The reaction force of the support is the designation of the unit of measurement. How to find the support reaction force

Reaction Force supports refers to elastic forces, and is always directed perpendicular to the surface. It opposes any force that causes the body to move perpendicular to the support. In order to calculate it, you need to identify and find out the numerical value of all the forces that act on a body standing on a support.

You will need

- scales;
- speedometer or radar;
- goniometer.

Instruction

Determine body weight using scales or in any other way. If the body is on a horizontal surface (and it does not matter whether it is moving or at rest), then the reaction force of the support is equal to the force of gravity acting on the body. In order to calculate it, multiply the mass of the body by the acceleration of gravity, which is equal to 9.81 m / s² N \u003d m g.
When a body moves along an inclined plane directed at an angle to the horizon, the support reaction force is at an angle in gravity. At the same time, it compensates only for the component of gravity that acts perpendicular to the inclined plane. To calculate the reaction force of the support, use a goniometer to measure the angle at which the plane is located to the horizon. Calculate strength support reactions by multiplying the body mass by the free fall acceleration and the cosine of the angle at which the plane is to the horizon N=m g Cos(α).
In the event that the body moves along the surface, which is a part of a circle with a radius R, for example, a bridge, a hillock, then the reaction force of the support takes into account the force acting in the direction from the center of the circle, with an acceleration equal to centripetal, acting on the body. To calculate the reaction force of the support at the highest point, subtract the ratio of the square of the speed to the radius of curvature of the trajectory from the acceleration of gravity.
Multiply the resulting number by the mass of the moving body N=m (g-v²/R). Speed should be measured in meters per second and radius in meters. At a certain speed, the value of acceleration directed from the center of the circle can equal and even exceed the acceleration of free fall, at which point the adhesion of the body to the surface will disappear, therefore, for example, motorists need to clearly control the speed on such sections of the road.
If the curvature is directed downwards and the body trajectory is concave, then calculate the reaction force of the support by adding the ratio of the square of the speed and the radius of curvature of the trajectory to the free fall acceleration, and multiply the resulting result by the body mass N=m (g+v²/R).
If the friction force and the friction coefficient are known, calculate the reaction force of the support by dividing the friction force by this coefficient N=Ftr/μ.

Uniform movement

S= v* t

S – path, distance [m] (meter)

v – speed [m/s] (meter per second)

t – time [ s ] (second)

Speed conversion formula:

x km/h= font-family:Arial"> m/s

average speed

vWednesdays= EN-US style="font-family:Arial"">s in – the whole way

t in - all time

Matter density

ρ= EN-US style="font-family:Arial"">ρ– density

m – mass [kg] (kilogram)

V – volume [m3] (cubic meter)

Gravity, Weight and Support Reaction Force

Gravityis the force of gravity towards the earth. attached to the body. Directed towards the center of the Earth.

The weight- the force with which the body presses on the support or stretches the suspension. attached to the body. Directed perpendicular to the support and parallel to the suspension down.

Support reaction force - the force with which the support or suspension resists pressure or tension. Attached to a support or suspension. Directed perpendicular to the support or parallel to the suspension up.

Ft=m*g; P=m*g*cosα; N=m*g*cosα

F t – gravity [N] (Newton)

P - weight [ N ]

N – support reaction force [N]

m – mass [kg] (kilogram)

α – angle between the horizon plane and the support plane [º,rad] (degree, radian)

g≈9.8 m/s2

Elastic Force (Hooke's Law)

Fex= k* x

F control - elastic force [N] (Newton)

k – stiffness factor [N/m] (Newton per meter)

x – extension / compression of the spring [m] (meter)

mechanical work

A=F*l*cosα

A – work [J] (Joule)

F – force [N] (Newton)

l – distance over which the force acts [m] (meter)

α is the angle between the direction of the force and the direction of motion [º,rad] (degree, radian)

Special cases:

1)α=0, i.e. the direction of the force coincides with the direction of motion

A=F*l;

