Why do you need a moment of power. Moment of power

Moment of power (synonyms: torque, torque, twisting moment, torque) is a vector physical quantity equal to the vector product of the radius vector drawn from the axis of rotation to the point of application of the force by the vector of this force. It characterizes the rotational action of a force on a solid.

The concepts of "rotating" and "torque" moments are generally not identical, since in technology the concept of "torque" is considered as an external force applied to an object, and "torque" is an internal force arising in an object under the action of applied loads (this the concept is operated in the strength of materials).

General information

Special cases

Lever torque formula

A very interesting special case is presented as the determination of the moment of force in the field:

$\backslash \backslash \; left\; |\; \backslash \backslash \; vec\; M\; \backslash \backslash \; right\; |\; \backslash u003d\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; (M)\; \_1\; \backslash \backslash \; right\; |\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; F\; \backslash \backslash \; right\; |$where: $\backslash \backslash \; left\; |\; \backslash \backslash \; vec\; (M)\; \_1\; \backslash \backslash \; right\; |$ - moment of the lever, $\backslash \backslash \; left\; |\; \backslash \backslash \; vec\; F\; \backslash \backslash \; right\; |$ - the magnitude of the acting force.

The problem with this view is that it does not give the direction of the moment of force, but only its magnitude. If the force is perpendicular to the vector $\backslash \backslash \; vec\; r$, the moment of the lever will be equal to the distance to the center and the moment of force will be maximum:

$\backslash \backslash \; left\; |\; \backslash \backslash \; vec\; (T)\; \backslash \backslash \; right\; |\; \backslash u003d\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; r\; \backslash \backslash \; right\; |\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; F\; \backslash \backslash \; right\; |$

Force at an angle

If strength $\backslash \backslash \; vec\; F$ directed at an angle $\backslash \backslash \; theta$ to lever r, then $M\; \backslash u003d\; r\; F\; \backslash \backslash \; sin\; \backslash \backslash \; theta$.

Static balance

In order for the object to be in equilibrium, not only the sum of all forces must be equal to zero, but also the sum of all moments of force around any point. For a two-dimensional case with horizontal and vertical forces: the sum of forces in two dimensions ΣH \u003d 0, ΣV \u003d 0 and the moment of force in the third dimension ΣM \u003d 0.

Moment of force as a function of time

$\backslash \backslash \; vec\; M\; \backslash u003d\; \backslash \backslash \; frac\; (d\; \backslash \backslash \; vec\; L)\; (dt)$,

where $\backslash \backslash \; vec\; L$ - angular momentum.

Let's take a solid. The movement of a rigid body can be represented as the movement of a specific point and rotation around it.

The moment of momentum relative to the point O of a rigid body can be described through the product of the moment of inertia and angular velocity relative to the center of mass and the linear motion of the center of mass.

$\backslash \backslash \; vec\; (L\_o)\; \backslash u003d\; I\_c\; \backslash \backslash ,\; \backslash \backslash \; vec\; \backslash \backslash \; omega\; +$

We will consider rotating motions in the Koenig coordinate system, since it is much more difficult to describe the motion of a rigid body in the world coordinate system.

Let's differentiate this expression in time. And if $I$ is a constant value in time, then

$\backslash \backslash \; vec\; M\; \backslash u003d\; I\; \backslash \backslash \; frac\; (d\; \backslash \backslash \; vec\; \backslash \backslash \; omega)\; (dt)\; \backslash u003d\; I\; \backslash \backslash \; vec\; \backslash \backslash \; alpha$,

The relationship between moment of power and work

$A\; \backslash u003d\; \backslash \backslash \; int\; \_\; (\backslash \backslash \; theta\_1)\; ^\; (\backslash \backslash \; theta\_2)\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; M\; \backslash \backslash \; right\; |\; \backslash \backslash \; mathrm\; (d)\; \backslash \backslash \; theta$

In the case of a constant torque, we get:

$A\; \backslash u003d\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; M\; \backslash \backslash \; right\; |\; \backslash \backslash \; theta$

The angular velocity is usually known $\backslash \backslash \; omega$ in radians per second and time of moment $t$.

Then the work done by the moment of force is calculated as:

$A\; \backslash u003d\; \backslash \backslash \; left\; |\; \backslash \backslash \; vec\; M\; \backslash \backslash \; right\; |\; \backslash \backslash \; omega\; t$

Moment of force relative to a point

If there is a material point $O\_F$to which the force is applied $\backslash \backslash \; vec\; F$, then the moment of force about the point $O$ is equal to the vector product of the radius vector $\backslash \backslash \; vec\; r$connecting the points $O$ and $O\_F$, on the force vector $\backslash \backslash \; vec\; F$:

$\backslash \backslash \; vec\; (M\_O)\; \backslash u003d\; \backslash \backslash \; left\; [\backslash \backslash \; vec\; r\; \backslash \backslash \; times\; \backslash \backslash \; vec\; F\; \backslash \backslash \; right]$.

Moment of force about the axis

The moment of force about the axis is equal to the algebraic moment of the projection of this force onto a plane perpendicular to this axis relative to the point of intersection of the axis with the plane, that is $M\_z\; (F)\; \backslash u003d\; M\_o\; (F\; ")\; \backslash u003d\; F"\; h\; "$.

Units

The moment of force is measured in newton meters... 1 Nm is the moment that a force of 1 N produces on a lever 1 m long, applied to the end of the lever and directed perpendicular to it.

