Finding the center of gravity of a flat body write down the experiment. Methods for determining the coordinates of the center of gravity. Test questions and assignments

Author: Let's take a body of arbitrary shape. Is it possible to hang it on a thread so that after hanging it retains its position (i.e. does not begin to turn) when any initial orientation (Fig. 27.1)?

In other words, is there a point relative to which the sum of the moments of gravity acting on various parts of the body would be equal to zero at any body orientation in space?

Reader: Yes, I think so. This point is called center of gravity of the body.

Proof. For simplicity, let us consider a body in the form of a flat plate of arbitrary shape, arbitrarily oriented in space (Fig. 27.2). Let's take the coordinate system X 0at with the beginning at the center of mass - point WITH, Then x C = 0, at C = 0.

Let us imagine this body as a collection of a large number of point masses m i, the position of each of which is specified by the radius vector.

By definition, the center of mass is , and the coordinate x C = .

Since in the coordinate system we adopted x C= 0, then . Let's multiply this equality by g and we get

As can be seen from Fig. 27.2, | x i| - this is the shoulder of strength. And if x i> 0, then the moment of force M i> 0, and if x j < 0, то Mj < 0, поэтому с учетом знака можно утверждать, что для любого x i the moment of force will be equal M i = m i gx i . Then equality (1) is equivalent to equality , where M i– moment of gravity. This means that with an arbitrary orientation of the body, the sum of the moments of gravity acting on the body will be equal to zero relative to its center of mass.

In order for the body we are considering to be in equilibrium, it is necessary to apply to it at the point WITH force T = mg, directed vertically upward. The moment of this force relative to the point WITH equal to zero.

Since our reasoning did not depend in any way on how exactly the body is oriented in space, we proved that the center of gravity coincides with the center of mass, which is what we needed to prove.

Problem 27.1. Find the center of gravity of a weightless rod of length l, at the ends of which two point masses are fixed T 1 and T 2 .

T 1 T 2 l	Solution. We will look not for the center of gravity, but for the center of mass (since these are the same thing). Let's introduce the axis X(Fig. 27.3). Rice. 27.3
x C =?

Answer: at a distance from the mass T 1 .

STOP! Decide for yourself: B1–B3.

Statement 1 . If a homogeneous flat body has an axis of symmetry, the center of gravity is on this axis.

Indeed, for any point mass m i, located to the right of the symmetry axis, there is the same point mass located symmetrically relative to the first one (Fig. 27.4). In this case, the sum of the moments of forces .

Since the entire body can be represented as divided into similar pairs of points, the total moment of gravity relative to any point lying on the axis of symmetry is equal to zero, which means that the center of gravity of the body is located on this axis. This leads to an important conclusion: if a body has several axes of symmetry, then the center of gravity lies at the intersection of these axes(Fig. 27.5).

Rice. 27.5

Statement 2. If two bodies have masses T 1 and T 2 are connected into one, then the center of gravity of such a body will lie on a straight line segment connecting the centers of gravity of the first and second bodies (Fig. 27.6).

Rice. 27.6 Rice. 27.7

Proof. Let us position the composite body so that the segment connecting the centers of gravity of the bodies is vertical. Then the sum of the moments of gravity of the first body relative to the point WITH 1 is equal to zero, and the sum of the moments of gravity of the second body relative to the point WITH 2 is equal to zero (Fig. 27.7).

notice, that shoulder gravity of any point mass t i the same with respect to any point lying on the segment WITH 1 WITH 2, and therefore the moment of gravity relative to any point lying on the segment WITH 1 WITH 2, the same. Consequently, the gravitational force of the entire body is zero relative to any point on the segment WITH 1 WITH 2. Thus, the center of gravity of the composite body lies on the segment WITH 1 WITH 2 .

An important practical conclusion follows from Statement 2, which is clearly formulated in the form of instructions.