2) α = π /2=90 º, i.e. the direction of the force is perpendicular to the direction of motion

A=0;

3) α = π \u003d 180 º, i.e. the direction of the force is opposite to the direction of movement

A=- F* l;

Power

N= EN-US" style="font-family:Arial">N– power [W] (Watt)

A – work [J] (Joule)

t – time [s] (second)

Pressure in liquids and solids

P= font-family:Arial">; P= ρ * g* h

P – pressure [Pa] (Pascal)

F – pressure force [N] (Newton)

s – base area [m2] (square meter)

ρ is the density of the material/liquid[kg/m3] (kilogram per cubic meter)

g – free fall acceleration [m/s2] (meter per second squared)

h – height of object/liquid column [m] (meter)

Strength of Archimedes

Strength of Archimedes- the force with which a liquid or gas tends to push out a body immersed in them.

FArch= ρ well* Vburial* g

F Arch – Archimedes force [N] (Newton)

ρ w – density liquid/gas [kg/m3] (kilogram per cubic meter)

V burial - volume submerged part body [m3] (cubic meter)

g – free fall acceleration [m/s2] (meter per second squared)

Bodies floating condition:

ρ well≥ρ t

ρ t is the density of the material of the body[kg/m3] (kilogram per cubic meter)

Lever rule

F1 * l1 = F2 * l2 (lever balance)

F 1.2 – force acting on the lever [N] (Newton)

l 1.2 – length of lever arm of corresponding force [m] (meter)

moment rule

M= F* l

M – moment of force [N*m] (Newton meter)

F - force [N] (Newton)

l – length (arm) [m] (meter)

M1=M2(equilibrium)

Friction force

Ftr=µ* N

F tr – friction force [N] (Newton)

µ - coefficient of friction[ , %]

N – support reaction force [N] (Newton)

body energy

Ekin= font-family:Arial">; EP= m* g* h

E kin – kinetic energy [J] (Joule)

m – body weight [kg] (kilogram)

v – body speed [m/s] (meter per second)

Ep - potential energy[J] (Joule)

g – free fall acceleration [m/s2] (meter per second squared)

h – height above ground [m] (meter)

Law of energy conservation: Energy does not disappear into nowhere and does not appear from nowhere, it only passes from one form to another.

Let's put a stone on a horizontal table top, standing on the ground (Fig. 104). Since the acceleration of a stone relative to the Earth is equal to a bullet, then according to Newton's second law, the sum of the forces acting on it is zero. Consequently, the action of the gravity force m · g on the stone must be compensated by some other forces. It is clear that under the action of the stone the table top is deformed. Therefore, from the side of the table, an elastic force acts on the stone. If we assume that the stone interacts only with the Earth and the table top, then the elastic force must balance the force of gravity: F control = -m · g. This elastic force is called support reaction force and are denoted by the Latin letter N. Since the acceleration of free fall is directed vertically downwards, the force N is directed vertically upwards - perpendicular to the surface of the table top.

Since the table top acts on the stone, then, according to Newton's third law, the stone also acts on the table top with the force P = -N (Fig. 105). This force is called weighing.

The weight of a body is the force with which this body acts on a suspension or support, being in a stationary state relative to the suspension or support.

It is clear that in the considered case the weight of the stone is equal to the force of gravity: P = m · g. This will be true for any body resting on a suspension (support) relative to the Earth (Fig. 106). Obviously, in this case, the attachment point of the suspension (or support) is stationary relative to the Earth.

For a body resting on a suspension (support) that is motionless relative to the Earth, the weight of the body is equal to the force of gravity.

The weight of the body will also be equal to the force of gravity acting on the body if the body and the suspension (support) move uniformly in a straight line relative to the Earth.

If the body and the suspension (support) move relative to the Earth with acceleration so that the body remains stationary relative to the suspension (support), then the weight of the body will not be equal to the force of gravity.