Torque measurement

Today, the measurement of the moment of force is carried out using strain gauge, optical and inductive load cells.

see also

Write a review on the article "Moment of Power"

Excerpt from the Moment of Power

But although by the end of the battle people felt all the horror of their act, although they would have been glad to stop, some incomprehensible, mysterious force still continued to guide them, and, sweating, covered in gunpowder and blood, remaining one by three, the artillerymen, although and stumbling and panting with fatigue, they brought charges, charged, directed, applied wicks; and the nuclei flew just as quickly and cruelly from both sides and flattened the human body, and the terrible deed continued, which is not done at the will of people, but at the will of the one who leads people and worlds.
Anyone who looked at the frustrated backsides of the Russian army would say that the French have to make one more small effort and the Russian army will disappear; and anyone who looked at the backs of the French would say that the Russians have to make one more little effort and the French will perish. But neither the French nor the Russians made this effort, and the flames of the battle were slowly dying out.
The Russians did not make this effort because they were not the ones who attacked the French. At the beginning of the battle, they only stood on the road to Moscow, blocking it, and in the same way they continued to stand at the end of the battle, as they stood at the beginning of it. But if even the goal of the Russians were to bring down the French, they could not make this last effort, because all the Russian troops were defeated, there was not a single part of the troops that did not suffer in the battle, and the Russians, remaining in their places , lost half of their troops.
It was easy for the French, with the memory of all the previous fifteen-year victories, with confidence in Napoleon's invincibility, with the knowledge that they had taken possession of part of the battlefield, that they had lost only one quarter of the people and that they still have a twenty-thousandth untouched guard, it was easy to make this effort. The French, who attacked the Russian army in order to knock it out of position, had to make this effort, because as long as the Russians, just as before the battle, blocked the road to Moscow, the French goal was not achieved and all their efforts and the losses were wasted. But the French did not make this effort. Some historians say that Napoleon should have given his pristine old guard in order for the battle to be won. To talk about what would have happened if Napoleon had given his guard is like talking about what would have happened if spring had come in the fall. It couldn't be. Napoleon did not give his guard, because he did not want it, but it could not be done. All the generals, officers, and soldiers of the French army knew that this could not be done, because the fallen spirit of the army did not allow it.
Not only Napoleon experienced that dreamlike feeling that the terrible sweep of the arm falls powerlessly, but all the generals, all the soldiers of the French army who participated and did not participate, after all the experiences of previous battles (where, after ten times less effort, the enemy fled), experienced the same feeling of horror in front of the enemy who, having lost half of the army, stood as menacing at the end as at the beginning of the battle. The moral strength of the French attacking army was exhausted. Not that victory, which is determined by the pieces of matter picked up on sticks called banners, and by the space on which the troops stood and are, but a moral victory, one that convinces the enemy of the moral superiority of his enemy and of his powerlessness, was won by the Russians under Borodin. The French invasion, like an angry beast that had received a mortal wound in its flight, felt its destruction; but it could not stop, just as it could not help but deviate the twice weaker Russian army. After this push, the French army could still reach Moscow; but there, without new efforts on the part of the Russian army, it had to die, bleeding from the mortal wound inflicted at Borodino. The direct consequence of the Battle of Borodino was the unreasonable flight of Napoleon from Moscow, the return along the old Smolensk road, the death of the five hundred thousandth invasion and the death of Napoleonic France, on which the hand of the strongest enemy was laid on for the first time at Borodino.

The absolute continuity of movement is incomprehensible to the human mind. The laws of any kind of movement become clear to a person only when he considers arbitrarily taken units of this movement. But at the same time, from this arbitrary division of continuous motion into discontinuous units, a large part of human delusions flows.
The so-called sophism of the ancients is known, which consists in the fact that Achilles will never catch up with a turtle walking in front, despite the fact that Achilles is ten times faster than a turtle: as soon as Achilles passes the space separating him from the turtle, the turtle will pass one tenth of this in front of him. space; Achilles will pass this tenth, the turtle will pass one hundredth, and so on, to infinity. This task seemed insoluble to the ancients. The senselessness of the decision (that Achilles would never catch up with the tortoise) stemmed only from the fact that discontinuous units of movement were arbitrarily allowed, while the movement of both Achilles and the tortoise was continuous.
Accepting smaller and smaller units of motion, we are only approaching the solution of the issue, but we never reach it. Only by admitting an infinitely small value and an ascending progression from it to one-tenth and taking the sum of this geometric progression, we achieve a solution to the problem. The new branch of mathematics, having achieved the art of handling infinitesimal quantities, and in other more complex questions of motion, now gives answers to questions that seemed insoluble.
This new, unknown to the ancients, branch of mathematics, when considering the issues of motion, admitting infinitely small quantities, that is, those under which the main condition of motion (absolute continuity) is restored, thereby correcting that inevitable mistake that the human mind cannot but make when considering instead of continuous movement, individual units of movement.
In the search for the laws of historical movement, exactly the same thing happens.
The movement of mankind, flowing from the countless number of human arbitrariness, is carried out continuously.
Comprehension of the laws of this movement is the goal of history. But in order to comprehend the laws of continuous movement of the sum of all the arbitrariness of people, the human mind allows arbitrary, discontinuous units. The first method of history is that, taking an arbitrary series of continuous events, consider it separately from the others, whereas there is and cannot be the beginning of any event, and always one event continuously follows from another. The second method is to consider the action of one person, a king, a commander, as the sum of the arbitrariness of people, while the sum of the arbitrariness of people is never expressed in the activity of one historical person.
Historical science in its movement constantly accepts smaller and smaller units for consideration and in this way strives to get closer to the truth. But no matter how small the units that history accepts, we feel that the assumption of a unit separated from another, the assumption of the beginning of some phenomenon and the assumption that the arbitrariness of all people is expressed in the actions of one historical person, are false in themselves.
Any conclusion of history, without the slightest effort on the part of criticism, disintegrates like dust, leaving nothing behind, only due to the fact that criticism chooses a larger or smaller discontinuous unit for the object of observation; to which it always has the right, since the historical unit taken is always arbitrary.
Only by admitting an infinitely small unit for observation - the differential of history, that is, homogeneous drives of people, and having achieved the art of integrating (taking the sums of these infinitely small ones), can we hope to comprehend the laws of history.
The first fifteen years of the 19th century in Europe represent an extraordinary movement of millions of people. People abandon their usual occupations, strive from one side of Europe to the other, rob, kill one another, triumph and despair, and the whole course of life changes for several years and represents an intensified movement, which first goes on increasing, then weakening. What is the reason for this movement or according to what laws did it take place? - the human mind asks.
Historians, answering this question, describe to us the actions and speeches of several dozen people in one of the buildings of the city of Paris, calling these actions and speeches the word revolution; then they give a detailed biography of Napoleon and some sympathetic and hostile persons to him, talk about the influence of some of these persons on others and say: this is why this movement took place, and these are its laws.
But the human mind not only refuses to believe in this explanation, but directly says that the method of explanation is not correct, because in this explanation the weakest phenomenon is taken as the cause of the strongest. The sum of human arbitrariness made both the revolution and Napoleon, and only the sum of these arbitrariness endured and destroyed them.

