When solving problems of moving objects, in some cases their spatial dimensions are neglected, introducing the concept of a material point. For another type of problems, in which bodies at rest or rotating bodies are considered, it is important to know their parameters and the points of application of external forces. In this case, we are talking about the moment of forces about the axis of rotation. Let's consider this issue in the article.
The concept of the moment of force
Before bringing about a fixed axis of rotation, it is necessary to clarify what phenomenon will be discussed. Below is a figure that shows a wrench of length d, a force F is applied to its end. It is easy to imagine that the result of its action will be the rotation of the wrench counterclockwise and unscrewing the nut.
According to the definition, the moment of force about the axis of rotation is the product of the shoulder (d in this case) and the force (F), that is, the following expression can be written: M = d * F. It should immediately be noted that the above formula is written in scalar form, that is, it allows you to calculate the absolute value of the moment M. As can be seen from the formula, the unit of measurement of the quantity under consideration is newtons per meter (N * m).
- vector quantity
As discussed above, the moment M is actually a vector. To clarify this statement, consider another figure.
Here we see a lever of length L, which is fixed on the axis (shown by the arrow). A force F is applied to its end at an angle Φ. It is not difficult to imagine that this force will cause the lever to rise. The formula for the moment in vector form in this case will be written as follows: M¯ = L¯*F¯, here the line over the symbol means that the quantity in question is a vector. It should be clarified that L¯ is directed from to the point of application of the force F¯.
The above expression is a vector product. Its resulting vector (M¯) will be perpendicular to the plane formed by L¯ and F¯. To determine the direction of the moment M¯, there are several rules ( right hand, gimlet). In order not to memorize them and not get confused in the order of multiplication of the vectors L¯ and F¯ (the direction of M¯ depends on it), you should remember one simple thing: the moment of force will be directed in such a way that if you look from the end of its vector, then the acting force F ¯ will rotate the lever counterclockwise. This direction of the moment is conditionally taken as positive. If the system rotates clockwise, then the resulting moment of forces has a negative value.
Thus, in the considered case with the lever L, the value of M¯ is directed upwards (from the figure to the reader).
In scalar form, the formula for the moment is written as: M = L*F*sin(180-Φ) or M = L*F*sin(Φ) (sin(180-Φ) = sin(Φ)). According to the definition of the sine, we can write the equality: M = d*F, where d = L*sin(Φ) (see the figure and the corresponding right triangle). The last formula is similar to the one given in the previous paragraph.
The above calculations demonstrate how to work with vector and scalar quantities of moments of forces in order to avoid errors.
The physical meaning of M¯
Since the two cases considered in the previous paragraphs are associated with rotational motion, one can guess what meaning the moment of force carries. If the force acting on a material point is a measure of the increase in the speed of the linear displacement of the latter, then the moment of force is a measure of its rotational ability in relation to the system under consideration.
Let's bring good example. Any person opens the door by holding its handle. This can also be done by pushing the door in the area of the handle. Why doesn't anyone open it by pushing in the hinge area? Very simple: the closer the force is applied to the hinges, the more difficult it is to open the door, and vice versa. The derivation of the previous sentence follows from the formula for the moment (M = d*F), which shows that for M = const, the values of d and F are in inverse relationship.
Moment of force - additive quantity
In all the cases considered above, there was only one acting force. When deciding real tasks the matter is much more complicated. Usually systems that rotate or are in equilibrium are subject to several torsional forces, each of which creates its own moment. In this case, the solution of problems is reduced to finding the total moment of forces relative to the axis of rotation.
The total moment is found by the usual sum of the individual moments for each force, however, remember to use the correct sign for each of them.
Problem solution example
To consolidate the acquired knowledge, it is proposed to solve the following problem: it is necessary to calculate the total moment of force for the system shown in the figure below.