Consider an example. Let a body of mass m lie on the floor of an elevator whose acceleration a is directed vertically upwards (Fig. 107). We will assume that only the gravity force m g and the floor reaction force N act on the body. (The weight of the body does not act on the body, but on the support - the floor of the elevator.) In a reference frame that is stationary relative to the Earth, the body on the floor of the elevator moves together with lift with acceleration a. In accordance with Newton's second law, the product of a body's mass and acceleration is equal to the sum of all forces acting on the body. Therefore: m a = N - m g.

Therefore, N = m a + m g = m (g + a). This means that if the elevator has an acceleration directed vertically upwards, then the modulus of force N of the floor reaction will be greater than the modulus of gravity. Indeed, the floor reaction force must not only compensate for the effect of gravity, but also give the body an acceleration in the positive direction of the X axis.

The force N is the force with which the elevator floor acts on the body. According to Newton's third law, the body acts on the floor with a force P, the modulus of which is equal to the modulus N, but the force P is directed in the opposite direction. This force is the weight of the body in the moving elevator. The modulus of this force is P = N = m (g + a). Thus, in an elevator moving with an upward acceleration relative to the Earth, the modulus of body weight is greater than the modulus of gravity.

Such a phenomenon is called overload.

For example, let the acceleration a of the elevator be directed vertically upwards and its value is equal to g, i.e. a = g. In this case, the modulus of body weight - the force acting on the floor of the elevator - will be equal to P = m (g + a) = m (g + g) = 2m g. That is, the weight of the body in this case will be twice as much as in the elevator, which is at rest relative to the Earth or moves uniformly in a straight line.

For a body on a suspension (or support) moving with an acceleration relative to the Earth, directed vertically upwards, the weight of the body is greater than the force of gravity.

The ratio of the weight of a body in an elevator moving at an accelerated rate relative to the Earth to the weight of the same body in an elevator at rest or moving uniformly in a straight line is called overload factor or, more briefly, overload.

The overload coefficient (overload) is the ratio of the body weight during overload to the force of gravity acting on the body.

In the case considered above, the overload is equal to 2. It is clear that if the acceleration of the elevator was directed upwards and its value was equal to a = 2g, then the overload coefficient would be equal to 3.

Now imagine that a body of mass m lies on the floor of an elevator whose acceleration a relative to the Earth is directed vertically downwards (opposite to the X axis). If the module a of the elevator acceleration is less than the module of the free fall acceleration, then the reaction force of the floor of the elevator will still be directed upwards, in the positive direction of the X axis, and its module will be equal to N = m (g - a). Consequently, the modulus of body weight will be equal to P = N = m (g - a), i.e., it will be less than the modulus of gravity. Thus, the body will press on the floor of the elevator with a force whose modulus is less than the modulus of gravity.

This feeling is familiar to anyone who has ridden a high-speed elevator or swung on a large swing. When moving down from the top point, you feel that your pressure on the support decreases. If the acceleration of the support is positive (the elevator and the swing begin to rise), you are pressed harder against the support.

If the acceleration of the elevator relative to the Earth is directed downward and is equal in absolute value to the free fall acceleration (the elevator falls freely), then the floor reaction force will become zero: N \u003d m (g - a) \u003d m (g - g) \u003d 0. B In this case, the floor of the elevator will no longer put pressure on the body lying on it. Therefore, according to Newton's third law, the body will not put pressure on the floor of the elevator, making a free fall together with the elevator. The weight of the body will become zero. Such a state is called weightlessness.

The state in which the weight of a body is zero is called weightlessness.

Finally, if the acceleration of the elevator towards the Earth becomes greater than the acceleration of free fall, the body will be pressed against the ceiling of the elevator. In this case, the weight of the body will change its direction. The state of weightlessness will disappear. This can be easily verified by pulling down the jar with the object in it sharply, closing the top of the jar with the palm of your hand, as shown in Fig. 108.

Results

The weight of a body is the force with which this body acts on a carrier or support, while being stationary relative to the suspension or support.