A moment of couple of forces

The moment of force relative to any point (center) is a vector numerically equal to the product of the modulus of the force by the shoulder, i.e. the shortest distance from the specified point to the line of action of the force, and directed perpendicular to the plane passing through the selected point and the line of action of the force in the direction from which the "rotation" made by the force around the point seems to occur counterclockwise. The moment of force characterizes its rotational action.

If a ABOUT - the point relative to which the moment of force is F, then the moment of force is indicated by the symbol M o (F)... Let us show that if the point of application of the force Fdefined by the radius vector r, then the relation

Mo (F) \u003d r × F. (3.6)

According to this relation the moment of force is equal to the vector product of the vectorr by vector F.

Indeed, the modulus of the vector product is

M about ( F)=rFsin \u003d Fh, (3.7)

where h - the shoulder of strength. Note also that the vector M o (F) directed perpendicular to the plane passing through the vectors rand F, in the direction from which the shortest rotation of the vector rto the direction of the vector Fappears to be going counterclockwise. Thus, formula (3.6) completely determines the modulus and direction of the moment of force F.

Sometimes it is useful to write formula (3.7) in the form

M about ( F)=2S, (3.8)

where S - area of \u200b\u200ba triangle OAV.

Let be x, y, z Are the coordinates of the point of application of the force, and F x, F y, F z - force projection on the coordinate axes. Then if the point ABOUT is at the origin, the moment of force is expressed as follows:

It follows that the projections of the moment of force on the coordinate axes are determined by the formulas:

M Ox(F)= yF z -zF y,

M Oy(F)= zF x -xF z ,

M Oy(F)= xF y -yF x. (3.10)

Let us now introduce the concept of the projection of a force onto a plane.

May strength be given Fand some plane. Let us drop perpendiculars from the beginning and end of the force vector to this plane.

Force projection on a planecalled vector , the beginning and end of which coincide with the projection of the beginning and the projection of the end of the force on this plane.

If we take the plane hoy, then the projection of the force Fon this plane there will be a vector F hu.

Moment of power F hu relative to point ABOUT(axis intersection points z with plane hoy) can be calculated by formula (3.9) if we take z=0, F z\u003d 0. We get

M O(F hu)=(xF y -yF x)k.

Thus, the moment is directed along the axis z, and its projection onto the axis z exactly coincides with the projection on the same axis of the moment of force Frelative to point ABOUT... In other words,

M Oz(F)=M Oz(F hu)= xF y -yF x. (3.11)

Obviously, the same result can be obtained if we design the force Fto any other plane parallel hoy... In this case, the point of intersection of the axis z the plane will be different (we denote the new intersection point by ABOUT 1). However, all quantities included in the right-hand side of equality (3.11) x, at, F x, F ywill remain unchanged, and, therefore, we can write

M Oz(F) \u003d M O 1 z ( F hu).

In other words, the projection of the moment of force relative to a point onto an axis passing through this point does not depend on the choice of a point on the axis ... Therefore, in what follows, instead of the symbol M Oz(F) we will use the symbol M z(F). This projection of the moment is called moment of force about the axis z... Calculating the moment of force about an axis is often more convenient to do by projecting the force F on a plane perpendicular to the axis, and calculating the quantity M z(F hu).

In accordance with formula (3.7) and taking into account the sign of the projection, we get:

M z(F)=M z(F hu)=± F xy h *. (3.12)

Here h * - shoulder strength F hu relative to point ABOUT... If the observer sees from the positive direction of the z-axis that the force F hu seeks to rotate the body around an axis z counterclockwise, the "+" sign is taken, and otherwise the "-" sign.

Formula (3.12) makes it possible to formulate the following rule for calculating the moment of force about the axis. For this you need:

· Select an arbitrary point on the axis and build a plane perpendicular to the axis;

· Project force on this plane;

· Determine the shoulder of the projection force h *.

The moment of force relative to the axis is equal to the product of the modulus of the projection of the force on its shoulder, taken with the corresponding sign (see the above rule).

It follows from formula (3.12) that the moment of force about the axis is zero in two cases:

When the projection of the force on the plane perpendicular to the axis is zero, i.e. when force and axis are parallel ;

When the shoulder is projection h * is equal to zero, i.e. when the line of action crosses the axis .

