleration; drive under the influence of traction force evenly, with acceleration. In the pictures, using two or three examples, we show what forces act on the body. Along the way, I introduce the concept of a rolling resultant. We write the equation of motion in vector form, then in it we replace the sum with the rolling resultant (label it as you like). We do this for two reasons: firstly, there is no need to project force vectors onto the axis and solve two equations; secondly, the balance of forces will be correctly shown based on the conditions of the problem.

I'll show you on specific examples. Example 1: a body moves uniformly under the influence of traction force (Figure 1).

Students must first learn the algorithm for constructing a drawing. We draw an inclined plane, in the middle of it there is a body in the form of a rectangle, and through the middle of the body we draw an axis parallel to the inclined plane. The direction of the axis is not significant, but in the case of uniformly accelerated motion, it is better to show it in the direction of the vector so that in algebraic form in the equation of motion there is a plus sign on the right side in front of it. Next we build strength. We draw the force of gravity vertically downwards of an arbitrary length (I require the drawings to be large so that everyone can understand everything). Then, from the point of application of gravity, a perpendicular to the axis, along which the support reaction force will go. Parallel to this perpendicular, draw a dotted line from the end of the vector until it intersects with the axis. From this point - a dotted line parallel to the intersection with the perpendicular - we obtain a vector of the correct length. Thus, we constructed a parallelogram on the vectors and , automatically indicating the correct magnitude of the support reaction force and constructing, according to all the rules of vector geometry, the resultant of these forces, which I call the rolling resultant (diagonal coinciding with the axis). At this point, using the method from the textbook, in a separate figure I show the reaction force of a support of arbitrary length: first shorter than necessary, and then longer than necessary. I show the resultant force of gravity and the support reaction force: in the first case, it is directed downward at an angle to the inclined plane (Figure 2), in the second case, upward at an angle to the inclined plane (Figure 3).

We draw a very important conclusion: the relationship between the force of gravity and the reaction force of the support must be such that the body, under their action (or under the action of the rolling resultant), in the absence of other forces, moves downwards along inclined plane. Next I ask: what other forces act on the body? The guys answer: traction force and friction force. I ask the following question: which strength will we show first, and which later? I am trying to achieve a correct and reasonable answer: first in this case it is necessary to show the traction force, and then the friction force, the modulus of which will be equal to the sum modules of traction force and rolling resultant: , because According to the conditions of the problem, the body moves uniformly, therefore, the resultant of all forces acting on the body must be equal to zero according to Newton’s first law. To control, I ask a provocative question: how much force is acting on the body? The guys must answer - four (not five!): gravity, ground reaction force, traction force and friction force. Now we write the equation of motion in vector form according to Newton’s first law:

We replace the sum of vectors with the rolling resultant:

We obtain an equation in which all vectors are parallel to the axis. Now let's write this equation through projections of vectors onto the axis:

You can skip this entry in the future. Let us replace in the equation the projections of vectors by their modules, taking into account the directions:

Example 2: a body, under the influence of traction force, moves onto an inclined plane with acceleration (Figure 4).

In this example, students must say that after constructing the force of gravity, the reaction force of the support and the rolling resultant, the next one must show the friction force, the last one is the vector of the traction force, which must be greater than the sum of the vectors, because the resultant of all forces must be directed in the same direction as the acceleration vector according to Newton's second law. The equation of motion of a body must be written according to Newton’s second law:

If there is an opportunity to consider other cases in class, then we do not neglect this opportunity. If not, then I give this task home. Some may consider all the remaining cases, others may consider the right to choose students. In the next lesson, we check, correct errors and move on to solving specific problems, having previously expressed from vector triangles and:

It is advisable to analyze equality (2) for various angles. At we have: as when moving horizontally under the influence of a horizontal traction force. As the angle increases, its cosine decreases, therefore, the support reaction force decreases and the force of gravity becomes less and less. At angle it is equal to zero, i.e. the body does not act on the support and the support, accordingly, “does not react.”

I foresee a question from opponents: how to apply this technique to cases where the traction force is horizontal or directed at an angle to an inclined plane? I will answer with specific examples.

a) The body is pulled with acceleration onto an inclined plane, applying a traction force horizontally (Figure 5).

We decompose the horizontal traction force into two components: along the axis - and perpendicular to the axis - (the opposite operation to constructing the resultant of perpendicular forces). We write down the equation of motion:

We replace the rolling resultant, and instead write:

From vector triangles we express: And : .

