Biology

Thrust force inclined plane. The body on the inclined plane

Let a small body be on the inclined plane with an angle of inclination A (Fig. 14.3, but). Find out: 1) What is the force of friction, if the body slides on the inclined plane; 2) what is the power of friction, if the body lies motionless; 3) With which the minimum value of the angle of inclination A, the body begins to sculpt from the inclined plane.

but) b)

The strength of friction will be hinder The movement, therefore, it will be directed upward on the inclined plane (Fig. 14.3, b.). In addition to the force of friction, there is still the power of gravity and the power of a normal reaction. We introduce the coordinate system Hou.As shown in the figure, and find the projection of all these forces on the coordinate axes:

H.: F. Tr. H. = –F. Tr, N X. = 0, mg x \u003d Mgsina;

Y.: F. Tr. Y. = 0, N y \u003d n, mg y \u003d -mgcOSA.

Since the body can accelerate only on the inclined plane, that is, along the axis X., it is obvious that the projection of the acceleration vector on the axis Y. It will always be zero: and Y. \u003d 0, which means that the amount of projections of all forces on the axis Y. Also should be zero:

F. Tr. Y. + N y + mg y\u003d 0 þ 0 + N - MG.cosa \u003d 0 þ

N \u003d Mg.cOSA. (14.4)

Then the grip friction force according to formula (14.3) is equal to:

F. Tr.sk \u003d M. N \u003d M. mG.cOSA. (14.5)

If the body rest, then the amount of projections of all forces acting on the body on the axis H. should be zero:

F. Tr. H. + N x + mg x= 0 Þ – F. Tr + 0. + Mg.sina \u003d 0 þ

F. Tr.P. \u003d Mg.sina. (14.6)

If we gradually increase the angle of inclination, then the value mG.sina will gradually increase, which means that the strength of the friction of the rest will increase, which is always "automatically adjusted" under the external impact and compensates for it.

But, as we know, the "possibilities" of the friction forces are not limitless. With some kind of corner a 0, the whole "resource" of the friction force of rest will be exhausted: it will reach its maximum value equal to the power of friction of sliding. Then equality will be fair:

F. T.Sk. \u003d Mg.sina 0.

Substituting into this equality F. T.Sk from formula (14.5), we get: m mG.cosa 0 \u003d. mG.sina 0.

Sharing both parts of the last equality on mG.cosa 0, we get:

Þ a 0 \u003d arctgm.

So, the angle A, in which the body slip begins on the inclined plane, is given by the formula:

a 0 \u003d arctgm. (14.7)

Note that if a \u003d a 0, then the body can or lie still (if you do not touch it), or slide with a constant speed down on the inclined plane (if it is a little bit to push it). If A.< a 0 , то тело «стабильно» неподвижно, и легкий толчок не произведет на него никакого «впечатления». А если a > a 0, the body will sculpt from an inclined plane with acceleration and without any jolts.

Task 14.1. The person is lucky of two sleds connected (Fig. 14.4, but), applying power F. At an angle A to the horizon. Mass sled are the same and equal t.. Friction coefficient of polozov in the snow m. Find Sun Acceleration and Tension Power T. Rope between Sanya, as well as power F. 1, with which the rope should pull the man in order for the sleigh moving evenly.

F. A M. m.	but)	b) Fig. 14.4.
but = ? T. = ? F. 1 = ?

Decision. We write Newton's second law for every sled in the projections on the axis h. and w. (Fig. 14.4, b.):

I. w.: N. 1 + F.sina - mG. = 0, (1)

x.: F.cOSA - T. - M. N. 1 = mA.; (2)

II. w.: 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) Find T \u003d.m. mg + + Ma. Substituting these values N. 1 I. T. in (2), get

Substituting but in (4), we get

T. \u003d M. N. 2 + mA.\u003d M. mG. + tA =

M. mG. + t. .

To find F. 1, we equate the expression for but to zero:

Answer: ; ;

STOP! Decide alone: \u200b\u200bB1, B6, C3.