Instructions,

how to find the center of gravity of a solid body if it can be broken

into parts, the positions of the centers of gravity of each of which are known

1. Each part should be replaced with a mass located at the center of gravity of that part.

2. Find center of mass(and this is the same as the center of gravity) of the resulting system of point masses, choosing a convenient coordinate system X 0at, according to the formulas:

In fact, let us arrange the composite body so that the segment WITH 1 WITH 2 was horizontal, and hang it on threads at points WITH 1 and WITH 2 (Fig. 27.8, A). It is clear that the body will be in equilibrium. And this balance will not be disturbed if we replace each body with point masses T 1 and T 2 (Fig. 27.8, b).

Rice. 27.8

STOP! Decide for yourself: C3.

Problem 27.2. Balls of mass are placed at two vertices of an equilateral triangle T every. A ball of mass 2 is placed at the third vertex T(Fig. 27.9, A). Triangle side A. Determine the center of gravity of this system.

T 2T A	Rice. 27.9
x C = ? at C = ?

Solution. Let us introduce the coordinate system X 0at(Fig. 27.9, b). Then

Answer: x C = A/2; ; center of gravity lies at half height AD.

Goal of the work – determine the center of gravity of a complex figure analytically and experimentally.

Theoretical background. Material bodies consist of elementary particles, the position of which in space is determined by their coordinates. The forces of attraction of each particle to the Earth can be considered a system of parallel forces, the resultant of these forces is called the force of gravity of the body or the weight of the body. The center of gravity of a body is the point of application of gravity.

The center of gravity is a geometric point that can be located outside the body (for example, a disk with a hole, a hollow ball, etc.). Determining the center of gravity of thin flat homogeneous plates is of great practical importance. Their thickness can usually be neglected and the center of gravity can be assumed to be located in a plane. If the coordinate plane xOy is combined with the plane of the figure, then the position of the center of gravity is determined by two coordinates:

where is the area of part of the figure, ();

– coordinates of the center of gravity of the parts of the figure, mm (cm).

Section of a figure	A, mm 2	X c ,mm	Yc, mm
	bh	b/2	h/2
	bh/2	b/3	h/3
	R 2a
At 2α = π πR 2 /2

Work procedure.

Draw a figure of complex shape, consisting of 3-4 simple figures (rectangle, triangle, circle, etc.) on a scale of 1:1 and indicate its dimensions.

Draw the coordinate axes so that they cover the entire figure, break the complex figure into simple parts, determine the area and coordinates of the center of gravity of each simple figure relative to the selected coordinate system.

Calculate the coordinates of the center of gravity of the entire figure analytically. Cut out this figure from thin cardboard or plywood. Drill two holes, the edges of the holes should be smooth, and the diameter of the holes should be slightly larger than the diameter of the needle for hanging the figure.

First hang the figure at one point (hole), draw a line with a pencil that coincides with the plumb line. Repeat the same when hanging the figure at another point. The center of gravity of the figure, found experimentally, must coincide.

Determine the coordinates of the center of gravity of a thin homogeneous plate analytically. Check experimentally

Solution algorithm

1. Analytical method.

a) Draw the drawing on a scale of 1:1.

b) Break a complex figure into simple ones

c) Select and draw coordinate axes (if the figure is symmetrical, then along the axis of symmetry, otherwise along the figure’s contour)

d) Calculate the area of simple figures and the entire figure

e) Mark the position of the center of gravity of each simple figure in the drawing

f) Calculate the coordinates of the center of gravity of each figure

(x and y axis)

g) Calculate the coordinates of the center of gravity of the entire figure using the formula

h) Mark the position of the center of gravity on drawing C (

2. Experimental determination.

The correctness of the solution to the problem can be verified experimentally. Cut out this figure from thin cardboard or plywood. Drill three holes, the edges of the holes should be smooth, and the diameter of the holes should be slightly larger than the diameter of the needle for hanging the figure.

First hang the figure at one point (hole), draw a line with a pencil that coincides with the plumb line. Repeat the same when hanging the figure at other points. The value of the coordinates of the center of gravity of the figure, found when hanging the figure at two points: . The center of gravity of the figure, found experimentally, must coincide.

3. Conclusion about the position of the center of gravity during analytical and experimental determination.

Exercise

Determine the center of gravity of a flat section analytically and experimentally.

Execution example

Task

Determine the coordinates of the center of gravity of a thin homogeneous plate.