Both of these cases can be combined into one: the moment of force about the axis is zero if and only if the line of action of the force and the axis are in the same plane .

Task 3.1.Calculate relative to a point ABOUT moment of power Fapplied to the point AND and diagonally directed cube face with side and.

When solving such problems, it is advisable to first calculate the moments of force Frelative to the coordinate axes x, y, z... Point coordinates AND force application F will be

Force projections Fon the coordinate axes:

Substituting these values \u200b\u200binto equalities (3.10), we find

, , .

The same expressions for the moments of force Frelative to the coordinate axes can be obtained using the formula (3.12). To do this, design the force F on a plane perpendicular to the axis x and at... It's obvious that ... Applying the above rule, we get, as one would expect, the same expressions:

, , .

The torque modulus is determined by the equality

.

Let us now introduce the concept of the moment of a pair. Let us first find what is the sum of the moments of forces that make up a pair, relative to an arbitrary point. Let be ABOUT Is an arbitrary point in space, and F and F "-the forces that make up the pair.

Then M o (F) \u003d OA × F, M o (F ") \u003d OV × F ",

M about (F) + M about (F ") \u003d OA × F+ OV × F ",

but since F \u003d -F "then

M about (F) + M about (F ") \u003d OA × F- OV × F=(OA-OV)× F.

Considering the equality OA-OV \u003d VA , we finally find:

M about (F) + M about (F ") \u003d VA × F.

Hence, the sum of the moments of forces that make up a pair does not depend on the position of the point relative to which the moments are taken .

Vector product VA × Fand called moment of couple ... The moment of the pair is indicated by the symbol M (F, F "), and

M (F, F ")= VA × F \u003d AB × F ",

or, in short,

M= VA × F \u003d AB × F ". (3.13)

Considering the right-hand side of this equality, we notice that the moment of a pair is a vector perpendicular to the plane of the pair, equal in magnitude to the product of the modulus of one force of the pair on the shoulder of the pair (i.e., the shortest distance between the lines of action of the forces that make up the pair) and directed in the direction from which the pair's "rotation" is seen to be happening counterclockwise ... If a h - pair's shoulder, then M (F, F ")=h × F.

From the definition itself, it is clear that the moment of a pair of forces is a free vector, the line of action of which is not defined (an additional justification of this remark follows from Theorems 2 and 3 of this chapter).

In order for a pair of forces to make up a balanced system (a system of forces equivalent to zero), it is necessary and sufficient that the moment of the pair is equal to zero. Indeed, if the moment of the pair is zero, M=h × Fthen either F\u003d 0, i.e. no strength, or a pair's shoulder h is equal to zero. But in this case, the forces of the pair will act in one straight line; since they are equal in absolute value and directed in opposite directions, then on the basis of axiom 1 they will form a balanced system. Conversely if two forces F 1and F 2that make up a pair are balanced, then on the basis of the same Axiom 1 they act along one straight line. But in this case, the shoulder of the pair h is equal to zero and, therefore, M=h × F=0.

Pair theorems

Let us prove three theorems by means of which equivalent transformations of pairs become possible. In all considerations, it should be remembered that they refer to vapors acting on any one solid body.

Theorem 1. Two pairs lying in the same plane can be replaced by one pair lying in the same plane, with a moment equal to the sum of the moments of these two pairs.

To prove this theorem, consider two pairs ( F 1,F "1) and ( F 2,F "2) and transfer the points of application of all forces along the lines of their action to the points AND and IN respectively. Adding forces according to axiom 3, we get

R \u003d F 1+F 2 and R "\u003d F" 1+F "2,

but F 1=-F "1 and F 2=-F "2.

Hence, R \u003d - R ", i.e. strength Rand R " form a pair. We find the moment of this pair using formula (3.13):

M \u003d M(R, R ")=VA × R \u003d VA × (F 1+F 2)=VA ×F 1+VA ×F 2. (3.14)

When the forces that make up a pair are transferred along the lines of their action, neither the shoulder nor the direction of rotation of the pairs change; therefore, the moment of the pair does not change either. Hence,

VA × F 1 \u003d M(F 1,F "1)=M 1, VA ×F 2 \u003d M(F 2,F "2)=M 2

and formula (3.14) takes the form

M \u003d M 1 + M 2, (3.15)

which proves the validity of the theorem stated above.

We make two remarks on this theorem.

1. The lines of action of the forces that make up the pair may turn out to be parallel. The theorem remains valid in this case, but to prove it, one should use the rule of addition of parallel forces.

2. After addition, it may turn out that M(R, R ") \u003d 0; Based on the remark made earlier, it follows from this that the set of two pairs ( F 1,F "1, F 2,F "2)=0.

Theorem 2. Two pairs with geometrically equal moments are equivalent.

Let the body in the plane I the pair ( F 1,F "1) with the moment M 1... Let us show that this pair can be replaced by another with a pair ( F 2,F "2) located in the plane IIif only her moment M 2is equal M 1(according to the definition (see 1.1) this will mean that the pairs ( F 1,F "1) and ( F 2,F "2) are equivalent). First of all, note that the planes I and II must be parallel, in particular they can be the same. Indeed, from the parallelism of the moments M 1and M 2(in our case M 1=M 2) it follows that the planes of action of pairs perpendicular to the moments are also parallel.

Let us introduce a new pair ( F 3,F "3) and apply it together with the pair ( F 2,F "2) to the body, placing both pairs in the plane II... To do this, according to axiom 2, you need to choose a pair ( F 3,F "3) with the moment M 3 so that the applied system of forces ( F 2,F "2, F 3,F "3) was balanced. This can be done, for example, as follows: put F 3=-F "1 and F "3 \u003d-F 1and combine the points of application of these forces with the projections AND 1 and IN 1 points AND and IN on the plane II... In accordance with the construction, we will have: M 3 \u003d -M 1or, given that M 1 \u003d M 2,

M 2 + M 3 \u003d0.

