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Ханс Кристиан Андерсен

Ханс Кристиан Андерсен

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урок

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учителю

ученику

физика

физика 7 класс

блоки

подвижный блок

неподвижный блок

«Золотое правило механики»

КПД.

00:00:00

Today's lesson we have devoted to

00:00:03

the study of blocks and we will also consider the

00:00:06

golden rule of mechanics. The

00:00:09

once famous Hungarian chemist of

00:00:11

solids and Vichy said that the thinking mind does not

00:00:15

feel happy until it

00:00:17

manages to connect together disparate

00:00:20

facts and observed

00:00:23

in previous lessons we

00:00:25

became acquainted with such a simple mechanism

00:00:27

as a lever, let's remember that a lever is

00:00:31

any solid body that can

00:00:33

rotate relative to a fixed

00:00:36

support or axis. We

00:00:38

divided the lever into two types: lever 1 and a

00:00:42

lever of the second kind.

00:00:44

Let's remember that a lever of the first kind is a

00:00:47

lever whose axis of rotation is located

00:00:50

between the points of application of forces and the forces themselves are

00:00:53

directed in one side

00:00:56

and a lever of the second type is a lever

00:00:59

whose axis of rotation is located on one

00:01:02

side of the points of application of forces and the

00:01:05

forces themselves are directed opposite to each

00:01:08

other, the

00:01:09

conditions for the equilibrium of a lever were derived

00:01:12

according to which the lever is in

00:01:14

equilibrium, provided that the forces applied to

00:01:17

it are inversely proportional to

00:01:20

the length of their arms,

00:01:23

introduced into considering the moment of force is a

00:01:25

physical quantity equal to the product of the

00:01:29

modulus of force of the rotating bodies by its shoulder and

00:01:33

formulated the conditions for equilibrium of the lever

00:01:35

through the rule of

00:01:37

moments according to which the lever under the

00:01:40

influence of two forces creating moments

00:01:43

is in equilibrium if the

00:01:46

moment of force rotating the lever clockwise is

00:01:50

equal to the moment of force rotating the lever

00:01:53

counterclockwise,

00:01:55

however, in addition to levers, a

00:01:59

simple block and a system of

00:02:02

blocks are often used to lift loads. Blocks are especially often used

00:02:05

on construction sites in ports and warehouses;

00:02:09

any block is a wheel with a

00:02:12

groove fixed in a cage along the block groove; a rope

00:02:15

is passed through a cable or chain;

00:02:20

what types of blocks and how do they convert

00:02:23

force

00:02:25

if the axis of the block is fixed and when lifting

00:02:27

loads it does not fall or

00:02:30

rise, then the block is called

00:02:32

motionless such a block can be

00:02:34

considered as equal to the arms of a lever

00:02:37

whose arms of forces are equal to the

00:02:39

radius of the wheel

00:02:43

does such a block give a gain in force we

00:02:47

will not answer this question experiment

00:02:50

let's take a load weighing 3 newtons and hang it

00:02:53

to one end of a thread thrown over a block

00:02:55

and attach a dynamometer to the other.

00:03:00

When the load is lifted uniformly, the dynamometer

00:03:03

will show a force equal to the weight of the load,

00:03:05

that is, 3 newtons,

00:03:08

let's now depict the schematic

00:03:10

forces acting on the blog,

00:03:13

this elastic force of the thread is equal to the weight of the

00:03:15

load

00:03:17

force the elasticity of the thread is equal to the force applied to the

00:03:19

dynamometer, the

00:03:21

force of gravity acting on the block and the

00:03:24

elastic forces of the axis of the block,

00:03:27

as can be seen from the figure, the shoulders of the forces of gravity and the

00:03:29

elasticity of the block are equal to zero, which means their

00:03:33

moments relative to the axis are

00:03:35

equal to zero; the

00:03:36

shoulders of the elastic forces of the thread 1 and 2 are equal to

00:03:40

each other as the radius and block in

00:03:43

a state of equilibrium of the block, the moments of forces

00:03:45

f1 and f2

00:03:47

must be equal,

00:03:49

and since the moments of these forces are equal, then the

00:03:52

forces themselves are equal to each other, in other words, the

00:03:55

applied force is equal to the weight of the load,

00:03:59

so we can conclude

00:04:02

that a stationary block does not provide a

00:04:04

gain in force

00:04:06

a only changes its direction,

00:04:09

you may have a logical question:

00:04:11

why use a fixed block if there

00:04:14

is no gain in strength, because with the same

00:04:17

success you

00:04:18

could use any

00:04:20

crossbar to lift the load, it is

00:04:22

indeed possible, but you will lose because you

00:04:25

will have to overcome the frictional force of the

00:04:27

rope sliding along the crossbar,

00:04:29

which is significant there is more

00:04:32

rolling friction force in the bearing of the block,

00:04:35

but can the block still give a gain in

00:04:38

force?

