Biology

Traction force on an inclined plane. Body on an inclined plane

Let a small body be on inclined plane with an angle of inclination a (Fig. 14.3, but). 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.

but) 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, but) 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 sleds, as well as force F 1, with which a person must pull the rope so that the sled moves evenly.

F a m m	but)	b) Rice. 14.4
but = ? 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 but in (4), we get

T= m N 2 + ma= m mg + that =

M mg + T .

To find F 1 , equate the expression for but 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, but. 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:
but = ? T = ? R = ?

If the loads are not moving, then .

Answer: 1) if T < mM, then but = 0, T = mg, ; 2) if T³m M, then , , .

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 one . 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 but>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 but>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 a rhombus ABCD. Because

Ð ADC\u003d 180 ° - 2,

where b = 90°– a, then by the cosine theorem

R 2 = .

From here .

Answer:

1) if , then , ;

2) if , then , ;

3) if , then but = 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, but). 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	but)	b) Rice. 14.10
but 1 = ? but 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

In our case F n \u003d m g, because the surface is horizontal. But, the normal force in magnitude does not always coincide with the force of gravity.

Normal force - the force of interaction between the surfaces of contacting bodies, the larger it is, the stronger the friction.

Normal force and friction force are proportional to each other:

F tr \u003d μF n

0 < μ < 1 - coefficient of friction, which characterizes the roughness of surfaces.

At μ=0 there is no friction (idealized case)

When μ=1, the maximum friction force is equal to the normal force.

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

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

1. Varieties of friction

Friction is of two types: static And kinetic.

Static friction (static friction) acts between contacting bodies that are at rest relative to each other. Static friction manifests itself at the microscopic level.

Kinetic friction (sliding friction) acts between bodies in contact and moving relative to each other. Kinetic friction manifests itself at the macroscopic level.

Static friction is greater than kinetic friction for the same bodies, or the coefficient of static friction is greater than the coefficient of sliding friction.

Surely you know this from personal experience: the cabinet is very difficult to move, but it is much easier to keep the cabinet moving. This is explained by the fact that when the surfaces of bodies move, they "do not have time" to switch to contact at the microscopic level.

Task #1: what force is required to lift a ball of mass 1 kg along an inclined plane located at an angle α=30° to the horizon. Friction coefficient μ = 0.1

We calculate the component of gravity. First we need to know the angle between the inclined plane and the gravity vector. We have already done a similar procedure when considering gravity. But repetition is the mother of learning :)

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

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

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

180° - 90° - α = 90° - α

Components of gravity along an inclined plane:

F g inc = F g cos(90° - α) = mgsinα

Required force to lift the ball:

F = F g inc + F friction = mgsinα + F friction

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

F friction = μF norm

Calculate the normal force F norms, which is equal to the component of gravity perpendicular to the inclined plane. We already know that the angle between the gravity vector and the inclined plane is 90° - α.

F norm = mgsin(90° - α) = mgcosα
F = mgsinα + μmgcosα

F = 1 9.8 sin30° + 0.1 1 9.8 cos30° = 4.9 + 0.85 = 5.75 N

We need to apply a force of 5.75 N to the ball in order to roll it to the top of the inclined plane.

Task #2: determine how far a ball of mass will roll m = 1 kg on a horizontal plane, rolling down an inclined plane with a length 10 meters with sliding friction coefficient μ = 0.05

The forces acting on a rolling ball are shown in the figure.

Component of gravity along an inclined plane:

F g cos(90° - α) = mgsinα

Normal strength:

F n \u003d mgsin (90 ° - α) \u003d mgcos (90 ° - α)

Sliding friction force:

F friction = μF n = μmgsin(90° - α) = μmgcosα

Resultant force:

F = F g - F friction = mgsinα - μmgcosα

F = 1 9.8 sin30° - 0.05 1 9.8 0.87 = 4.5 N

F=ma; a = F/m = 4.5/1 = 4.5 m/s 2

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

V 2 \u003d 2as; V = 2as = 2 4.5 10 = 9.5 m/s

The ball finishes moving along an inclined plane and starts moving along a horizontal straight line at a speed of 9.5 m/s. Now only the friction force acts on the ball in the horizontal direction, and the gravity component is equal to zero.

Total Strength:

F = μF n = μF g = μmg = 0.05 1 9.8 = -0.49 N

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

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

Ball stopping distance:

V 1 2 - V 0 2 \u003d 2as; s \u003d (V 1 2 - V 0 2) / 2a

Since we are determining the path of the ball to a complete stop, then V1=0:

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

Our ball rolled in a straight line as much as 92 meters!