Under the influence of horizontal force, the body not only rises up the inclined plane, but is also additionally pressed against it. Therefore, an additional pressure force arises equal to the vector modulus and, according to Newton’s third law, an additional support reaction force: . Then the friction force will be: .

The equation of motion will take the form:

Now we have completely deciphered the equation of motion. Now it remains to express the desired value from it. Try to solve this problem the traditional way and you will get the same equation, only the solution will be more cumbersome.

b) The body is pulled evenly from the inclined plane, applying a traction force horizontally (Figure 6).

In this case, the traction force, in addition to pulling the body down along the inclined plane, also tears it away from the inclined plane. So the final equation is:

c) The body is dragged evenly onto the inclined plane, applying a traction force at an angle to the inclined plane (Figure 7).

I propose to consider specific problems in order to further convincingly advertise my methodological approach to solving such problems. But first, I draw attention to the solution algorithm (I think all physics teachers draw the attention of students to it, and my entire story was subordinated to this algorithm):

1) after carefully reading the problem, find out how the body moves;
2) make a drawing with the correct image of the forces, based on the conditions of the problem;
3) write down the equation of motion in vector form according to Newton’s first or second law;
4) write this equation through the projections of force vectors onto the x-axis (this step can be omitted later, when the ability to solve problems in dynamics is brought to automaticity);
5) express the projections of vectors through their modules, taking into account directions and write the equation in algebraic form;
6) express the force modules using formulas (if necessary);
7) express the required value.

Task 1. How long does it take for a body of mass to slide down an inclined plane with height and angle of inclination if it moves uniformly along an inclined plane with an angle of inclination?

What would it be like to solve this problem in the usual way!

Task 2. What is easier: to hold the body on an inclined plane or to move it evenly upward along it?

Here, in the explanation, it is impossible to do without the rolling resultant.

As can be seen from the figures, in the first case, the friction force helps to hold the body (directed in the same direction as the holding force), in the second case, it, together with the rolling resultant, is directed against the movement. In the first case, in the second case.

The movement of a body along 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 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 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 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 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 at a constant speed (In this case uniform motion only downwards possible).

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.

Projection of forces. Movement on an inclined plane

Dynamics problems.

Newton's I and II laws.

Input and direction of axes.

Non-collinear forces.

Projection of forces on the axes.

Solving systems of equations.

The most typical problems in dynamics

Let's start with Newton's I and II laws.

Let's open a physics textbook and read it. Newton's first law: there are such inertial frames of reference in which... Let's close this tutorial, I don't understand either. Okay, I’m joking, I understand, but I’ll explain it more simply.

Newton's first law: if a body stands still or moves uniformly (without acceleration), the sum of the forces acting on it is zero.

Conclusion: If a body moves at a constant speed or stands still, the vector sum of forces will be zero.

Newton's II law: if a body moves uniformly accelerated or uniformly decelerated (with acceleration), the sum of the forces acting on it is equal to the product of mass and acceleration.

Conclusion: If a body moves with varying speed, then the vector sum of the forces that somehow influence this body (traction force, friction force, air resistance force) is equal to the mass of this body times the acceleration.

In this case, the same body most often moves differently (uniformly or with acceleration) in different axes. Let's consider just such an example.

Task 1. Determine the coefficient of friction of the tires of a car weighing 600 kg if an engine traction force of 4500 N causes an acceleration of 5 m/s².

Let's make a drawing and show the forces that act on the car.

On X Axis: movement with acceleration

On the Y axis: no movement (here the coordinate, as it was zero, will remain so, the movement will only be along the X axis)

Those forces whose direction coincides with the direction of the axes will be plus, in the opposite case - minus.

Ftr = μN, where N - ground reaction force. On the Y axis: N = mg, then in this problem Ftr = μmg.

We get that:

The friction coefficient is a dimensionless quantity. Therefore, there are no units of measurement.

Problem 2. A mass of 5 kg, tied to a weightless inextensible thread, is lifted upward with an acceleration of 3 m/s². Determine the tension of the thread.

Let's make a drawing and show the forces that act on the load

T - thread tension force

Let's figure out the direction of forces on the Y axis:

Let's express T and substitute the numerical values:

The most important thing is not to get confused with the direction of forces (along the axis or against), everything elsemake a calculator or everyone’s favorite column.