We see that three forces (F1, F2, F3) act on a lever 7 m long, and they have different points applications relative to the axis of rotation. Since the direction of forces is perpendicular to the lever, there is no need to use a vector expression for the moment of torsion. It is possible to calculate the total moment M using the scalar formula and remembering the statement desired sign. Since the forces F1 and F3 tend to turn the lever counterclockwise, and F2 - clockwise, the moment of rotation for the first will be positive, and for the second - negative. We have: M \u003d F1 * 7-F2 * 5 + F3 * 3 \u003d 140-50 + 75 \u003d 165 N * m. That is, the total moment is positive and directed upwards (at the reader).
The moment of force about the axis of rotation is called physical quantity equal to the product of the force on its shoulder.
The moment of force is determined by the formula:
M - FI, where F is the force, I is the arm of the force.
The shoulder of the force is the shortest distance from the line of action of the force to the axis of rotation of the body.
On fig. 1.33, a shows a rigid body that can rotate around an axis. The axis of rotation of this body is perpendicular to the plane of the figure and passes through the point denoted by the letter O. The shoulder of the force F here is the distance 1X from the axis of rotation to the line of action of the force. Find it in the following way. First draw the line of action of the force. Then, from the point O, through which the axis of rotation of the body passes, a perpendicular is lowered to the line of action of the force. The length of this perpendicular is the arm of the given force.
The moment of force characterizes the rotating action of the force. This action depends on both strength and leverage. The larger the shoulder, the less force must be applied in order to obtain the desired result, i.e., the same moment of force (see (1.33)). That is why it is much more difficult to open the door by pushing it near the hinges than by holding the handle, and it is much easier to unscrew the nut with a long wrench than with a short wrench.
The unit of moment of force in SI is taken to be a moment of force of 1 N, the arm of which is 1 m - a newton meter (N m).
moment rule
A rigid body capable of rotating around a fixed axis is in equilibrium if the moment of force M, which rotates it clockwise, is equal to the moment of force M2, which rotates it counterclockwise:
M1 \u003d -M2 or F 1 ll \u003d - F 2 l 2.
The rule of moments is a consequence of one of the theorems of mechanics, formulated by the French scientist P. Varignon in 1687.
If two equal and oppositely directed forces that do not lie on one straight line act on a body, then such a body is not in equilibrium, since the resulting moment of these forces relative to any axis is not equal to zero, since both forces have moments directed in the same direction . Two such forces simultaneously acting on a body are called a pair of forces. If the body is fixed on an axis, then under the action of a pair of forces it will rotate. If a pair of forces is applied to a free body, then it will rotate around an axis passing through the center of gravity of the body, Fig. 1.33, b.
The moment of a pair of forces is the same about any axis perpendicular to the plane of the pair. The total moment M of a pair is always equal to the product of one of the forces F and the distance I between the forces, which is called the arm of the pair, regardless of the segments and /2 into which the position of the axis of the arm of the pair is divided:
M = Fll + Fl2=F(l1 + l2) = Fl.
The moment of several forces, the resultant of which is equal to zero, will be the same with respect to all axes parallel to each other, so the action of all these forces on the body can be replaced by the action of one pair of forces with the same moment.
Moment of power (synonyms: torque, torque, torque, 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 force application by the vector of this force. Characterizes the rotational action of force on a rigid body.
The concepts of "rotating" and "torque" moments are generally not identical, since in technology the concept of "rotating" moment is considered as an external force applied to an object, and "torque" is an internal force that occurs in an object under the action of applied loads (this the concept is used in the resistance of materials).