Task 14.2. Two bodies masses t. and M. Related thread, as shown in Fig. 14.5, but. What acceleration the body is moving M.if the friction coefficient is about the surface of the table m. What is the tension of the thread T.? What is the power of pressure on the axis of the block?

t. M. M.	Decision. We write Newton's second law in the projections on the axis h. 1 I. h. 2 (Fig. 14.5, b.), given that: h. 1: T - M. MG. = Ma., (1) h. 2: mG - T \u003d MA. (2) solving the system of equations (1) and (2), we find:
but = ? T. = ? R. = ?

If loads do not move, then.

Answer: 1) if t. < mM.T. but = 0, T. = mG.; 2) if T. ³ M. M.then , .

STOP! Decide alone: \u200b\u200bB9-B11, C5.

Task 15.3. Two bodies masses t. 1 I. t. 2 are linked to the thread, perched through the block (Fig. 14.6). Body t. 1 is located on an inclined plane with an angle of inclination a. The coefficient of friction on the plane m. Body mass t. 2 hangs on the thread. Find the acceleration of bodies, the strength of the thread tension and the power of the block pressure on the axis provided when t. 2 < t. one . Read TGA\u003e m.

Fig. 14.7

We write Newton's second law in the projections on the axis h. 1 I. h. 2, considering that and:

h. 1: t. 1 g.sina - T - M. m. 1 g. cosa \u003d. m. 1 a.,

h. 2: T - M. 2 g \u003d M. 2 a..

, .

As but \u003e 0, then

If the inequality (1) is not executed, the load t. 2 exactly not moving up! Then two more options are possible: 1) the system is fixed; 2) Cargo t. 2 moves down (and cargo t. 1, respectively, up).

Suppose that car t. 2 moves down (Fig. 14.8).

Fig. 14.8.

Then the equations of the second law of Newton on the axis h. 1 I. h. 2 will look:

h. 1: T - T. 1 g.sina. – M. m. 1 g.cosa \u003d. m. 1 a.,

h. 2: m. 2 g - T \u003d M 2 a..

Solving this system of equations, we find:

, .

As but \u003e 0, then

So, if inequality is performed (1), then the cargo t. 2 goes up, and if inequality is performed (2), then down. Consequently, if none of these conditions is performed, i.e.

the system is fixed.

It remains to find the pressure force on the axis of the block (Fig. 14.9). Pressure force on the axis of the block R. in this case can be found like diagonal rhombus Assd.. As

Ð ADC. \u003d 180 ° - 2,

where b \u003d 90 ° - a, then on the cosine theorem

R. 2 = .

From here .

Answer:

1) if T. , ;

2) if T. , ;

3) if T. but = 0; T. = t. 2 g..

In all cases .

STOP! Decide alone: \u200b\u200bB13, B15.

Task 14.4. On a truck mass M. Horizontal power acts F. (Fig. 14.10, but). Friction coefficient between cargo t. and the cart is m. Determine the acceleration of goods. What should be the minimum force F. 0 to cargo t. Began to slide on the cart?

M., t. F. M.	but)	b) Fig. 14.10
but 1 = ? but 2 = ? F. 0 = ?

Decision. First, we note that the force leading the cargo t. In motion, this is the strength of friction of rest, with which the cart is acting on the cargo. The maximum possible value of this force is M mG..

According to the third law of Newton, the cargo acts on the trolley with the same largest force - (Fig. 14.10, b.). Slock begins at the moment when it has already reached its maximum value, but the system is still moving as one body mass t.+M. with acceleration. Then according to the second law of Newton

In our case F H \u003d M · Gbecause Surface horizontal. But, the normal power is not always coincided with the strength of gravity.

Normal force is the strength of the interaction of surfaces of contacting bodies than it is more - the stronger the friction.

Normal strength and friction force are proportional to each other:

F Tr \u003d μF n

0 < μ < 1 - The coefficient of friction that characterizes the roughness of the surfaces.

When μ \u003d 0, friction is missing (idealized case)

With μ \u003d 1, the maximum friction force is equal to normal.

The friction force does not depend on the area of \u200b\u200bcontact of the two surfaces (if their masses do not change).