Not always all forces acting on a body are directed along the axes.

A simple example: a boy pulling a sled

If we also construct the X and Y axes, then the tension (traction) force will not lie on any of the axes.

To project the traction force onto the axes, recall a right triangle.

The ratio of the opposite side to the hypotenuse is the sine.

The ratio of the adjacent leg to the hypotenuse is the cosine.

Traction force on the Y axis - segment (vector) BC.

The traction force on the X axis is a segment (vector) AC.

If this is not clear, look at problem #4.

The longer the rope and, accordingly, the smaller the angle α, the easier it will be to pull the sled. Ideal when the rope is parallel to the ground, because the force that acts on the X axis is Fнcosα. The larger this leg is, the stronger the horizontal force.

Task 3. The block is suspended by two threads. The tension force of the first is 34N, the second- 21Н, θ1 = 45°, θ2 = 60°. Find the mass of the block.

Let's introduce the axes and project the forces:

We get two right triangle. Hypotenuses AB and KL are tension forces. LM and BC are tensile forces projected onto the X axis, AC and KM - onto the Y axis.

Task 4. A block with a mass of 5 kg (mass is not needed in this problem, but to make everything known in the equations, let’s take a specific value) slides off a plane that is inclined at an angle of 45°, with a coefficient of friction μ = 0.1. Find the acceleration of the block?

When there is an inclined plane, it is best to direct the axes (X and Y) in the direction of movement of the body. Some forces in this case (here it is mg) will not lie on any of the axes. This force must be projected so that it has the same direction as the taken axes.
ΔABC is always similar to ΔKOM in such problems (by right angle and the angle of inclination of the plane).

Let's take a closer look at ΔKOM:

We obtain that KO lies on the Y axis, and the projection of mg onto the Y axis will be with a cosine. And the vector MK is collinear (the segment MK is parallel) to the X axis, the projection mg onto the X axis will be with a sine, and the vector MK is directed against the X axis (that is, it will be with a minus).

Do not forget that if the directions of the axis and the force do not coincide, it must be taken with a minus!

From the Y axis we express N and substitute it into the equation of the X axis, we find the acceleration:

As you can see, the mass in the numerator can be taken out of brackets and reduced with the denominator. Then it is not necessary to know it; it is possible to get an answer without it.
Yes, yes, under ideal conditions (when there is no air resistance, etc.), both the feather and the weight will roll (fall) at the same time.

Task 5. A bus slides down a hill at a slope of 60° with an acceleration of 8 m/s² and a traction force of 8 kN. The coefficient of friction between tires and asphalt is 0.4. Find the mass of the bus.

Let's make a drawing with forces:

Let's introduce the X and Y axes. Let's project mg onto the axes:

Let's write Newton's second law for X and Y:

Task 6. A train moves along a curve with a radius of 800 m at a speed of 72 km/h. Determine how much the outer rail should be higher than the inner one. The distance between the rails is 1.5 m.

The most difficult thing is to understand which forces act where, and how the angle affects them.

Remember, when you drive in a circle in a car or on a bus, where does it push you? This is why the tilt is needed so that the train does not fall on its side!

Corner α specifies the ratio of the difference in the height of the rails to the distance between them (if the rails were horizontal)

Let's write down what forces act on the axis:

The acceleration in this problem is centripetal!

Let's divide one equation by another:

Tangent is the ratio of the opposite side to the adjacent side:

As we found out, solving such problems comes down to arranging the directions of forces, projecting them onto axes and solving systems of equations, which is almost a mere trifle.

To reinforce the material, solve several similar problems with hints and answers.

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Simple machines - This name refers to the following mechanisms, a description and explanation of the operation of which can be found in all elementary courses in physics and mechanics: lever, blocks, pulleys, gates, inclined plane, wedge and screw. The blocks and gates are based on the lever principle, the wedge and screw are based on the inclined plane principle.

Lever- the simplest mechanical device, which is solid(crossbar) rotating around a fulcrum. The sides of the crossbar on either side of the fulcrum are called lever arms.

The lever is used to obtain more force on the short arm with less force on the long arm (or to obtain more movement on the long arm with less movement on the short arm). By making the lever arm long enough, theoretically, any force can be developed.