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General information
Special occasions
Lever Moment Formula
A very interesting special case is presented as the definition of the moment of force in the field:
| M → | = | M → 1 | | F → | (\displaystyle \left|(\vec (M))\right|=\left|(\vec (M))_(1)\right|\left|(\vec (F))\right|), where: | M → 1 | (\displaystyle \left|(\vec (M))_(1)\right|)- moment of the lever, | F → | (\displaystyle \left|(\vec (F))\right|)- the magnitude of the acting force.The problem with this representation 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 r → (\displaystyle (\vec (r))), the moment of the lever will be equal to the distance to the center and the moment of force will be maximum:
| T → | = | r → | | F → | (\displaystyle \left|(\vec (T))\right|=\left|(\vec (r))\right|\left|(\vec (F))\right|)Force at an angle
If strength F → (\displaystyle (\vec (F))) directed at an angle θ (\displaystyle \theta ) to lever r, then M = r F sin θ (\displaystyle M=rF\sin \theta ).
Static balance
In order for an 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=0, ΣV=0 and the moment of force in the third dimension ΣM=0.
Moment of force as a function of time
M → = d L → d t (\displaystyle (\vec (M))=(\frac (d(\vec (L)))(dt))),
where L → (\displaystyle (\vec (L)))- angular momentum.
Let's take a rigid body. The motion of a rigid body can be represented as the motion of a specific point and rotation around it.
The angular momentum relative to the point O of a rigid body can be described through the product of the moment of inertia and the angular speed relative to the center of mass and the linear motion of the center of mass.
L o → = I c ω → + [ M (r o → − r c →) , v c → ] (\displaystyle (\vec (L_(o)))=I_(c)\,(\vec (\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 with respect to time. And if I (\displaystyle I) is a constant in time, then
M → = I d ω → d t = I α → (\displaystyle (\vec (M))=I(\frac (d(\vec (\omega )))(dt))=I(\vec (\alpha ))),where α → (\displaystyle (\vec (\alpha )))- angular acceleration, measured in radians per second per second (rad / s 2). Example: A uniform disk is rotating.
If the inertia tensor changes with time, then the motion about the center of mass is described using the Euler dynamic equation:
M c → = I c d ω → d t + [ w → , I c w → ] (\displaystyle (\vec (M_(c)))=I_(c)(\frac (d(\vec (\omega ))) (dt))+[(\vec (w)),I_(c)(\vec (w))]).
In physics, the consideration of problems with rotating bodies or systems that are in equilibrium is carried out using the concept of "moment of force". This article will consider the formula for the moment of force, as well as its use to solve this type of problem.
in physics
As noted in the introduction, this article will focus on systems that can rotate either around an axis or around a point. Consider an example of such a model, shown in the figure below.
We see that the lever gray color fixed on the axis of rotation. At the end of the lever there is a black cube of some mass, on which a force acts (red arrow). It is intuitively clear that the result of this force will be the rotation of the lever around the axis counterclockwise.
The moment of force is a quantity in physics, which is equal to the vector product of the radius connecting the axis of rotation and the point of application of the force (green vector in the figure), and the external force itself. That is, the force relative to the axis is written as follows:
The result of this product will be the vector M¯. Its direction is determined based on the knowledge of multiplier vectors, that is, r¯ and F¯. According to the definition of the cross product, M¯ must be perpendicular to the plane, formed by vectors r¯ and F¯, and is directed in accordance with the right hand rule (if four fingers of the right hand are placed along the first multiplied vector towards the end of the second, then the thumb set aside will indicate where the desired vector is directed). In the figure, you can see where the vector M¯ is directed (blue arrow).
Scalar notation M¯
In the figure in the previous paragraph, the force (red arrow) acts on the lever at an angle of 90 o. In the general case, it can be applied at absolutely any angle. Consider the image below.
Here we see that the force F is already acting on the lever L at a certain angle Φ. For this system, the formula for the moment of force relative to a point (shown by an arrow) in scalar form takes the form:
M = L * F * sin(Φ)
It follows from the expression that the moment of force M will be the greater, the closer the direction of action of the force F is to the angle of 90 o with respect to L. Conversely, if F acts along L, then sin(0) = 0, and the force does not create any moment ( M = 0).