Please note: equation F Tr \u003d μF n It is not a relationship between vectors, since they are directed in different directions: the normal force is perpendicular to the surface, and the friction force is parallel.

1. Precision of friction

Friction is two species: static and kinetic.

Static friction (friction of rest) acts between contacting bodies, located in peace relative to each other. Static friction is manifested on a microscopic level.

Kinetic friction (sliding friction) acts between contacting and moving on each other with bodies. Kinetic friction is manifested in a macroscopic level.

Static friction is greater than the kinetic for the same bodies, or the coefficient of peace friction is greater than the slip coefficient.

Surely you know this from personal experience: the wardrobe is very difficult to move off the place, but the cabinet movement is much easier. This is due to the fact that when the surface moves the body, "do not have time to" prohibit contact on the microscopic level.

Task number 1: what power will be required to raise a ball weighing 1 kg along the inclined plane located at an angle α \u003d 30 ° to the horizon. Friction coefficient μ \u003d 0.1

Calculate the component of gravity. To begin with, we need to know the angle between the inclined plane and the gravity vector. We have already done a similar procedure, considering gravity. But, the repetition is the mother of the exercise :)

Gravity is directed vertically down. The sum of the angles of any triangle is 180 °. Consider a triangle formed by three forces: a vector of gravity; inclined plane; The base of the plane (in the figure it is highlighted in red).

The angle between the gravity strength vector and the base plane is 90 °.
The angle between the inclined plane and its base is α

Therefore, the remaining angle is the angle between the inclined plane and the gravity strength vector:

180 ° - 90 ° - α \u003d 90 ° - α

The components of gravity along the inclined plane:

F G None \u003d F G COS (90 ° - α) \u003d MgSinα

The necessary force to raise the ball:

F \u003d F G slot + F friction \u003d mgsinα + f friction

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

F friction \u003d μF norms

Calculate normal power F normswhich is equal to the gravity component, perpendicular to the directional plane. We already know that the angle between the gravity vector and the inclined plane is 90 ° - α.

F norms \u003d MgSin (90 ° - α) \u003d mgcosα
F \u003d mgsinα + μmgcosα

F \u003d 1 · 9.8 · sin30 ° + 0.1 · 1 · 9.8 · cos30 ° \u003d 4,9 + 0.85 \u003d 5.75

We will need to apply the ball into the ball in 5.75 n in order to roll it on the top of the inclined plane.

Task number 2: determine how far rolled the ball mass m \u003d 1 kg Along the horizontal plane, having risen on the inclined plane length 10 meters with the coefficient of sliding μ \u003d 0.05

Forces acting on the rolling ball are shown in the picture.

Gravity component along the inclined plane:

F G COS (90 ° - α) \u003d MgSinα

Normal strength:

F H \u003d MgSin (90 ° - α) \u003d MgCOS (90 ° - α)

Slip friction force:

F friction \u003d μF H \u003d μmgsin (90 ° - α) \u003d μmgcosα

Revulting force:

F \u003d F G - F friction \u003d MgSinα - μmgcosα

F \u003d 1 · 9.8 · sin30 ° - 0.05 · 1 · 9.8 · 0.87 \u003d 4.5

F \u003d MA; a \u003d f / m \u003d 4.5 / 1 \u003d 4.5 m / s 2

Determine the ball speed at the end of the inclined plane:

V 2 \u003d 2as; V \u003d 2as \u003d 2 · 4,5 · 10 \u003d 9.5 m / s

The ball ends the movement along the inclined plane and starts the movement along the horizontal direct at a speed of 9.5 m / s. Now in the horizontal direction on the ball only the friction force is valid, and the component of gravity is zero.

Total power:

F \u003d μF H \u003d μF G \u003d μmg \u003d 0.05 · 1 · 9.8 \u003d -0.49

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

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

Bowl brake path:

V 1 2 - V 0 2 \u003d 2as; S \u003d (V 1 2 - V 0 2) / 2A

Since we define the path of the ball until the complete stop, V 1 \u003d 0:

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

Our ball rolled in a straight line 92 meters!

Proceeding forces. Motion on the inclined plane

Tasks on dynamics.

I and II Law Newton.