Taking into account the second remark to the previous theorem, we obtain ( F 2,F "2, F 3,F "3) \u003d 0. Thus, pairs ( F 2,F "2) and ( F 3,F "3) are mutually balanced and their attachment to the body does not violate its state (axiom 2), so that

(F 1,F "1)= (F 1,F "1, F 2,F "2, F 3,F "3). (3.16)

On the other hand, strength F 1 and F 3, and F "1 and F "3 can be folded according to the rule of addition of parallel forces directed in one direction. In modulus, all these forces are equal to each other, so their resultant R and R " must be attached at the intersection of the diagonals of the rectangle ABB 1 AND 1 ; moreover, they are equal in magnitude and directed in opposite directions. This means that they constitute a system equivalent to zero. So,

(F 1,F "1, F 3,F "3)=(R, R ")=0.

Now we can write

(F 1,F "1, F 2,F "2, F 3,F "3)=(F 3,F "3). (3.17)

Comparing relations (3.16) and (3.17), we obtain ( F 1,F "1)=(F 2,F "2), as required.

From this theorem it follows that a pair of forces can be moved in the plane of its action, transferred to a parallel plane; finally, in a pair it is possible to change simultaneously the forces and the shoulder, keeping only the direction of rotation of the pair and the modulus of its moment ( F 1 h 1 = F 2 h 2).

In what follows, we will make extensive use of such equivalent transformations of a pair.

Theorem 3. Two pairs lying in intersecting planes are equivalent to one pair, the moment of which is equal to the sum of the moments of the two given pairs.

Let pairs ( F 1,F "1) and ( F 2,F "2) are located in intersecting planes I and II respectively. Using the corollary of Theorem 2, we bring both pairs to the shoulder ABlocated on the line of intersection of the planes I and II... We denote the transformed pairs by ( Q 1,Q "1) and ( Q 2,Q "2). In this case, the equalities must be satisfied

M 1 \u003d M(Q 1,Q "1)=M(F 1,F "1) and M 2 \u003d M(Q 2,Q "2)=M(F 2,F "2).

Let us sum the forces applied at the points according to axiom 3 AND and IN respectively. Then we get R \u003d Q 1 + Q 2and R "\u003d Q" 1 + Q "2... Considering that Q "1 \u003d -Q 1and Q "2 \u003d -Q 2, we get R \u003d -R "... Thus, we have proved that a system of two pairs is equivalent to one pair ( R,R ").

Find the moment Mthis pair. Based on formula (3.13), we have

M(R,R ")=VA × (Q 1 + Q 2)=VA ×Q 1 + VA ×Q 2=

=M(Q 1,Q "1)+M(Q 2,Q "2)=M(F 1,F "1)+M(F 2,F "2)

M \u003d M 1 + M 2,

those. the theorem is proved.

Note that the result obtained is also valid for pairs lying in parallel planes. By Theorem 2, such pairs can be reduced to one plane, and by Theorem 1 they can be replaced by one pair, the moment of which is equal to the sum of the moments of the constituent pairs.

The pair theorems proved above allow us to draw an important conclusion: the moment of a pair is a free vector and completely determines the action of a pair on an absolutely rigid body ... Indeed, we have already proved that if two pairs have the same moments (hence, lie in the same plane or in parallel planes), then they are equivalent to each other (Theorem 2). On the other hand, two pairs lying in intersecting planes cannot be equivalent, because this would mean that one of them and the pair opposite to the other are equivalent to zero, which is impossible, since the sum of the moments of such pairs is nonzero.

Thus, the introduced concept of the moment of a pair is extremely useful, since it fully reflects the mechanical action of a pair on a body. In this sense, we can say that the moment in an exhaustive way represents the action of a pair on a solid.

For deformable bodies, the above theory of pairs is inapplicable. Two opposite pairs, acting, for example, along the ends of a bar, are equivalent to zero from the point of view of the statics of a rigid body. Meanwhile, their action on a deformable rod causes its torsion, and the greater, the greater the modules of moments.

Let's move on to solving the first and second problems of statics, when only pairs of forces act on the body.

In this lesson, the topic of which is "Moment of Power", we will talk about the force with which you need to act on the body to change its speed, as well as the point of application of this force. Consider examples of the rotation of different bodies, for example, a swing: at which point you need to act with force so that the swing starts moving or remains in balance.

Imagine that you are a football player and there is a soccer ball in front of you. To make it fly, you need to hit it. It's simple: the harder you hit, the faster and farther it will fly, and you will most likely hit the center of the ball (see Fig. 1).

And so that the ball rotates in flight and flies along a curved trajectory, you will not hit the center of the ball, but from the side, which is what the players do to deceive the opponent (see Fig. 2).

Figure: 2. Curve trajectory of the ball

Here it is already important which point to hit.

Another simple question: where do you need to take the stick so that it does not overturn when lifting? If the stick is uniform in thickness and density, then we will take it in the middle. And if it is more massive on one edge? Then we will take it closer to the massive edge, otherwise it will outweigh (see Fig. 3).

Figure: 3. Lifting point

Imagine: dad sat down on a swing-balancer (see fig. 4).

Figure: 4. Swing-balancer

To outweigh it, you sit on the swing closer to the opposite end.

In all the examples given, it was important for us not only to act on the body with some force, but also important in what place, on which point of the body to act. We chose this point at random, using life experience. What if there are three different weights on the stick? And if you lift it together? And if we are talking about a crane or cable-stayed bridge (see Fig. 5)?