Two other simple mechanisms are also special cases of a lever: a gate and a block. The principle of operation of the lever is a direct consequence of the law of conservation of energy. For levers, as for other mechanisms, a characteristic is introduced showing the mechanical effect that can be obtained due to the lever. This characteristic is the gear ratio; it shows how the load and the applied force relate:

There are levers of the 1st class, in which the fulcrum is located between the points of application of forces, and levers of the 2nd class, in which the points of application of forces are located on one side of the support.

Block- a simple mechanical device that allows you to adjust the force, the axis of which is fixed when lifting loads, does not rise or fall. It is a wheel with a groove around its circumference, rotating around its axis. The groove is intended for a rope, chain, belt, etc. The axis of the block is placed in cages attached to a beam or wall; such a block is called stationary; if a load is attached to these clips, and the block can move with them, then such a block is called movable.

A fixed block is used to lift small loads or to change the direction of force.

Block equilibrium condition:

F is the applied external force, m is the mass of the load, g is the acceleration of gravity, f is the resistance coefficient in the block (for chains approximately 1.05, and for ropes - 1.1). In the absence of friction, lifting requires a force equal to the weight of the load.

The moving block has a free axis and is designed to change the amount of force applied. If the ends of the rope clasping the block make equal angles with the horizon, then the force acting on the load is related to its weight, as the radius of the block is to the chord of the arc clasped by the rope; hence, if the ropes are parallel (that is, when the arc encircled by the rope is equal to a semicircle), then lifting the load will require a force half as much as the weight of the load, that is:

In this case, the load will travel a distance half as large as that traveled by the point of application of force F; accordingly, the gain in the force of the moving block is equal to 2.

In fact, any block is a lever, in the case of a fixed block - equal arms, in the case of a moving one - with a ratio of arms of 1 to 2. As for any other lever, the rule is true for a block: The number of times we win in an effort, the same number of times we lose in the distance. In other words, the work done when moving a load a certain distance without using a block is equal to the work expended when moving a load the same distance using a block, provided there is no friction. In a real block there is always some loss.

Inclined plane- this is a flat surface installed at an angle other than straight and/or zero, to horizontal surface. An inclined plane allows you to overcome significant resistance by applying relatively little force over a greater distance than the load needs to be lifted.

The inclined plane is one of the well-known simple mechanisms. Examples of inclined planes are:

ramps and ladders;
tools: chisel, axe, hammer, plow, wedge and so on;

The most canonical example of an inclined plane is an inclined surface, such as the entrance to a bridge with a difference in height.

§ tr - where m is the mass of the body, is the acceleration vector, is the reaction force (impact) of the support, is the free fall acceleration vector, tr is the friction force.

§ a = g(sin α + μcos α) - when climbing an inclined plane and in the absence of additional forces;

§ a = g(sin α − μcos α) - when descending from an inclined plane and in the absence of additional forces;

here μ is the coefficient of friction of the body on the surface, α is the angle of inclination of the plane.

The limiting case is when the angle of inclination of the plane is 90o degrees, that is, the body falls, sliding along the wall. In this case: α = g, that is, the friction force does not affect the body in any way, it is in free fall. Another limiting case is the situation when the angle of inclination of the plane is zero, i.e. the plane is parallel to the ground; in this case, the body cannot move without the application of an external force. It should be noted that, following from the definition, in both situations the plane will no longer be inclined - the angle of inclination should not be equal to 90o or 0o.

The type of movement of the body depends on the critical angle. The body is at rest if the angle of inclination of the plane is less than the critical angle, is at rest or moves uniformly if the angle of inclination of the plane is equal to the critical angle, and moves uniformly accelerated, provided that the angle of inclination of the plane is greater than the critical angle.

§ or α< β - тело покоится;

§ or α = β - the body is at rest or moving uniformly;

§ or α > β - the body moves with uniform acceleration;

Wedge- a simple mechanism in the form of a prism, the working surfaces of which converge at an acute angle. Used for moving apart and dividing the object being processed into parts. The wedge is one of the varieties of the mechanism called “inclined plane”. When a force acts on the base of the prism, two components appear, perpendicular to the working surfaces. The ideal gain in force given by a wedge is equal to the ratio of its length to the thickness at the blunt end - the wedging action of the wedge gives a gain in force at a small angle and a large length of the wedge. The actual gain of the wedge depends greatly on the frictional force, which changes as the wedge moves.

; where IMA is the ideal gain, W is the width, L is the length. The wedge principle is used in such tools and tools as an axe, chisel, knife, nail, needle, and stake.