I Analytical method

1. The drawing is drawn to scale (dimensions are usually given in mm)

2. We break a complex figure into simple ones.

1- Rectangle

2- Triangle (rectangle)

3- Area of the semicircle (it doesn’t exist, minus sign).

We find the position of the center of gravity of simple figures of points, and

3. Draw the coordinate axes as convenient and mark the origin of coordinates.

4. Calculate the areas of simple figures and the area of the entire figure. [size in cm]

(3. no, sign -).

Area of the entire figure

5. Find the coordinate of the central point. , and in the drawing.

6. Calculate the coordinates of points C 1, C 2 and C 3

7. Calculate the coordinates of point C

8. Mark a point on the drawing

II Experienced

Coordinates of the center of gravity experimentally.

Control questions.

1. Is it possible to consider the force of gravity of a body as a resultant system of parallel forces?

2. Can the center of gravity of the entire body be located?

3. What is the essence of the experimental determination of the center of gravity of a flat figure?

4. How is the center of gravity of a complex figure consisting of several simple figures determined?

5. How should a figure of complex shape be rationally divided into simple figures when determining the center of gravity of the entire figure?

6. What sign does the area of the holes have in the formula for determining the center of gravity?

7. At the intersection of which lines of the triangle is its center of gravity located?

8. If a figure is difficult to break down into a small number of simple figures, what method of determining the center of gravity can provide the fastest answer?

Practical work No. 6

“Solving complex problems”

Goal of the work: be able to solve complex problems (kinematics, dynamics)

Theoretical background: Velocity is a kinematic measure of the movement of a point, characterizing the speed of change in its position. The speed of a point is a vector characterizing the speed and direction of movement of a point at a given moment in time. When specifying the motion of a point by equations, the velocity projections on the Cartesian coordinate axes are equal to:

The velocity modulus of a point is determined by the formula

The direction of the speed is determined by the direction cosines:

The characteristic of the speed of change of speed is acceleration a. The acceleration of a point is equal to the time derivative of the velocity vector:

When specifying the motion of a point, the equations for the projection of acceleration onto the coordinate axes are equal to:

Acceleration module:

Full acceleration module

The tangential acceleration module is determined by the formula

The normal acceleration modulus is determined by the formula

where is the radius of curvature of the trajectory at a given point.

The direction of acceleration is determined by the direction cosines

The equation of rotational motion of a rigid body around a fixed axis has the form

Angular velocity of the body:

Sometimes angular velocity is characterized by the number of revolutions per minute and is designated by the letter . The dependence between and has the form

Angular acceleration of the body:

A force equal to the product of the mass of a given point by its acceleration and the direction in the direction directly opposite to the acceleration of the point is called the force of inertia.

Power is the work done by a force per unit time.

Basic dynamics equation for rotational motion

– the moment of inertia of the body relative to the axis of rotation, is the sum of the products of the masses of material points by the square of their distances to this axis

Exercise

A body of mass m, with the help of a cable wound on a drum of diameter d, moves up or down along an inclined plane with an angle of inclination α. Equation of body motion S=f(t), equation of drum rotation, where S is in meters; φ - in radians; t – in seconds. P and ω are, respectively, the power and angular velocity on the drum shaft at the moment of the end of acceleration or the beginning of braking. Time t 1 – acceleration time (from rest to a given speed) or braking (from a given speed to a stop). The coefficient of sliding friction between the body and the plane is –f. Neglect friction losses on the drum, as well as the mass of the drum. When solving problems, take g=10 m/s 2

No. var	α, deg	Law of motion	For example, movement	m, kg	t 1 , s	d, m	P, kW	, rad/s	f	Def. quantities
		S=0.8t 2	Down	-	-	0,20	4,0		0,20	m,t 1
		φ=4t 2	Down		1,0	0,30	-	-	0,16	P,ω
		S=1.5t-t 2	up	-	-	-	4,5		0,20	m, d
		ω=15t-15t 2	up	-	-	0,20	3,0	-	0,14	m,ω
		S=0.5t 2	Down		-	-	1,76		0,20	d,t 1
		S=1.5t 2	Down	-	0,6	0,24	9,9	-	0,10	m,ω
		S=0.9t 2	Down		-	0,18	-		0,20	P, t 1
		φ=10t 2	Down		-	0,20	1,92	-	0,20	P, t 1
		S=t-1.25t 2	up		-	-	-		0,25	P,d
		φ=8t-20t 2	up		-	0,20	-	-	0,14	P, ω

Execution example

Problem 1(picture 1).