Enter and direction axes.

Nonollyline forces.

Projecting forces on the axis.

Solution of systems of equations.

The most type tasks on the dynamics

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

Let's open the textbook of physics and read. I Newton's Law: There are such inertial systems reference in which ...Close such a textbook, I also do not understand. Okay, kidding, I understand, but I will explain easier.

I Newton's law: if the body is on the spot or moves evenly (without acceleration), the sum of the forces acting on it is zero.

Conclusion: If the body moves at a constant speed or stands on the spot the vector amount of strength will be zero.

II Newton's Law: If the body moves equally or equifiable (with acceleration), the sum of the forces acting on it is equal to the mass of the mass to acceleration.

Conclusion: If the body moves with a changing rate, then the vector sum of the forces that somehow affect this body (the force of the thrust, the force of friction, the strength of air resistance) is equal to the mass of this body multiply to acceleration.

At the same time, the same body is most often moving differently (evenly or with acceleration) in different axes. Consider such an example.

Task 1. Determine the coefficient of friction of the car tires with a mass of 600 kg, if the engine of engine thrust is 4500 H accelerates 5 m / s².

Be sure to make a drawing in such tasks, and show the forces that are measured by car:

On the x axis: movement with acceleration

On the Y axis: there is no movement (here the coordinate, as it was zero and remains, the car does not raise into the mountains or descends down)

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

According to the X axis: the thrust force is directed to the right, as well as the X axis, the acceleration is also directed to the right.

FTR \u003d μN, where n is the power of the support reaction. On the Y axis: n \u003d mg, then in this task FTR \u003d μmg.

We get that:

The friction coefficient is a dimensionless value. Consequently, there are no units of measurement.

Answer: 0.25

Task 2. Cargo Weight 5kg, tied to the weightless unprofitable thread, raise up with 3m / C² acceleration. Determine the strength of the tension of the thread.

Let's make a drawing, show the forces that are mad for the cargo

T - thread tension force

On the x axis: no strength

We will deal with the direction of forces on the Y axis:

Express T (tension force) and substitute numerical values:

Answer: 65 N

The most important thing is not to get confused with the direction of forces (on the axis or against), everything else Make a calculator or all the favorite column.

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

Simple example: boy pulls sledge

If we also construct x and y axes, then the strength of the tension (traction) will not lie on any of the axes.

To predict the force of thrust on the axis, remember the rectangular triangle.

The attitude of the opposite category for hypotenuse is sinus.

The ratio of the adjacent category for hypotenuse is a cosine.

Thrust force on the Y axis - cut (vector) BC.

Press force on the X axis - cut (vector) AC.

If it is not clear, look at the task number 4.

The longer than the version and, correspondingly, less than the angle α, the easier it will pull the sledge. Perfect optionwhen the rope is parallel to the earthAfter all, the force that act on the X axis is FNCOSα. With what corner of the cosine is maximized? The more this catt will, the stronger the horizontal force.

Task 3. The bar is suspended on two threads. The force of tension is the first to 34 N, the second- 21N, θ1 \u003d 45 °, θ2 \u003d 60 °. Find a lot of bar.

We introduce the axis and proper strength:

We get two rectangular triangles. AB and KL hypotenuses - tension force. LM and BC - projections on the X, AC and KM axis - on the Y axis.

Answer: 4.22 kg

Task 4. A bit of 5 kg (weight in this problem is not needed, but so that everything is known in the equations, we take a specific value) slides from the plane, which is tilted at an angle of 45 °, with the friction coefficient μ \u003d 0.1. Find the acceleration of the BROUSE movement?

When there is an inclined plane, the axis (x and y) is best to send in the direction of the body movement. Some forces in this case (here it is MG) will not lie on any of the axes. This force needs to be promoted to have the same direction as the axis taken.
Always ΔABC is similar to Δkom in such tasks (by straight corner and angle of inclination of the plane).

Consider in more detail Δkom:

We get that Ko lies on the Y axis, and the projection of Mg on the Y axis will be with a cosine. A vector MK collineaire (parallel) X axis, MG projection on the X axis will be with sinus, and the MK vector 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 forces do not coincide, it needs to be taken with a minus!