The weight of a body in an elevator moving with an upward acceleration relative to the Earth is greater in modulus than the modulus of gravity. Such a phenomenon is called overload.

The overload coefficient (overload) is the ratio of the weight of a body during overload to the force of gravity acting on this body.

If the weight of the body is zero, then this state is called weightlessness.

Questions

What force is called the support reaction force? What is body weight?
What is the weight of the body?
Give examples when the weight of a body: a) is equal to the force of gravity; b) is equal to zero; c) more gravity; d) less gravity.
What is called overload?
What state is called weightlessness?

Exercises

Seventh grader Sergei is standing on the floor scales in the room. The arrow of the device was set opposite the division of 50 kg. Determine the modulus of Sergey's weight. Answer the other three questions about this power.
Find the g-force experienced by an astronaut who is in a rocket rising vertically with an acceleration a = 3g.
With what force does an astronaut of mass m = 100 kg act on the rocket indicated in exercise 2? What is the name of this force?
Find the weight of an astronaut with mass m = 100 kg in a rocket, which: a) stands motionless on the launcher; b) rises with an acceleration a = 4g directed vertically upwards.
Determine the moduli of forces acting on a weight of mass m = 2 kg, which hangs motionless on a light thread attached to the ceiling of the room. What are the modules of the elastic force acting from the side of the thread: a) on the weight; b) on the ceiling? What is the weight of the kettlebell? Hint: use Newton's laws to answer the questions.
Find the weight of a load of mass m = 5 kg, suspended on a thread from the ceiling of a high-speed elevator, if: a) the elevator rises uniformly; b) the elevator descends evenly; c) the elevator going up with a speed v = 2 m/s started braking with an acceleration a = 2 m/s 2 ; d) descending down with a speed v = 2 m / s, the elevator began braking with an acceleration a = 2 m / s 2; e) the elevator started moving up with an acceleration a = 2 m/s 2; f) the elevator started moving down with an acceleration a = 2 m/s 2 .

The force acting on the body from the side of the support (or suspension) is called the reaction force of the support. When the bodies come into contact, the reaction force of the support is directed perpendicular to the contact surface. If the body lies on a horizontal fixed table, the reaction force of the support is directed vertically upwards and balances the force of gravity:

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See what the "Normal Support Reaction Force" is in other dictionaries:

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Instruction

Case 1. The formula for sliding: Ftr = mN, where m is the coefficient of sliding friction, N is the reaction force of the support, N. For a body sliding along a horizontal plane, N = G = mg, where G is the weight of the body, N; m – body weight, kg; g is the free fall acceleration, m/s2. The values of the dimensionless coefficient m for a given pair of materials are given in the reference. Knowing the mass of the body and a couple of materials. sliding relative to each other, find the force of friction.

Case 2. Consider a body sliding on a horizontal surface and moving with uniform acceleration. Four forces act on it: the force that sets the body in motion, the force of gravity, the reaction force of the support, the force of sliding friction. Since the surface is horizontal, the reaction force of the support and the force of gravity are directed along one straight line and balance each other. The displacement describes the equation: Fdv - Ftr = ma; where Fdv is the modulus of force that sets the body in motion, N; Ftr is the friction force modulus, N; m – body weight, kg; a is acceleration, m/s2. Knowing the values of the mass, acceleration of the body and the force acting on it, find the force of friction. If these values are not set directly, see if there is data in the condition from which to find these values.

Example of problem 1: a 5 kg bar lying on the surface is subjected to a force of 10 N. As a result, the bar moves with uniform acceleration and passes 10 for 10. Find the force of sliding friction.

The equation for the movement of the bar: Fdv - Ftr \u003d ma. The path of the body for uniformly accelerated motion is given by the equation: S = 1/2at^2. From here you can determine the acceleration: a = 2S/t^2. Substitute these conditions: a \u003d 2 * 10 / 10 ^ 2 \u003d 0.2 m / s2. Now find the resultant of the two forces: ma = 5 * 0.2 = 1 N. Calculate the friction force: Ftr = 10-1 = 9 N.