00:04:39

To answer this question, let’s

00:04:41

look at another type of block - a

00:04:43

movable block, that is, a block whose rotation axis

00:04:49

moves together with the load when lifting a load;

00:04:52

hang the load from such a block weighing 6

00:04:54

newtons, we will fasten one end of the thread thrown over the

00:04:57

block and

00:05:00

we will evenly lift the load behind the other

00:05:03

using a dynamometer;

00:05:05

look, the dynamometer shows that the

00:05:08

force applied to the end of the rope is equal to

00:05:11

three newtons, that is, 2 times less than the

00:05:14

weight of the load,

00:05:16

which means the movable block gives a gain in

00:05:19

strength of approximately Let's try

00:05:22

to explain this result twice,

00:05:24

so the block is acted upon by the weight of the load, the

00:05:28

elastic forces of the thread, which are equal to each other,

00:05:31

and the gravity of the block,

00:05:33

while most often the gravity of the block is

00:05:36

neglected since it is usually

00:05:38

much less than the weight of the load,

00:05:41

when the load moves, the movable block

00:05:43

rotates relative to point d,

00:05:47

which means the movable block is a lever of the second

00:05:51

kind,

00:05:52

we write the equilibrium condition for it

00:05:54

through the rule of moments.

00:05:56

From the figure it can be seen that the arm of the weight of the load is

00:05:59

equal to the radius of the block

00:06:01

and the arm 2 of the force is equal to the two radii of the block,

00:06:05

taking into account the fact that the force f 2 is equal to the force

00:06:09

applied to the end of the rope and using the

00:06:12

basic property of proportion we obtain the force

00:06:15

applied to the end of the rope

00:06:17

is equal to half the weight of the load,

00:06:21

so we can conclude

00:06:23

that the movable block

00:06:25

gives a double gain in force.

00:06:29

Now you and I can draw the main

00:06:32

conclusion that when using

00:06:34

simple mechanisms

00:06:35

we can get a gain in force,

00:06:38

then the logical one arises The question is, is it possible

00:06:43

to obtain a gain in work using a simple mechanism?

00:06:46

If the applied force is less than the weight of the load,

00:06:49

then will the work done by it be less than the

00:06:52

work of lifting the load without

00:06:54

using a mechanism?

00:06:56

To answer this question, we will conduct an experiment;

00:07:00

we will raise the load uniformly to a

00:07:02

certain height using a movable

00:07:04

block we neglect the gravity of the block and the friction force,

00:07:10

so the work of the force applied to the thread is equal to the

00:07:13

product of the force applied to the thread and the

00:07:16

height of the lift and the point of application,

00:07:20

as can be seen from the figure, the height of the lift of the

00:07:22

point of application of the force is twice the

00:07:25

height of the lift of the load, the

00:07:27

work of lifting the load is equal in modulus to

00:07:30

the product the weight of the load and the height of lifting the

00:07:34

load,

00:07:35

now we compare the two works, taking into account

00:07:38

that the force applied to the end of the rope is

00:07:40

approximately two times less than the weight of the load,

00:07:45

taking this fact into account, we obtain

00:07:47

that the work of lifting the load is equal to the work of the

00:07:51

force applied to the thread,

00:07:53

so the use of a moving

00:07:56

block does not gain in work,

00:07:59

look, winning twice in strength, we

00:08:02

lost twice on the way,

00:08:06

similarly, you can approach the consideration of a

00:08:08

lever;

00:08:10

for this, two forces of different magnitudes are balanced on the lever

00:08:13

and the lever is set

00:08:15

in motion

00:08:17

if you measure the distance traveled by both

00:08:19

more and less forces and the modules of these forces,

00:08:22

then we get that the paths traversed by the points of

00:08:25

application of forces on the lever are inversely

00:08:28

proportional to the forces.

