Guide

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Case 1. The formula for the force of friction: Ftr = Mn, where M - the coefficient of sliding friction, N - reaction force, N. For the body, moving on a horizontal plane, N= G = mg, where G - body weight, H;m - weight in kilograms;g - acceleration of gravity, m / s2.The dimensionless coefficient m for a given pair of materials are given in reference literature.Knowing the mass of the body and a pair of materials.moving relative to each other, get friction.

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Case 2. Consider a body moving on a level surface and move with uniform acceleration.On it are four forces: the force that leads the body in motion, gravity, ground reaction f
orce, the force of friction.Since the surface is horizontal, reaction force and gravity is directed along a straight line and balance each other.Moving described by the equation: Fdv - Ftr = ma;where Fdv - power module, resulting in a body in motion, H;FTP - module of the friction force, N;m - weight in kilograms;a - Acceleration, m / s2.Knowing the mass, acceleration of a body and the force exerted on him, get friction.If these values are not set right, look, if there is provided in the data from which these values can be found.

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Example Problem 1: the bar on the 5 kg, which lies on the surface impact force of 10 N. As a result, the bar moves uniformly accelerated and runs 10 meters in 10 seconds.Find the force of friction.

equation for the motion of the bar: Fdv - Ftr = ma.The path for the body of uniformly accelerated motion is given by: S = 1 / 2at ^ 2.From here you can define the acceleration: a = 2S / t ^ 2.Put these conditions: a = 2 * 10/10 ^ 2 = 0.2 m / s2.Now find the resultant of two forces: ma = 5 * 0.2 = 1 N. Calculate the friction force: Ftr = 10-1 = 9 N.

equation for the motion of the bar: Fdv - Ftr = ma.The path for the body of uniformly accelerated motion is given by: S = 1 / 2at ^ 2.From here you can define the acceleration: a = 2S / t ^ 2.Put these conditions: a = 2 * 10/10 ^ 2 = 0.2 m / s2.Now find the resultant of two forces: ma = 5 * 0.2 = 1 N. Calculate the friction force: Ftr = 10-1 = 9 N.

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Case 3. If the body on a horizontal surface is in a staterest, or moves uniformly on Newton's second law of force are in equilibrium: Ftr = Fdv.

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Example Problem 2: 1 kg bar, located on a flat surface, said momentum as a result of which he traveled 10 meters in 5 seconds and stopped.Determine the force of friction.

As in the first example, the sliding bar impact strength of the movement and friction.As a result of this impact body stops, i.e.the balance comes.The equation of motion of the bar: Ftr = Fdv.Or: N * m = ma.The bar slides uniformly accelerated.Calculate its acceleration similar to Problem 1: a = 2S / t ^ 2.Substitute the values of the condition: a = 2 * 10/5 ^ 2 = 0.8 m / s2.Now find the frictional force: Ftr = ma = 0,8 * 1 = 0.8 N.

As in the first example, the sliding bar impact strength of the movement and friction.As a result of this impact body stops, i.e.the balance comes.The equation of motion of the bar: Ftr = Fdv.Or: N * m = ma.The bar slides uniformly accelerated.Calculate its acceleration similar to Problem 1: a = 2S / t ^ 2.Substitute the values of the condition: a = 2 * 10/5 ^ 2 = 0.8 m / s2.Now find the frictional force: Ftr = ma = 0,8 * 1 = 0.8 N.

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Case 4. On the body spontaneously sliding downhill, there are three forces: the force of gravity (G), reaction force (N) and the frictional force (Ftr).Gravity can be written in this form: G = mg, n, where m - weight in kilograms;g - acceleration of gravity, m / s2.Since these forces are not directed along the same line, write down the equation of motion in vector form.

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Adding the rule of a parallelogram force N and mg, you get the resultant force F '.From the figure we can conclude: N = mg * cosα;F '= mg * sinα.Where α - the angle of the plane.The frictional force can be written formula Ftr = m * N = m * mg * cosα.The equation for the motion takes the form: F'-Ftr = ma.Or Ftr = mg * sinα-ma.