Case 3. If a body on a horizontal surface is at rest or moves uniformly, according to Newton's second law, the forces are in equilibrium: Ftr = Fdv.

Problem 2 example: a 1 kg bar on a flat surface is told , as a result of which it travels 10 meters in 5 seconds and stops. Determine the force of sliding friction.

As in the first example, the sliding of the bar is affected by the force of motion and the force of friction. As a result of this action, the body stops, i.e. balance comes. The equation of motion of the bar: Ftr = Fdv. Or: N*m = ma. The block slides with uniform acceleration. Calculate its acceleration similarly to problem 1: a = 2S/t^2. Substitute the values of the quantities from the condition: a \u003d 2 * 10 / 5 ^ 2 \u003d 0.8 m / s2. Now find the friction force: Ftr \u003d ma \u003d 0.8 * 1 \u003d 0.8 N.

Case 4. Three forces act on a body spontaneously sliding along an inclined plane: gravity (G), support reaction force (N) and friction force (Ftr). The force of gravity can be written as follows: G = mg, N, where m is the body weight, kg; g is the free fall acceleration, m/s2. Since these forces are not directed along a single straight line, write the equation of motion in vector form.

By adding the forces N and mg according to the parallelogram rule, you get the resultant force F'. The following conclusions can be drawn from the figure: N = mg*cosα; F' = mg*sinα. Where α is the angle of inclination of the plane. The friction force can be written by the formula: Ftr = m*N = m*mg*cosα. The equation for motion takes the form: F’-Ftr = ma. Or: Ftr = mg*sinα-ma.

Case 5. If an additional force F is applied to the body, directed along an inclined plane, then the friction force will be expressed: Ftr = mg * sinα + F-ma, if the direction of movement and force F are the same. Or: Ftr \u003d mg * sinα-F-ma, if the force F opposes the movement.

Problem 3 Example: A 1 kg block slid down the top of an inclined plane in 5 seconds after traveling a distance of 10 meters. Determine the force of friction if the angle of inclination of the plane is 45o. Consider also the case where the block was subjected to an additional force of 2 N applied along the angle of inclination in the direction of motion.

Find the acceleration of the body in the same way as in examples 1 and 2: a = 2*10/5^2 = 0.8 m/s2. Calculate the friction force in the first case: Ftr \u003d 1 * 9.8 * sin (45o) -1 * 0.8 \u003d 7.53 N. Determine the friction force in the second case: Ftr \u003d 1 * 9.8 * sin (45o) +2-1*0.8= 9.53 N.

Case 6. A body moves uniformly along an inclined surface. So, according to Newton's second law, the system is in equilibrium. If the sliding is spontaneous, the motion of the body obeys the equation: mg*sinα = Ftr.

If an additional force (F) is applied to the body, which prevents uniformly accelerated movement, the expression for motion has the form: mg*sinα–Ftr-F = 0. From here, find the friction force: Ftr = mg*sinα-F.

Sources:

slip formula

The coefficient of friction is a combination of the characteristics of two bodies that are in contact with each other. There are several types of friction: static friction, sliding friction and rolling friction. Resting friction is the friction of a body that was at rest and was set in motion. Sliding friction occurs when the body moves, this friction is less than static friction. Rolling friction occurs when a body rolls on a surface. Friction is designated depending on the type, as follows: μsk - sliding friction, μ - static friction, μroll - rolling friction.

Instruction

When determining the coefficient of friction during the experiment, the body is placed on a plane at an inclination and the angle of inclination is calculated. At the same time, take into account that when determining the coefficient of static friction, the given body moves, and when determining the coefficient of sliding friction, it moves at a speed that is constant.

The coefficient of friction can also be calculated during the experiment. It is necessary to place the object on an inclined plane and calculate the angle of inclination. Thus, the coefficient of friction is determined by the formula: μ=tg(α), where μ is the friction force, α is the angle of inclination of the plane.