Figure: 5. Examples from life

To solve such problems, intuition and experience are not enough. They cannot be solved without a clear theory. Today we will talk about solving such problems.

Usually in tasks we have a body to which forces are applied, and we solve them, as always before, without thinking about the point of application of the force. It is enough to know that the force is applied simply to the body. Such tasks are encountered often, we know how to solve them, but it happens that it is not enough to apply force simply to the body - it becomes important at what point.

An example of a problem in which body size is not important

For example, there is a small iron ball on the table, which is subject to a gravity force of 1 N. What force must be applied to lift it? The ball is attracted by the Earth, we will act upward on it, applying some force.

The forces acting on the ball are directed in opposite directions, and in order to lift the ball, you need to act on it with a force greater in magnitude than the force of gravity (see Fig. 6).

Figure: 6. Forces acting on the ball

The force of gravity is equal, which means that the ball must be acted upon upward with the force:

We didn't think exactly how we take the ball, we just pick it up and pick it up. When we show how we raised the ball, we may well draw a point and show: we acted on the ball (see Fig. 7).

Figure: 7. Action on the ball

When we can do this with a body, show it in a drawing when explained in the form of a point and not pay attention to its size and shape, we consider it a material point. This is a model. In reality, the ball has a shape and size, but we did not pay attention to them in this task. If the same ball needs to be made to rotate, then it is no longer possible to simply say that we are acting on the ball. It is important here that we pushed the ball from the edge and not to the center, making it spin. In this problem, the same ball can no longer be considered a point.

We already know examples of problems in which the point of application of force must be taken into account: the problem with a soccer ball, with a non-uniform stick, with a swing.

The point of application of the force is also important in the case of a lever. Using a shovel, we act on the end of the cutting. Then it is enough to apply a small force (see fig. 8).

Figure: 8. The action of low force on the handle of the shovel

What is common between the examples considered, where it is important for us to take into account body size? And the ball, and the stick, and the swing, and the shovel - in all these cases, we were talking about the rotation of these bodies around some axis. The ball rotated around its axis, the swing rotated around the mount, the stick around the place where we held it, the shovel around the fulcrum (see Fig. 9).

Figure: 9. Examples of rotating bodies

Consider the rotation of bodies around a fixed axis and see what makes the body rotate. We will consider rotation in one plane, then we can assume that the body rotates around one point O (see Fig. 10).

Figure: 10. Point of rotation

If we want to balance the swing, in which the beam is glass and thin, then it can simply break, and if the beam is made of soft metal and is also thin, then it can bend (see Fig. 11).

We will not consider such cases; we will consider the rotation of strong rigid bodies.

It would be wrong to say that rotational motion is determined only by force. Indeed, on a swing, the same force can cause them to rotate, or maybe not, depending on where we sit. It is not only about strength, but also about the location of the point on which we act. Everyone knows how difficult it is to lift and hold a load on an outstretched arm. To determine the point of application of force, the concept of the shoulder of force is introduced (by analogy with the shoulder of the arm, which is used to lift the load).

The shoulder of force is the minimum distance from a given point to a straight line along which the force acts.

From geometry, you probably already know that this is a perpendicular dropped from point O to a straight line along which a force acts (see Fig. 12).

Figure: 12. Graphical representation of shoulder strength

Why is the shoulder of force the minimum distance from point O to a straight line along which the force acts

It may seem strange that the shoulder of the force is measured from point O not to the point of application of the force, but to the straight line along which this force acts.

Let's do this experiment: tie a thread to the lever. Let's act on the lever with some force at the point where the thread is tied (see fig. 13).

Figure: 13. The thread is tied to the lever

If enough moment of force is created to turn the lever, it will turn. The thread will show a straight line along which the force is directed (see Fig. 14).

Let's try to pull the lever with the same force, but now grabbing the thread. Nothing will change in the impact on the lever, although the point of application of the force will change. But the force will act along the same straight line, its distance to the axis of rotation, that is, the shoulder of the force, will remain the same. Let's try to act on the lever at an angle (see fig. 15).

Figure: 15. Action on the lever at an angle

Now the force is applied to the same point, but acts along a different line. Its distance to the axis of rotation has become small, the moment of force has decreased, and the lever may no longer turn.

The body is influenced by rotation, the rotation of the body. This effect depends on the strength and on her shoulder. The quantity characterizing the rotational effect of the force on the body is called moment of power, sometimes it is also called torque or torque.

The meaning of the word "moment"

We are accustomed to using the word "moment" in the meaning of a very short period of time, as a synonym for the word "moment" or "moment". Then it is not entirely clear what relation moment has to power. Let's turn to the origin of the word "moment".

The word comes from the Latin momentum, which means "driving force, push." The Latin verb movēre means "to move" (like the English word move, and movement means "movement"). It is now clear to us that the torque is what makes the body rotate.

A moment of force is the product of force on her shoulder.

The unit of measurement is the newton multiplied by the meter:.

If you increase the shoulder of force, you can decrease the force and the moment of force will remain the same. We use it very often in our daily life: when we open the door, when we use pliers or a wrench.

The last point of our model remains - we need to figure out what to do if several forces act on the body. We can calculate the moment of each force. It is clear that if the forces rotate the body in one direction, then their action will add up (see Fig. 16).

Figure: 16. The action of the forces adds up

If in different directions - the moments of forces will balance each other and it is logical that they will need to be subtracted. Therefore, the moments of forces that rotate the body in different directions will be written with different signs. For example, let's write down if the force supposedly rotates the body around the axis clockwise, and - if counterclockwise (see Fig. 17).