00:08:30

Thus, as in the case of the moving

00:08:33

block, we can conclude that when we act

00:08:36

on the long arm of the lever, we gain in

00:08:39

strength, but at the same time

00:08:41

we lose the same number of times in the path

00:08:44

since the product of force there is work on the way, then

00:08:47

in this case we do not get a gain in work,

00:08:51

as centuries-old practice has shown, not a

00:08:54

single mechanism gives a gain in work,

00:08:57

this statement is called the

00:08:59

golden rule of mechanics,

00:09:03

if with the help of any simple

00:09:05

mechanism we win by force, then

00:09:08

we lose by the same amount on the way,

00:09:12

but is it possible, when comparing works, to establish

00:09:14

strict equality between them? Is it

00:09:17

by chance that we always draw one or

00:09:19

another conclusion? We warned that the force of

00:09:22

gravity acting on the block and the force

00:09:24

of friction must be neglected because three does not

00:09:27

exist; it is present in all

00:09:30

mechanisms and the force of gravity that

00:09:32

acts on the block itself, even if it is

00:09:34

small, there is also even if there is no

00:09:37

lifting of a simple mechanism

00:09:39

or its parts,

00:09:41

as in the case of a stationary block, we must

00:09:44

apply additional force to

00:09:46

bring it into motion, that is, to

00:09:49

overcome the inertia of the mechanism, therefore the

00:09:52

force applied to the mechanism must

00:09:54

actually do more work than the

00:09:57

useful one the work of lifting a load,

00:10:01

the work of force applied to the mechanism

00:10:03

is called

00:10:04

expended or full work, and

00:10:07

useful work is the work of lifting

00:10:10

only the load itself,

00:10:13

no matter what mechanism we take,

00:10:15

useful work always constitutes only a

00:10:18

certain part of the total work; we denote

00:10:21

useful work by the letter a with the index p

00:10:24

a expended a with index d

00:10:28

then the ratio of useful work to work

00:10:31

expended is called the

00:10:33

efficiency of the mechanism,

00:10:36

abbreviated as efficiency, the

00:10:39

efficiency coefficient is

00:10:41

denoted by a small Greek letter

00:10:43

and is most often expressed as a percentage

00:10:48

since useful work is always

00:10:50

emphasized as always less than perfect,

00:10:53

then the efficiency of

00:10:55

the mechanism is always less than 100 percent

00:11:00

and now to secure the new

00:11:02

material, let's consider solving

00:11:04

several problems of

00:11:06

the condition of problem 1 such

00:11:09

as what minimum force needs to be applied to the

00:11:11

end of the rope to lift a bag of cement

00:11:14

weighing 50 kilograms using a movable

00:11:18

block to what height the bag will be raised

00:11:21

when applying this force and work of two

00:11:24

and a half thousand joules we

00:11:27

will write down given

00:11:28

so from the conditions of the problem we know

00:11:31

the mass of the bag of cement and the work

00:11:34

done by the force to lift this bag and

00:11:38

for the convenience of calculations we

00:11:40

will take the coefficient equal to 10 newtons per kilogram

00:11:44

to determine we need the minimum

00:11:46

force that needs to be applied to the rope and the

00:11:49

height of lifting the bag of cement let's

00:11:53

proceed to the solution, we already know that the

00:11:56

movable block gives a double gain in strength,

00:12:00

then the minimum force that needs to be

00:12:02

applied to the end of the rope to lift a

00:12:04

bag of cement is equal to half the weight of this

00:12:07

bag, I

00:12:09

believe that the bag rises evenly,

00:12:11

its weight will be determined as the product of

00:12:14

its mass and the coefficient,

00:12:18

substituting numerical data we get that

00:12:20

the weight of a bag of cement is 500

00:12:23

newtons

00:12:24

and the minimum force that needs to be

00:12:27

applied to the end of the rope is 250 newtons.

00:12:31

Now let’s calculate the height of the rise of the bear and

00:12:34

so the work of force is equal to the product of the

00:12:37

modulus of force and the path it has passed.

00:12:41

application of this force,

00:12:43

we already know that if we gain in

00:12:45

strength, then we lose by the same amount on the

00:12:48

way,

00:12:50

since the moving block gives us a gain

00:12:52

in strength twice, then on the way we will

00:12:55

also lose twice,

00:12:58

then the lifting height of the bag of cement will be

00:13:01

twice less than the height of the rise of the point of

00:13:03

application of force,

00:13:08

we define the height of rise of the point of application of force as the ratio of the work done by the

00:13:11

perfect force to the modulus of this force,

00:13:15

we obtain that it is 10 meters; the

00:13:19

height to which the bag of cement was raised

00:13:21

is 5 meters.