When considering the moment of force in scalar form, the concept of "lever of force" is often used. This value is the distance between the axis (rotation point) and the vector F. Applying this definition to the figure above, we can say that d = L * sin(Φ) is the lever of force (the equality follows from the definition trigonometric function"sinus"). Through the lever of force, the formula for the moment M can be rewritten as follows:
The physical meaning of the quantity M
The considered physical quantity determines the ability of the external force F to exert a rotational effect on the system. To bring the body into rotational motion, it needs to impart some moment M.
A prime example of this process is opening or closing a door to a room. Holding the handle, the person makes an effort and turns the door on its hinges. Everyone can do it. If you try to open the door by acting on it near the hinges, then you will need to make great efforts to move it.
Another example is loosening a nut with a wrench. The shorter this key is, the more difficult it is to complete the task.
These features are demonstrated by the formula for the moment of force over the shoulder, which was given in the previous paragraph. If M is considered a constant value, then the smaller d, the greater F must be applied to create given moment strength.
Several acting forces in the system
The cases were considered above when only one force F acts on a system capable of rotation, but what if there are several such forces? Indeed, this situation is more frequent, since forces of various nature (gravitational, electrical, friction, mechanical, and others) can act on the system. In all these cases, the resulting moment of force M¯ can be obtained using the vector sum of all moments M i ¯, i.e.:
M¯ = ∑ i (M i ¯), where i is the number of the force F i
An important conclusion follows from the property of the additivity of moments, which is called the Varignon theorem, named after the mathematician of the late 17th and early 18th centuries, the Frenchman Pierre Varignon. It reads: "The sum of the moments of all forces acting on the system under consideration can be represented as a moment of one force, which is equal to the sum of all the others and is applied to a certain point." Mathematically, the theorem can be written as follows:
∑ i (M i ¯) = M¯ = d * ∑ i (F i ¯)
This important theorem is often used in practice to solve problems on the rotation and balance of bodies.
Does the moment of force do work?
Analyzing the above formulas in scalar or vector form, we can conclude that the value of M is some work. Indeed, its dimension is N * m, which in SI corresponds to the joule (J). In fact, the moment of force is not work, but only a quantity that is capable of doing it. For this to happen, it is necessary to have a circular motion in the system and a long-term action M. Therefore, the formula for the work of the moment of force is written as follows:
In this expression, θ is the angle through which the moment of force M was rotated. As a result, the unit of work can be written as N * m * rad or J * rad. For example, a value of 60 J * rad indicates that when rotated by 1 radian (approximately 1/3 of the circle), the force F that creates the moment M did 60 joules of work. This formula is often used when solving problems in systems where friction forces act, which will be shown below.
Moment of force and moment of impulse
As was shown, the action of the moment M on the system leads to the appearance of rotational motion in it. The latter is characterized by a quantity called "momentum". It can be calculated using the formula:
Here I is the moment of inertia (a value that plays the same role during rotation as the mass during the linear motion of the body), ω is the angular velocity, it is related to the linear velocity by the formula ω = v/r.
Both moments (momentum and force) are related to each other by the following expression:
M = I * α, where α = dω / dt is the angular acceleration.
Here is another formula that is important for solving problems for the work of moments of forces. Using this formula, you can calculate the kinetic energy of a rotating body. She looks like this:
Equilibrium of several bodies
The first problem is related to the equilibrium of a system in which several forces act. The figure below shows a system that is subject to three forces. It is necessary to calculate what mass the object must be suspended from this lever and at what point it must be done so that this system is in equilibrium.
From the conditions of the problem, it can be understood that to solve it, one should use the Varignon theorem. The first part of the problem can be answered immediately, since the weight of the object to be hung from the lever will be equal to:
P \u003d F 1 - F 2 + F 3 \u003d 20 - 10 + 25 \u003d 35 N
The signs here are chosen taking into account that the force that rotates the lever counterclockwise creates a negative moment.
