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". In this article, we will consider the formula for the moment of force, as well as its use for solving the indicated type of problems.
Moment of power in physics
As noted in the introduction, this article will discuss 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 gray lever is fixed on the axis of rotation. At the end of the lever there is a black cube of some mass, on which the force acts (red arrow). Intuitively, the result of this force will be the rotation of the lever around the axis counterclockwise.
The moment of force is the quantity in physics, which is equal to the vector product of the radius connecting the axis of rotation and the point of application of force (the green vector in the figure), and the external force itself. That is, the formula of the moment of force relative to the axis is written as follows:
M¯ = r¯ * F¯
The result of this product is the vector M¯. Its direction is determined on the basis of knowledge of the factor vectors, that is, r¯ and F¯. According to the definition of a vector product, M¯ must be perpendicular to the plane formed by the vectors r¯ and F¯ and directed in accordance with the rule of the right hand (if four fingers of the right hand are placed along the first multiplied vector towards the end of the second, then the thumb is laid up will indicate where the desired vector is directed). In the figure, you can see where the vector M¯ (blue arrow) is directed.
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 any angle. Consider the image below.
Here we see that the force F already acts on the lever L at a certain angle Φ. For this system, the formula of the moment of force relative to the point (shown by the arrow) in a scalar form will take the form:
M = L * F * sin (Φ)
From the expression it follows that the moment of force M will be the greater, the closer the direction of action of the force F to the angle 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, they often use the concept of “lever of force”. This value represents the distance between the axis (the pivot 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 (equality follows from the definition of the trigonometric function "sine"). Through the lever of force, the formula for moment M can be rewritten as follows:
M = d * F
The physical meaning of the quantity M
The physical quantity under consideration determines the ability of an external force F to exert a rotational effect on the system. To bring the body into rotational motion, it needs to be informed of a certain moment M.
A striking example of this process is opening or closing a door to a room. Holding the handle, a man makes an effort and turns the door on the hinges. Everyone can do it. If you try to open the door, acting on it near the hinges, you will need to make great efforts to move it from its place.
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 of the moment of force over the shoulder, which was given in the previous paragraph. If M is assumed to be a constant, then the smaller d, the greater F should be applied to create a given moment of force.
Several forces in the system
The cases above were considered when only one force F acts on a system capable of rotation, but what about when 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 ¯, that is:
M¯ = ∑ i (M i ¯), where i is the number of force F i
An important conclusion follows from the property of additivity of moments, which was called the Varignon theorem, named after the mathematician of the late XVII - early XVIII centuries - the Frenchman Pierre Varignon. It reads: "The sum of the moments of all the forces that affect the system in question can be represented as the moment of one force, which is equal to the sum of all the others and is applied to some 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 of rotation and equilibrium of bodies.
Does the work make a moment of force?
By analyzing the above formulas in a scalar or vector form, we can conclude that the quantity M is some work. Indeed, its dimension is N * m, which in SI corresponds to the joule (J). In fact, the moment of power is not work, but only a quantity that is capable of accomplishing it. For this to happen, there must be 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:
A = M * θ
In this expression, θ is the angle at 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 means that when you turn 1 radian (approximately 1/3 of the circumference), the moment-generating force M F has done 60 joules. This formula is often used in solving problems in systems where friction forces act, which will be shown below.
Moment of force and angular momentum
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 value called the "moment of momentum". It can be calculated using the formula:
L = I * ω
Here I is the moment of inertia (a quantity that plays the same role during rotation as 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.
We give one more formula that is important for solving problems on the work of moments of forces. Using this formula, you can calculate the kinetic energy of a rotating body. It looks like this:
E k = 1/2 * I * ω 2
Next, we present two problems with solutions, where we show how to use the considered physical formulas.
The balance of several bodies
The first task is related to the equilibrium of a system in which several forces act. The figure below shows a system that is affected by three forces. It is necessary to calculate how much mass the object needs to be suspended from this lever and at what point it should be done so that the given system is in equilibrium.
From the conditions of the problem, we can understand that to solve it one should use the Varignon theorem. The first part of the task can be answered right away, since the weight of the item that should be hung on the lever will be equal to:
P = F 1 - F 2 + F 3 = 20 - 10 + 25 = 35 N
The signs here are selected taking into account the fact that the force that rotates the lever counterclockwise creates a negative moment.
The position of point d, where this weight should be suspended, 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 the three forces. In order for the system to be in equilibrium, it is necessary to suspend a body weighing 35 N at a point of 4.714 m from the axis on the other side of the lever.
Moving disk task
The solution to the following problem is based on the use of the formula of the moment of friction force and the kinetic energy of the body of revolution. Task: a disk of radius r = 0.3 meters is given, which rotates at a speed of ω = 1 rad / s. It is necessary to calculate how much distance he is able to walk on the surface if the rolling friction coefficient is μ = 0.001.
This problem is most easily solved by using the law of conservation of energy. We have the initial kinetic energy of the disk. When it begins to roll, then 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 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 desired distance that the disk will go 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 in any way.