Solution 1. Rectilinear movement (Figure 1, a). A point moving uniformly at some point in time received a new law of motion, and after a certain period of time stopped. Determine all kinematic characteristics of the point’s movement for two cases; a) movement along a straight path; b) movement along a curved path of constant radius of curvature r=100cm

Figure 1(a).

Law of change of point speed

We find the initial speed of the point from the condition:

We find the braking time to stop from the condition:

at , from here .

Law of motion of a point during a period of uniform motion

The distance traveled by the point along the trajectory during the braking period is

Law of change in tangential acceleration of a point

whence it follows that during the period of braking the point moved equally slow, since the tangential acceleration is negative and constant in value.

The normal acceleration of a point on a straight trajectory is zero, i.e. .

Solution 2. Curvilinear movement (Figure 1, b).

Figure 1(b)

In this case, compared to the case of rectilinear motion, all kinematic characteristics remain unchanged, with the exception of normal acceleration.

Law of change in normal acceleration of a point

Normal acceleration of a point at the initial moment of braking

The numbering of point positions on the trajectory accepted in the drawing: 1 – current position of the point in uniform motion before the start of braking; 2 – position of the point at the moment of braking; 3 – current position of the point during the braking period; 4 – final position of the point.

Task 2.

The load (Fig. 2, a) is lifted using a drum winch. The diameter of the drum is d=0.3m, and the law of its rotation is .

The acceleration of the drum lasted until the angular velocity. Determine all kinematic characteristics of the movement of the drum and load.

Solution. Law of change in drum angular velocity. We find the initial angular velocity from the condition: ; therefore, acceleration began from a state of rest. We will find the acceleration time from the condition: . Drum rotation angle during acceleration period.

The law of change in the angular acceleration of the drum, it follows that during the acceleration period the drum rotated with uniform acceleration.

The kinematic characteristics of the load are equal to the corresponding characteristics of any point of the traction rope, and therefore point A lying on the rim of the drum (Fig. 2, b). As is known, the linear characteristics of a point of a rotating body are determined through its angular characteristics.

The distance traveled by the load during the acceleration period, . Velocity of the load at the end of acceleration.

Acceleration of cargo.

Law of cargo movement.

The distance, speed and acceleration of the load could be determined differently, through the found law of motion of the load:

Task 3. The load, moving uniformly upward along an inclined support plane, at some point in time received braking in accordance with the new law of motion , where s is in meters and t is in seconds. Mass of the load m = 100 kg, coefficient of sliding friction between the load and the plane f = 0.25. Determine the force F and power on the traction rope for two moments of time: a) uniform movement before braking begins;

b) initial moment of braking. When calculating, take g=10 m/.

Solution. We determine the kinematic characteristics of the movement of the load.

Law of change in speed of load

Initial speed of the load (at t=0)

Cargo acceleration

Since the acceleration is negative, the movement is slow.

1. Uniform movement of the load.

To determine the driving force F, we consider the equilibrium of the load, which is acted upon by a system of converging forces: the force on the cable F, the gravitational force of the load G=mg, the normal reaction of the supporting surface N and the friction force directed towards the movement of the body. According to the law of friction, . We choose the direction of the coordinate axes, as shown in the drawing, and draw up two equilibrium equations for the load:

The power on the cable before braking begins is determined by the well-known formula

Where is m/s.

2. Slow movement of cargo.

As is known, with uneven translational movement of a body, the system of forces acting on it in the direction of movement is not balanced. According to d'Alembert's principle (kinetostatic method), the body in this case can be considered to be in conditional equilibrium if we add to all the forces acting on it an inertial force, the vector of which is directed opposite to the acceleration vector. The acceleration vector in our case is directed opposite to the velocity vector, since the load moves slowly. We create two equilibrium equations for the load:

Power on the cable at the start of braking

Control questions.