The position of the point d, where this weight should be hung, is calculated by the formula:
M 1 - M 2 + M 3 = d * P = 7 * 20 - 5 * 10 + 3 * 25 = d * 35 => d = 165/35 = 4.714 m
Note that using the formula for the moment of gravity, we calculated the equivalent value M of the one created by three forces. In order for the system to be in equilibrium, it is necessary to suspend a body weighing 35 N at a point 4.714 m from the axis on the other side of the lever.
Moving disk problem
The solution of the following problem is based on the use of the formula for the moment of friction force and the kinetic energy of a body of revolution. Task: Given a disk with radius r = 0.3 meters, which rotates at a speed of ω = 1 rad/s. It is necessary to calculate how far it can travel on the surface if the rolling friction coefficient is μ = 0.001.
This problem is easiest to solve using the law of conservation of energy. We have the initial kinetic energy of the disk. When it starts to roll, all this energy is spent on heating the surface due to the action of the friction force. Equating both quantities, we obtain the expression:
I * ω 2 /2 = μ * N/r * r * θ
The first part of the formula is the kinetic energy of the disk. The second part is the work of the moment of the friction force F = μ * N/r applied to the edge of the disk (M=F * r).
Given that N = m * g and I = 1/2m * r 2 , we calculate θ:
θ = m * r 2 * ω 2 / (4 * μ * m * g) = r 2 * ω 2 / (4 * μ * g) = 0.3 2 * 1 2 / (4 * 0.001 * 9.81 ) = 2.29358 rad
Since 2pi radians correspond to a length of 2pi * r, then we get that the required distance that the disk will cover is:
s = θ * r = 2.29358 * 0.3 = 0.688 m or about 69 cm
Note that the mass of the disk does not affect this result.
Moment of force relative to an arbitrary center in the plane of action of the force, the product of the modulus of force and the arm is called.
Shoulder- the shortest distance from the center O to the line of action of the force, but not to the point of application of the force, because force-sliding vector.
Moment sign:
Clockwise-minus, anti-clockwise-plus;
The moment of force can be expressed as a vector. This is a perpendicular to the plane according to Gimlet's rule.
If several forces or a system of forces are located in the plane, then the algebraic sum of their moments will give us main point force systems.
Consider the moment of force about the axis, calculate the moment of force about the Z axis;
Project F onto XY;
F xy =F cosα= ab
m 0 (F xy)=m z (F), i.e. m z =F xy * h= F cosα* h
The moment of force about the axis is equal to the moment of its projection onto a plane perpendicular to the axis, taken at the intersection of the axes and the plane
If the force is parallel to the axis or crosses it, then m z (F)=0
Expression of the moment of force as a vector expression
Draw r a to point A. Consider OA x F.
This is the third vector m o perpendicular to the plane. The cross product modulus can be calculated using twice the area of the shaded triangle.
Analytical expression of force relative to the coordinate axes.
Suppose that the Y and Z, X axes are associated with point O with unit vectors i, j, k Considering that:
r x = X * Fx ; r y = Y * F y ; r z =Z * F y we get: m o (F)=x =
Expand the determinant and get:
m x = YF z - ZF y
m y =ZF x - XF z
m z =XF y - YF x
These formulas make it possible to calculate the projection of the moment vector on the axis, and then the moment vector itself.
Varignon's theorem on the moment of the resultant
If the system of forces has a resultant, then its moment relative to any center is equal to the algebraic sum of the moments of all forces relative to this point
If we apply Q= -R, then the system (Q,F 1 ... F n) will be equally balanced.
The sum of the moments about any center will be equal to zero.
Analytical equilibrium condition for a plane system of forces
This is a flat system of forces, the lines of action of which are located in the same plane.
Purpose of task calculation of this type- determination of reactions of external links. For this, the basic equations in a flat system of forces are used.
2 or 3 moment equations can be used.
Example
Let's make an equation for the sum of all forces on the X and Y axis.