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

Answer: 6.36 m / s²

As can be seen, the mass in the numerator can be taken out of brackets and reduce with the banner. Then it is not necessary to know it, get the answer really without it.
Yes Yes,in ideal conditions (when there is no strength of air resistance, etc.), that the pen is that the weight will be accurate (fall) in the same time.

Task 5. The bus moves from a slide under the slope of 60 ° with the acceleration of 8 m / s² and with the force of 8 kN traction. The coefficient of friction of tires on asphalt is 0.4. Find a lot of buses.

Let's make a drawing with forces:

We introduce x and y axes. Sprogit MG on the axis:

We write Newton's second law on x and y:

Answer: 6000 kg

Task 6. The train moves to round the radius of 800 m at a speed of 72 km / h. Determine how much the external rail should be higher than internal. Distance between rails 1.5 m.

The most difficult thing is to understand what strengths where they act, and how an angle affects them.

Remember when you go in a circle on the car or in the bus, where does it push you out? To do this, you need a slope so that the train does not fall the soup!

Angle α sets the ratio of the difference in the height of rails to the distance between them (if the rails were horizontally)

We write out what forces act on the axis:

Acceleration in this task centripetal!

We divide one equation to another:

Tangent is the attitude of an opposing category to the adjacent:

Answer: 7.5 cm

As we found out, the solution of such tasks is reduced to the placement of the directions of forces, projecting them on the axis and to solve the systems of equations, almost the only trifle.

As a consolidation of the material, solve several similar tasks with prompts and answers.

On the surface of the Earth gravity (gravitis) is constant and equal to the product of the mass of the incident body to accelerate the free fall: F G \u003d MG

It should be noted that the acceleration of the free incidence is permanent: G \u003d 9.8 m / s 2, and is directed towards the center of the Earth. Based on this, we can say that the bodies with different mass will fall on the ground equally quickly. How so? If you throw a piece of cotton wool and brick from the same height, the latter will do your way to the ground faster. Do not forget about air resistance! For cotton, it will be essential because its density is very small. In airless space, brick and wool will fall at the same time.

The ball moves along the inclined plane of 10 meters long, the angle of inclination of the plane is 30 °. What will be the speed of the ball at the end of the plane?

Only the strength of gravity F G acts on the ball, directional perpendicular to the base of the plane. Under the action of this force (component directed along the surface of the plane), the ball will move. What will be the component of gravity, acting along the inclined plane?

To determine the component, it is necessary to know the angle between the power vector F G and the inclined plane.

Determine the angle is quite simple:

the sum of the angles of any triangle is 180 °;
the angle between the power vector F G and the base of the inclined plane is 90 °;
the angle between the inclined plane and its base is α

Based on the foregoing, the desired angle will be equal to: 180 ° - 90 ° - α \u003d 90 ° - α

From trigonometry:

F G None \u003d F G · COS (90 ° -α)

SINα \u003d COS (90 ° -α)

F G None \u003d F G · SINα

This is true:

at α \u003d 90 ° (vertical plane) f g totted \u003d f g
at α \u003d 0 ° (horizontal plane) F G None \u003d 0

We define the acceleration of the ball from the famous formula:

F g · sinα \u003d m · a

A \u003d f g · sinα / m

A \u003d m · g · sinα / m \u003d g · sinα

The acceleration of the ball along the inclined plane does not depend on the mass of the ball, but only on the angle of the plane.

Determine the ball speed at the end of the plane:

V 1 2 - V 0 2 \u003d 2 · A · s

(V 0 \u003d 0) - the ball starts moving from the spot

V 1 2 \u003d √2 · A · s

V \u003d 2 · G · sinα · s \u003d √2 · 9.8 · 0.5 · 10 \u003d √98 \u003d 10 m / s

Pay attention to the formula! The body velocity at the end of the inclined plane will depend only on the angle of the plane and its length.

In our case, the speed of 10 m / s at the end of the plane will have a billiard ball and a passenger car, and a dump truck, and a schoolboy on sledding. Of course, we do not consider friction.