Force projection. Movement on an inclined plane

Problems in dynamics.

I and II Newton's law.

Input and direction of the axes.

non-collinear forces.

Projection of forces on the axis.

Solution of systems of equations.

The most typical tasks in dynamics

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

Let's open a physics textbook and read. Newton's first law: there are such inertial systems countdown in which... Let's close this textbook, I don't understand either. Okay, I'm kidding, I understand, but I'll explain it easier.

I Newton's law: if the body is stationary or moving uniformly (without acceleration), the sum of the forces acting on it is equal to zero.

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

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

Conclusion: If a body moves at a variable speed, then the vector sum of the forces that somehow affect 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 with a mass of 600 kg if the engine traction force of 4500 N causes an acceleration of 5 m/s².

It is obligatory in such tasks to make a drawing and show the forces that act on the machine:

On the X-axis: movement with acceleration

On the Y-axis: no movement (here the coordinate, as it was zero, will remain, the car does not lift into the mountains or goes down)

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

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

We get that:

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

Answer: 0.25

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

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

T - thread tension force

On the X axis: no force

Let's deal with the direction of forces on the Y axis:

We express T (tension force) and substitute the numerical values:

Answer: 65 N

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

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

A simple example: a boy pulling a sled

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

To project the traction force on the axles, recall a right triangle.

The ratio of the opposite leg to the hypotenuse is a sine.

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

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

Traction force on the X axis - segment (vector) AC.

If this is not clear, see problem number 4.

The longer the rope is and, accordingly, the smaller the angle α, the easier it will be to pull the sled. Perfect option when the rope is parallel to the ground, because the force that acts on the X-axis is Fнcosα. At what angle is the cosine maximum? The larger this leg, the stronger the horizontal force.

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

We introduce the axes and project the forces:

We get two right triangle. Hypotenuses AB and KL - tension forces. LM and BC - projections on the X axis, AC and KM - on the Y axis.

Answer: 4.22 kg

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

When there is an inclined plane, the axes (X and Y) are best directed 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 axes taken.
Always ΔABC is similar to ΔKOM in such problems (according to right angle and angle of inclination of the plane).

Let's take a closer look at ΔKOM:

We get that KO lies on the Y axis, and the projection mg on the Y axis will be with a cosine. And the vector MK is collinear (parallel) to the X axis, the projection mg on 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!

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

Answer: 6.36 m/s²

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, to get an answer is real and without it.
Yes Yes, under ideal conditions (when there is no air resistance force, etc.), that the feather, that the weight will roll (fall) in the same time.

Task 5. A bus is driving down a hill at a slope of 60° with an acceleration of 8 m/s² and a traction force of 8 kN. The friction coefficient of tires on 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:

Answer: 6000 kg

Task 6. The 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? That's what the tilt is for, so the train doesn't fall on its side!

Injection α sets 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 leg to the adjacent one:

Answer: 7.5 cm

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

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

On the surface of the earth the force of gravity (gravity) is constant and equal to the product of the mass of the falling body and the free fall acceleration: F g = mg

It should be noted that the free fall acceleration is a constant value: g=9.8 m/s 2 , and is directed towards the center of the Earth. Based on this, we can say that bodies with different masses will fall to the Earth equally quickly. How so? If you throw a piece of cotton wool and a brick from the same height, the latter will make its way to the ground faster. Don't forget air resistance! For cotton wool, it will be significant, since its density is very low. In an airless space, brick and cotton wool will fall at the same time.

The ball moves along an inclined plane 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?

The ball is affected only by gravity F g , directed downward perpendicular to the base of the plane. Under the action of this force (a 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 force vector F g and the inclined plane.

Determining the angle is quite simple:

the sum of the angles of any triangle is 180°;
the angle between the force 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 required angle will be equal to: 180° - 90° - α = 90° - α

From trigonometry:

F g inc = F g cos(90°-α)

Sinα = cos(90°-α)

F g inc = F g sinα

It's really like this:

at α=90° (vertical plane) F g tilt = F g
at α=0° (horizontal plane) F g tilt = 0

Let's determine the acceleration of the ball from the well-known formula:

F g sinα = m a

A = F g sinα/m

A = m g sinα/m = g sinα

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

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

V 1 2 - V 0 2 = 2 a s

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

V 1 2 = √2 a s

V = 2 g sinα S = √2 9.8 0.5 10 = √98 = 10 m/s

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

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