Figure: 17. Definition of signs

Then we can write down one important thing: for the body to be in equilibrium, the sum of the moments of the forces acting on it must be equal to zero.

Lever formula

We already know the principle of the lever: two forces act on the lever, and how many times the lever arm is, the force is less so many times:

Consider the moments of forces that act on the lever.

Choose a positive direction of rotation of the lever, for example counterclockwise (see Fig. 18).

Figure: 18. Selecting the direction of rotation

Then the moment of force will be with a plus sign, and the moment of force will be with a minus sign. For the lever to be in balance, the sum of the moments of forces must be equal to zero. Let's write:

Mathematically, this equality and the ratio written above for the lever are one and the same, and what we obtained experimentally was confirmed.

For example, determine whether the lever shown in the figure will be in equilibrium. Three forces act on him(see fig. 19) . , and... Shoulders of forces are equal, and.

Figure: 19. Figure for the condition of problem 1

For the lever to be in equilibrium, the sum of the moments of the forces that act on it must be equal to zero.

By condition, the lever is acted upon by three forces:, and. Their shoulders are respectively equal, and.

The direction of rotation of the lever clockwise will be considered positive. In this direction, the lever rotates the force, its moment is equal to:

Forces and rotate the lever counterclockwise, we write down their moments with a minus sign:

It remains to calculate the sum of the moments of forces:

The total moment is not equal to zero, which means that the body will not be in equilibrium. The total moment is positive, which means that the lever will rotate clockwise (in our problem, this is a positive direction).

We solved the problem and got the result: the total moment of forces acting on the lever is equal to. The lever will begin to turn. And when it turns, if the forces do not change direction, the shoulders of the forces will change. They will decrease until they are equal to zero when the lever is rotated vertically (see Fig. 20).

Figure: 20. Shoulders forces are equal to zero

And with further rotation, the forces will be directed so as to rotate it in the opposite direction. Therefore, having solved the problem, we determined in which direction the lever would start rotating, not to mention what would happen next.

Now you have learned to determine not only the force with which you need to act on the body in order to change its speed, but also the point of application of this force so that it does not turn (or turn, as we need).

How to push a cabinet so that it does not turn over?

We know that when we push the cabinet with force at the top, it flips over, and to prevent this from happening, we push it lower. Now we can explain this phenomenon. The axis of its rotation is located on the edge on which it stands, while the shoulders of all forces, except for the force, are either small or equal to zero, so the cabinet falls under the action of the force (see Fig. 21).

Figure: 21. Action on the top of the cabinet

By applying the force below, we reduce its shoulder, which means that both the moment of this force and overturning does not occur (see Fig. 22).

Figure: 22. Force applied below

The wardrobe, as a body, the dimensions of which we take into account, obeys the same law as a wrench, a doorknob, bridges on supports, etc.

This concludes our lesson. Thanks for your attention!

List of references

1. Sokolovich Yu.A., Bogdanova G.S. Physics: Handbook with examples of problem solving. - 2nd edition redistribution. - X .: Vesta: Ranok Publishing House, 2005. - 464 p.
2. Peryshkin A.V. Physics. 7th grade: textbook. for general education. institutions - 10th ed., add. - M .: Bustard, 2006 .-- 192 p .: ill.

1. Abitura.com ().
2. Solverbook.com ().

Homework

The best definition of torque is the tendency of a force to rotate an object about an axis, fulcrum, or pivot point. The torque can be calculated using the force and the shoulder of the moment (the perpendicular distance from the axis to the line of action of the force), or using the moment of inertia and angular acceleration.

Steps

Using force and shoulder moment

1. Determine the forces acting on the body and the corresponding moments. If the force is not perpendicular to the moment arm in question (i.e. it acts at an angle), then you may need to find its components using trigonometric functions such as sine or cosine.

   • The force component considered will depend on the equivalent perpendicular force.
   • Imagine a horizontal bar to which you need to apply a force of 10 N at an angle of 30 ° above the horizontal plane in order to rotate it around the center.
   • Since you need to use a force that is not perpendicular to the shoulder of the moment, you need the vertical component of the force to rotate the rod.
   • Therefore, you need to consider the y-component, or use F \u003d 10sin30 ° H.

2. Use the torque equation, τ \u003d Fr, and simply replace the variables with given or received data.

   • A simple example: Imagine a 30kg child sitting on one end of a board swing. The length of one side of the swing is 1.5 m.
   • Since the swing's axis of rotation is in the center, you don't need to multiply the length.
   • You need to determine the force exerted by the child using mass and acceleration.
   • Since the mass is given, you need to multiply it by the acceleration due to gravity, g, which is 9.81 m / s 2. Hence:
   • Now you have all the data you need to use the equation of moment:

3. Use the signs (plus or minus) to show the direction of the moment. If the force rotates the body clockwise, then the moment is negative. If the force rotates the body counterclockwise, then the moment is positive.

   • In case of multiple applied forces, simply add up all the moments in the body.
   • Since each force tends to induce different directions of rotation, it is important to use the rotation sign in order to keep track of the direction of action of each force.
   • For example, two forces were applied to the rim of a wheel having a diameter of 0.050 m, F 1 \u003d 10.0 N in a clockwise direction and F 2 \u003d 9.0 N in a counterclockwise direction.
   • Since this body is a circle, the fixed axis is its center. You need to divide the diameter and get the radius. The size of the radius will serve as the shoulder of the moment. Therefore, the radius is 0.025 m.
   • For clarity, we can solve separate equations for each of the moments arising from the corresponding force.
   • For force 1, the action is directed clockwise, therefore, the moment it creates is negative:
   • For force 2, the action is directed counterclockwise, therefore, the moment it creates is positive:
   • Now we can add all the moments to get the resulting torque:

   Using moment of inertia and angular acceleration

   1. To start solving the problem, understand how the moment of inertia of the body works. The moment of inertia of a body is the body's resistance to rotational motion. The moment of inertia depends both on the mass and on the nature of its distribution.