1. How to determine the numerical value and direction of the speed of a point at a given moment?

2. What characterizes the normal and tangential components of total acceleration?

3. How to move from expressing angular velocity in min -1 to expressing it in rad/s?

4. What is body weight called? Name the unit of measurement of mass

5. At what motion of a material point does the force of inertia arise? What is its numerical value and what is its direction?

6. State d'Alembert's principle

7. Does the force of inertia arise during uniform curvilinear motion of a material point?

8. What is torque?

9. How is the relationship between torque and angular velocity expressed for a given transmitted power?

10. Basic dynamics equation for rotational motion.

Practical work No. 7

"Calculation of structures for strength"

Goal of the work: determine strength, cross-sectional dimensions and permissible load

Theoretical background.

Knowing the force factors and geometric characteristics of the section during tensile (compression) deformation, we can determine the stress using the formulas. And to understand whether our part (shaft, gear, etc.) will withstand external load. It is necessary to compare this value with the permissible voltage.

So, the static strength equation

Based on it, 3 types of problems are solved:

1) strength test

2) determination of section dimensions

3) determination of permissible load

So, the equation of static stiffness

Based on it, 3 types of problems are also solved

Equation of static tensile (compressive) strength

1) First type - strength test

i.e., we solve the left-hand side and compare it with the permissible stress.

2) Second type - determination of section dimensions

from the right side the cross-sectional area

Section circle

hence the diameter d

Rectangle section

Section square

A = a² (mm²)

Semicircle section

Sections: channel, I-beam, angle, etc.

Area values - from the table, accepted according to GOST

3) The third type is determining the permissible load;

taken to the smaller side, integer

EXERCISE

Task

A) Strength check (test calculation)

For a given beam, construct a diagram of longitudinal forces and check the strength in both sections. For timber material (steel St3) accept

Option No.
	12,5	5,3	-			-
		2,3			-	-
		4,2		-	-

B) Selection of section (design calculation)

For a given beam, construct a diagram of longitudinal forces and determine the cross-sectional dimensions in both sections. For timber material (steel St3) accept

Option No.
	1,9	2,5

	2,8	1,9
	3,2

B) Determination of permissible longitudinal force

For a given beam, determine the permissible values of loads and ,

construct a diagram of longitudinal forces. For timber material (steel St3) accept . When solving the problem, assume that the type of loading is the same on both sections of the beam.

Option No.
	-	-
			-	-
	-	-

Example of completing a task

Problem 1(picture 1).

Check the strength of a column made of I-profiles of a given size. For the column material (steel St3), accept the permissible tensile stresses and during compression . In the event of overloading or significant underloading, select I-beam sizes that ensure optimal column strength.

Solution.

The given beam has two sections 1, 2. The boundaries of the sections are the sections in which external forces are applied. Since the forces loading the beam are located along its central longitudinal axis, only one internal force factor arises in the cross sections - longitudinal force, i.e. there is tension (compression) of the beam.

To determine the longitudinal force, we use the section method. Conducting a mental section within each section, we will discard the lower fixed part of the beam and leave the upper part for consideration. In section 1, the longitudinal force is constant and equal to

The minus sign indicates that the beam is compressed in both sections.

We build a diagram of longitudinal forces. Having drawn the base (zero) line of the diagram parallel to the axis of the beam, we plot the obtained values perpendicular to it on an arbitrary scale. As you can see, the diagram turned out to be outlined by straight lines parallel to the base one.

We check the strength of the timber, i.e. We determine the design stress (for each section separately) and compare it with the permissible one. To do this, we use the compressive strength condition

where area is a geometric characteristic of the strength of the cross section. From the table of rolled steel we take:

for I-beam
for I-beam

Strength test:

The values of longitudinal forces are taken in absolute value.

The strength of the beam is ensured, however, there is a significant (more than 25%) underload, which is unacceptable due to excessive consumption of material.