00:13:26

1 problem: a

00:13:27

slab of mass 120 kilograms was

00:13:30

uniformly lifted using a movable

00:13:33

block to a height of 16 meters in a period of

00:13:36

time equal to

00:13:37

40 seconds and considering the

00:13:41

efficiency equal to eighty percent and the

00:13:44

mass of the block is 10 kilograms, determine the

00:13:47

total work and the developed power we

00:13:51

write down

00:13:53

from the condition we know the mass of the slab and the

00:13:55

block, the height to which the slab was raised

00:13:58

and the time of its rise

00:14:02

as well as the efficiency of the

00:14:04

lifting mechanism, let’s

00:14:06

take the coefficient equal to 9.8 newton

00:14:10

per kilogram;

00:14:12

determine; we need the total or

00:14:15

expended work and the

00:14:17

power developed by the lifting mechanism; let’s

00:14:20

proceed to the solution;

00:14:22

write down the formula for determining the

00:14:24

efficiency of

00:14:25

the mechanism;

00:14:27

in it, two unknown quantities are

00:14:30

useful and the work expended, the

00:14:33

useful work of the mechanism is performed

00:14:36

against the gravity forces of the load and the block,

00:14:39

as we know, gravity is defined

00:14:41

as the product of mass by a coefficient,

00:14:45

then the useful work will be determined

00:14:48

as the product of the sum of the masses of the plate and the

00:14:51

block, the lifting height and the coefficient, and

00:14:55

substituting the numerical data, we obtain that the

00:14:58

useful work is equal to 20,000 384 joules

00:15:05

now from the formula for the efficiency factor

00:15:08

we express the work expended by

00:15:11

substituting the numerical data and we obtain that the

00:15:13

expended work is 25 thousand

00:15:16

480 joules. All we

00:15:20

have to do is find the power that the

00:15:22

lifting mechanism develops;

00:15:24

this is the expended power,

00:15:27

by definition, power is the work

00:15:30

done per unit of time,

00:15:32

then we substitute the numerical data and

00:15:34

we get that the power of this lifting

00:15:37

mechanism

00:15:38

is 637 watts

00:15:42

and now it’s time to sum up

00:15:44

our lesson.

00:15:46

Today we got acquainted with

00:15:48

another simple mechanism - a block. A block is

00:15:53

one of the types of levers

00:15:54

that are a wheel with a

00:15:57

groove fixed in a cage;

00:15:59

there are movable and fixed blocks; a

00:16:03

fixed block is a block the axis of rotation of

00:16:06

which is fixed and when lifting loads

00:16:09

it does not rise or fall,

00:16:13

and the movable block is a block whose axis of rotation

00:16:15

rises and falls along

00:16:18

with the load.

00:16:20

We found out that a stationary block does not give a

00:16:23

gain in force, but only changes its

00:16:25

direction. The movable block, if we

00:16:28

neglect friction and the weight of the block itself

00:16:31

gives a gain in strength twice as much;

00:16:34

we also learned that not a single mechanism

00:16:37

gives a gain in work; this statement

00:16:39

is called the golden rule of mechanics,

00:16:43

according to which the number of times we

00:16:45

gain in strength is the same number of times we

00:16:48

lose along the way; we

00:16:51

introduced the concept of coefficient into consideration

00:16:52

useful action of a

00:16:54

mechanism, which shows what part

00:16:58

of the perfect applied force and work

00:17:00

is useful work;

00:17:03

also remember that due to the action of

00:17:06

friction forces and other losses,

00:17:07

useful work is always less than

00:17:09

perfect, therefore the

00:17:12

efficiency of any mechanism is

00:17:14

less than 100 percent

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This feature is available in the UDL Helper extension. Make sure that "Show the video snapshot button" is checked in the settings. A camera icon should appear in the lower right corner of the player to the left of the "Settings" icon. When you click on it, the current frame from the video will be saved to your computer in JPEG format.

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It costs nothing. Our services are absolutely free for all users. There are no PRO subscriptions, no restrictions on the number or maximum length of downloaded videos.