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Case 5. If the body is attached to the additional force F, directed along the inclined plane, the friction force is expressed: Ftr = mg * sinα + F-ma, if the direction of motionand force F coincide.Or Ftr = mg * sinα-F-ma, if the force F counteracts the movement.

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Example Problem 3: 1 kg bar slipped from the top of the inclined plane for 5 seconds, having risen 10 meters.Determine the effect of friction, if the angle of inclination of the plane of 45 °.Consider also the case when the bar impacted additional force 2 N is applied along the angle of the direction of travel.

Find the acceleration of the body as in Examples 1 and 2: a = 2 * 10/5 ^ 2 = 0.8 m / s2.Calculate the friction force in the first case: Ftr = 1 * 9,8 * sin (45 °) -1 * 0.8 = 7.53 N. Determine the frictional force in the second case: Ftr = 1 * 9,8 * sin (45 °)+ 2-1 * 0.8 = 9.53 N.

Find the acceleration of the body as in Examples 1 and 2: a = 2 * 10/5 ^ 2 = 0.8 m / s2.Calculate the friction force in the first case: Ftr = 1 * 9,8 * sin (45 °) -1 * 0.8 = 7.53 N. Determine the frictional force in the second case: Ftr = 1 * 9,8 * sin (45 °)+ 2-1 * 0.8 = 9.53 N.

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Case 6. The body moves along an inclined surface evenly.Hence, by Newton's second law the system is in equilibrium.If slipping spontaneous movement of the body obeys the equation: mg * sinα = Ftr.

If the body is applied additional force (F), which prevents uniformly accelerated movement, expression of motion is: mg * sinα-Ftr-F = 0. Hence, find the friction force: Ftr = mg * sinα-F.

If the body is applied additional force (F), which prevents uniformly accelerated movement, expression of motion is: mg * sinα-Ftr-F = 0. Hence, find the friction force: Ftr = mg * sinα-F.

# Tip 2: How to determine the friction

When the relative motion of two bodies there is friction between them.It may also occur when driving in a gaseous or liquid medium.Friction can both hinder and facilitate normal movement.As a result of this phenomenon on the interacting bodies the force of friction

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Guide

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The most common case is considered

**force****sliding friction when one of the bodies is fixed and at rest, and the other slips over its surface.On the part of the body, in which the moving body slides, the last valid reaction force directed perpendicular to the slip plane.This force is designated by the letter N.Telo may also rest with respect to the fixed body.Then the force of friction****, acting on it Ftr & lt;? N.?- A dimensionless coefficient of friction****.It depends on the friction surfaces of materials, degree of otshlifovki and other factors.** 2

In the case of motion of the body relative to the surface of the fixed body strength

**sliding friction becomes equal to the product of the coefficient of friction****on****floor reaction force: Ftr =? N.** 3

If the surface is horizontal, the reaction force of the module is equal to the force of gravity acting on a body, that is, N = mg, where m - mass of the moving body, g - gravitational acceleration equal to about 98 m / (s ^ 2) in the world.Hence, Ftr =? Mg.

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Suppose now that the body of a constant force F & gt; Ftr =? N, parallel to the surface of the bodies in contact.When the sliding body, the resultant force component in the horizontal direction is equal to F-Ftr.Then, according to Newton's second law, the acceleration of the body is due to the resultant force of the formula: a = (F-Ftr) / m.Hence, Ftr = F-ma.Acceleration of the body can be found from kinematic considerations.

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Often considered a special case of the friction force

**shown at slipping body with a fixed ramp.Let?- The angle of the plane, and let the body slides off evenly, ie without acceleration.Then the equations of motion of the body will look like this: N = mg * cos ?, mg * sin?Ftr = =? N.Then from the first equation of motion****force****friction can be expressed as Ftr =? Mg * cos? .If The body moves along an inclined plane with an acceleration a, the second equation of motion has the form: mg * sin? -Ftr = Ma.Then Ftr = mg * sin? -ma.** Sources:

- glide formula