From the strength condition, we determine the new dimensions of the I-beam for each section of the beam:
Hence the required area

According to the GOST table, we select I-beam No. 16, for which;

Hence the required area

According to the GOST table, we select I-beam No. 24, for which ;

With the selected I-beam sizes, underload also occurs, but it is insignificant (less than 5%)

Task No. 2.

For a beam with given cross-sectional dimensions, determine the permissible load values and . For timber material (steel St3), accept permissible tensile stresses and during compression .

Solution.

The given beam has two sections 1, 2. There is tension (compression) of the beam.

Using the method of sections, we determine the longitudinal force, expressing it through the required forces and. Carrying out a section within each section, we will discard the left part of the beam and leave the right part for consideration. In section 1, the longitudinal force is constant and equal to

In section 2, the longitudinal force is also constant and equal to

The plus sign indicates that the beam is stretched in both sections.

We build a diagram of longitudinal forces. The diagram is outlined by straight lines parallel to the base one.

From the condition of tensile strength, we determine the permissible load values and having previously calculated the areas of the given cross sections:

Control questions.

1. What internal force factors arise in the section of a beam during tension and compression?

2. Write down the tensile and compressive strength conditions.

3. How are the signs of longitudinal force and normal stress assigned?

4. How will the voltage change if the cross-sectional area increases by 4 times?

5. Are the strength conditions different for tensile and compressive calculations?

6. In what units is voltage measured?

7. Which mechanical characteristic is chosen as the limiting stress for ductile and brittle materials?

8. What is the difference between limiting and permissible stress?

Practical work No. 8

“Solving problems to determine the main central moments of inertia of flat geometric figures”

Goal of the work: determine analytically the moments of inertia of flat bodies of complex shape

Theoretical background. The coordinates of the center of gravity of the section can be expressed through the static moment:

where relative to the Ox axis

relative to the Oy axis

The static moment of the area of a figure relative to an axis lying in the same plane is equal to the product of the area of the figure and the distance of its center of gravity to this axis. The static moment has a dimension. The static moment can be positive, negative or equal to zero (relative to any central axis).

The axial moment of inertia of a section is the sum of the products or integral of elementary areas taken over the entire section by the squares of their distances to a certain axis lying in the plane of the section under consideration.

The axial moment of inertia is expressed in units - . The axial moment of inertia is a quantity that is always positive and not equal to zero.

The axes passing through the center of gravity of the figure are called central. The moment of inertia about the central axis is called the central moment of inertia.

The moment of inertia about any axis is equal to the center

Physics lesson notes, grade 7

Topic: Determining the center of gravity

Physics teacher, Argayash Secondary School No. 2

Khidiyatulina Z.A.

Laboratory work:

"Determination of the center of gravity of a flat plate"

Target : finding the center of gravity of a flat plate.

Theoretical part:

All bodies have a center of gravity. The center of gravity of a body is the point relative to which the total moment of gravity acting on the body is zero. For example, if you hang an object by its center of gravity, it will remain at rest. That is, its position in space will not change (it will not turn upside down or on its side). Why do some bodies tip over while others don't? If you draw a line perpendicular to the floor from the center of gravity of the body, then if the line goes beyond the boundaries of the body’s support, the body will fall. The larger the area of support, the closer the center of gravity of the body is to the central point of the area of support and the center line of the center of gravity, the more stable the position of the body will be. For example, the center of gravity of the famous Leaning Tower of Pisa is located just two meters from the middle of its support. And the fall will happen only when this deviation is about 14 meters. The center of gravity of the human body is approximately 20.23 centimeters below the navel. An imaginary line drawn vertically from the center of gravity passes exactly between the feet. For a tumbler doll, the secret also lies in the center of gravity of the body. Its stability is explained by the fact that the center of gravity of the tumbler is at the very bottom; it actually stands on it. The condition for maintaining the balance of a body is the passage of the vertical axis of its common center of gravity within the area of the body’s support. If the vertical center of gravity of the body leaves the support area, the body loses balance and falls. Therefore, the larger the area of support, the closer the center of gravity of the body is located to the central point of the area of support and the central line of the center of gravity, the more stable the position of the body will be. The area of support when a person is in a vertical position is limited by the space that is under the soles and between the feet. The center point of the vertical line of the center of gravity on the foot is 5 cm in front of the heel tubercle. The sagittal size of the support area always prevails over the frontal one, therefore the displacement of the vertical line of the center of gravity occurs more easily to the right and left than backward, and is especially difficult forward. In this regard, stability during turns during fast running is significantly less than in the sagittal direction (forward or backward). A foot in shoes, especially with a wide heel and a hard sole, is more stable than without shoes, as it acquires a larger area of support.