      • To clearly understand this, imagine two cylinders of the same diameter, but different mass.
      • Imagine that you need to rotate both cylinders around their central axis.
      • Obviously, a cylinder with a higher mass will be more difficult to rotate than another cylinder because it is “heavier”.
      • Now imagine two cylinders of different diameters, but the same mass. To look cylindrical and have different masses, but at the same time have different diameters, the shape, or mass distribution of both cylinders must be different.
      • A cylinder with a larger diameter will look like a flat, rounded plate, while a smaller cylinder will look like a one-piece fabric tube.
      • A cylinder with a larger diameter will be more difficult to rotate as you need to apply more force to overcome the longer arm of the moment.

   2. Select the equation you will use to calculate the moment of inertia. There are several equations that can be used for this.

      • The first equation is the simplest: the sum of the masses and moment arms of all particles.
      • This equation is used for material points, or particles. An ideal particle is a body that has mass, but does not occupy space.
      • In other words, the only significant characteristic of this body is mass; you don't need to know its size, shape, or structure.
      • The idea of \u200b\u200ba material particle is widely used in physics with the aim of simplifying calculations and using ideal and theoretical schemes.
      • Now imagine an object like a hollow cylinder or a solid uniform sphere. These objects have a clear and definite shape, size and structure.
      • Therefore, you cannot consider them as a material point.
      • Fortunately, you can use formulas that apply to some common objects:

   3. Find the moment of inertia. To start calculating the torque, you need to find the moment of inertia. Use the following example as a guide:

      • Two small “weights” weighing 5.0 kg and 7.0 kg are installed at a distance of 4.0 m from each other on a light rod (whose mass can be neglected). The axis of rotation is in the middle of the bar. The rod is unwound from rest to an angular velocity of 30.0 rad / s in 3.00 s. Calculate the torque produced.
      • Since the axis of rotation is in the middle of the rod, the moment arm of both weights is equal to half of its length, i.e. 2.0 m.
      • Since the shape, size and structure of the “cargo” is not specified, we can assume that the cargo is material particles.
      • The moment of inertia can be calculated as follows:

   4. Find the angular acceleration, α. To calculate the angular acceleration, you can use the formula α \u003d at / r.

      • The first formula, α \u003d at / r, can be used if tangential acceleration and radius are given.
      • Tangential acceleration is acceleration tangential to the direction of travel.
      • Imagine an object moving along a curved path. Tangential acceleration is simply its linear acceleration at any point along the path.
      • In the case of the second formula, it is easiest to illustrate it by connecting it with concepts from kinematics: displacement, linear velocity and linear acceleration.
      • Displacement is the distance traveled by an object (SI unit - meters, m); linear velocity is an indicator of the change in displacement per unit of time (SI unit - m / s); linear acceleration is a measure of the change in linear speed per unit of time (SI unit - m / s 2).
      • Now let's consider analogs of these quantities during rotational motion: angular displacement, θ is the angle of rotation of a certain point or segment (SI unit is rad); angular velocity, ω - change in angular displacement per unit of time (SI unit - rad / s); and angular acceleration, α is the change in angular velocity per unit of time (SI unit - rad / s 2).
      • Returning to our example - we were given data for angular momentum and time. Since the rotation started from rest, the initial angular velocity is 0. We can use the equation to find:
   5. If you find it difficult to imagine how rotation occurs, then take a pen and try to recreate the problem. For more accurate reproduction, do not forget to copy the position of the axis of rotation and the direction of the applied force.

The moment of force about the axis or simply the moment of force is called the projection of the force onto a straight line that is perpendicular to the radius and drawn at the point of application of the force multiplied by the distance from this point to the axis. Or the product of force on the shoulder of its application. The shoulder in this case is the distance from the axis to the point of application of the force. The moment of force characterizes the rotational action of the force on the body. The axis in this case is the place of attachment of the body, relative to which it can rotate. If the body is not fixed, then the center of mass can be considered the axis of rotation.

Formula 1 - Moment of Power.

F - Force acting on the body.

r - Shoulder strength.

Figure 1 - Moment of force.

As can be seen from the figure, the shoulder of the force is the distance from the axis to the point of application of the force. But this is if the angle between them is 90 degrees. If this is not the case, then it is necessary to draw a line along the action of the force and lower a perpendicular to it from the axis. The length of this perpendicular will be equal to the shoulder of the force. And moving the point of application of the force along the direction of the force does not change its moment.

It is considered to be a positive moment of force that causes the body to turn clockwise relative to the observation point. And negative, respectively, causing rotation against it. The moment of force is measured in Newtons per meter. One Newtonometer is a force of 1 Newton acting on a shoulder of 1 meter.

If the force acting on the body passes along a line passing through the axis of rotation of the body, or the center of mass, if the body does not have an axis of rotation. The moment of force in this case will be equal to zero. Since this force will not cause the body to rotate, but will simply move it translationally along the line of application.

Picture 2 - The moment of force is equal to zero.

If several forces act on the body, then the moment of force will be determined by their resultant. For example, two forces, equal in magnitude and directed oppositely, can act on the body. In this case, the total moment of force will be equal to zero. Since these forces will compensate each other. To put it simply, imagine a children's carousel. If one boy pushes it clockwise, and another with the same force against, then the carousel will remain motionless.