Practical part:

Purpose of the work: Using the proposed equipment, experimentally find the position of the center of gravity of two figures made of cardboard and a triangle.

Equipment:Tripod, thick cardboard, triangle from a school kit, ruler, tape, thread, pencil...

Task 1: Determine the position of the center of gravity of a flat figure of arbitrary shape

Using scissors, cut out a random shape from cardboard. Attach the thread to it at point A with tape. Hang the figure by the thread to the tripod leg. Using a ruler and pencil, mark the vertical line AB on the cardboard.

Move the thread attachment point to position C. Repeat the above steps.

Point O of the intersection of lines AB andCDgives the desired position of the center of gravity of the figure.

Task 2: Using only a ruler and pencil, find the position of the center of gravity of a flat figure

Using a pencil and ruler, divide the shape into two rectangles. By construction, find the positions O1 and O2 of their centers of gravity. It is obvious that the center of gravity of the entire figure is on the O1O2 line

Divide the figure into two rectangles in another way. By construction, find the positions of the centers of gravity O3 and O4 of each of them. Connect points O3 and O4 with a line. The intersection point of lines O1O2 and O3O4 determines the position of the figure’s center of gravity

Task 2: Determine the position of the center of gravity of the triangle

Using tape, secure one end of the thread at the top of the triangle and hang it from the tripod leg. Using a ruler, mark the direction AB of the gravity line (make a mark on the opposite side of the triangle)

Repeat the same procedure, hanging the triangle from vertex C. On the opposite vertex C side of the triangle, make a markD.

Using tape, attach pieces of thread AB andCD. Point O of their intersection determines the position of the center of gravity of the triangle. In this case, the center of gravity of the figure is outside the body itself.

III . Solving quality problems

1.For what purpose do circus performers hold heavy poles in their hands when walking on a tightrope?

2. Why does a person carrying a heavy load on his back lean forward?

3. Why can’t you get up from a chair unless you tilt your body forward?

4.Why does the crane not tip towards the load being lifted? Why, without a load, does the crane not tip towards the counterweight?

5. Why do cars and bicycles, etc. Is it better to put brakes on the rear wheels rather than the front wheels?

6. Why does a truck loaded with hay overturn more easily than the same truck loaded with snow?

Draw a diagram of the system and mark the center of gravity on it. If the found center of gravity is outside the object system, you received an incorrect answer. You may have measured distances from different reference points. Repeat the measurements.

For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, and not to the right or left of the swing. Also, the center of gravity will never coincide with the point where the child is sitting.
These arguments are valid in two-dimensional space. Draw a square that will contain all the objects of the system. The center of gravity should be inside this square.

Check your math if you get a small result. If the reference point is at one end of the system, a small result places the center of gravity near the end of the system. This may be the correct answer, but in the vast majority of cases this result indicates an error. When you calculated the moments, did you multiply the corresponding weights and distances? If instead of multiplying you added the weights and distances, you would get a much smaller result.

Correct the error if you found multiple centers of gravity. Each system has only one center of gravity. If you found multiple centers of gravity, you most likely did not add up all the moments. The center of gravity is equal to the ratio of the “total” moment to the “total” weight. There is no need to divide “every” moment by “every” weight: this way you will find the position of each object.

Check the reference point if the answer differs by some integer value. In our example, the answer is 3.4 m. Let's say you got the answer 0.4 m or 1.4 m, or another number ending in ".4". This is because you did not choose the left end of the board as your starting point, but a point that is located a whole amount to the right. In fact, your answer is correct no matter which reference point you choose! Just remember: the reference point is always at position x = 0. Here's an example:

In our example, the reference point was at the left end of the board and we found that the center of gravity was 3.4 m from this reference point.
If you choose as a reference point a point that is located 1 m to the right from the left end of the board, you will get the answer 2.4 m. That is, the center of gravity is 2.4 m from the new reference point, which, in turn, is located 1 m from the left end of the board. Thus, the center of gravity is at a distance of 2.4 + 1 = 3.4 m from the left end of the board. It turned out to be an old answer!
Note: when measuring distances, remember that the distances to the “left” reference point are negative, and to the “right” reference point are positive.

Measure distances in straight lines. Suppose there are two children on a swing, but one child is much taller than the other, or one child is hanging under the board rather than sitting on it. Ignore this difference and measure the distances along the straight line of the board. Measuring distances at angles will give close but not entirely accurate results.

For the see-saw board problem, remember that the center of gravity is between the right and left ends of the board. Later, you will learn to calculate the center of gravity of more complex two-dimensional systems.

Rectangle. Since a rectangle has two axes of symmetry, its center of gravity is at the intersection of the axes of symmetry, i.e. at the point of intersection of the diagonals of the rectangle.

Triangle. The center of gravity lies at the point of intersection of its medians. From geometry it is known that the medians of a triangle intersect at one point and are divided in a ratio of 1:2 from the base.

Circle. Since a circle has two axes of symmetry, its center of gravity is at the intersection of the axes of symmetry.

Semicircle. A semicircle has one axis of symmetry, then the center of gravity lies on this axis. Another coordinate of the center of gravity is calculated by the formula: .

Many structural elements are made from standard rolled products - angles, I-beams, channels and others. All dimensions, as well as geometric characteristics of rolled profiles, are tabular data that can be found in reference literature in tables of normal assortment (GOST 8239-89, GOST 8240-89).

Example 1. Determine the position of the center of gravity of the figure shown in the figure.

Solution:

We select the coordinate axes so that the Ox axis runs along the bottommost overall dimension, and the Oy axis goes along the leftmost overall dimension.

We break a complex figure into a minimum number of simple figures:

rectangle 20x10;

triangle 15x10;

circle R=3 cm.

We calculate the area of each simple figure and its coordinates of the center of gravity. The calculation results are entered into the table

Figure No.	Area of figure A,	Center of gravity coordinates
Figure No.	Area of figure A,

Answer: C(14.5; 4.5)

Example 2 . Determine the coordinates of the center of gravity of a composite section consisting of a sheet and rolled sections.

Solution.

We select the coordinate axes as shown in the figure.

Let's designate the figures by numbers and write out the necessary data from the table:

Figure No.	Area of figure A,	Center of gravity coordinates
Figure No.	Area of figure A,

We calculate the coordinates of the center of gravity of the figure using the formulas:

Answer: C(0; 10)

Laboratory work No. 1 “Determination of the center of gravity of composite flat figures”

Target: Determine the center of gravity of a given flat complex figure using experimental and analytical methods and compare their results.

Work order

Draw your flat figure in your notebooks in size, indicating the coordinate axes.

Determine the center of gravity analytically.

Divide the figure into the minimum number of figures whose centers of gravity we know how to determine.
Indicate the area numbers and coordinates of the center of gravity of each figure.
Calculate the coordinates of the center of gravity of each figure.
Calculate the area of each figure.
Calculate the coordinates of the center of gravity of the entire figure using the formulas (the position of the center of gravity is plotted on the drawing of the figure):

The installation for experimentally determining the coordinates of the center of gravity using the hanging method consists of a vertical stand 1 (see figure) to which the needle is attached 2 . Flat figure 3 Made of cardboard, which is easy to punch holes in. Holes A And IN pierced at randomly located points (preferably at the furthest distance from each other). A flat figure is suspended on a needle, first at a point A , and then at the point IN . Using a plumb line 4 , attached to the same needle, draw a vertical line on the figure with a pencil corresponding to the thread of the plumb line. Center of gravity WITH the figure will be located at the intersection point of the vertical lines drawn when hanging the figure at the points A And IN .