The branch of physics that studies bodies at rest in terms of mechanics is called static. The key points of statics are understanding the conditions of equilibrium of bodies in the system and the ability to apply these conditions to solve practical problems.
Active forces
The reason for the rotation, translational movement or complex movement of bodies along curved paths is the effect of an external nonzero force on these bodies. In physics, a force is called a quantity, which, acting on the body, can give it acceleration, that is, change the momentum. This value has been studied since ancient times, however, the laws of statics and dynamics finally took shape in a harmonious physical theory only with the advent of a new time. A large role in the formation of the mechanics of movement was played by the work of Isaac Newton, in whose honor the unit of measurement of force is called Newton.
When considering the equilibrium conditions of bodies in physics, it is important to know several parameters of the acting forces. These include the following:
- direction of action;
- absolute value;
- point of application;
- the angle between the force in question and other forces applied to the system.
The totality of these parameters allows you to clearly say whether this system will move or rest.
The first condition for the equilibrium of the system
When the system of solids will not progressively move in space? The answer to this question will become clear if we recall the second Newtonian law. According to him, the system will not make translational movement if and only if the sum of the forces external to the system is zero. That is, the first condition for the equilibrium of solids mathematically looks like this:
∑ i = 1 n F i ¯ = 0.
Here n is the number of external forces in the system. The above expression assumes vector summation of forces.
Consider a simple case. Suppose that two forces are acting on the body that are the same in magnitude, but directed in different directions. As a result, one of them will tend to give acceleration to the body along the positive direction of an arbitrarily chosen axis, and the other along the negative. The result of their action will be a resting body. The vector sum of these two forces will be zero. In fairness, we note that the described example will lead to the appearance of tensile stresses in the body, but this fact does not apply to the topic of the article.
To facilitate verification of the recorded condition for the equilibrium of bodies, you can use the geometric image of all the forces in the system. If their vectors are arranged so that each subsequent force starts from the end of the previous one, then the recorded equality will be fulfilled when the beginning of the first force coincides with the end of the last. Geometrically, this looks like a closed loop of force vectors.
Moment of power
Before proceeding to the description of the following equilibrium condition for a solid, it is necessary to introduce an important physical concept of statics - the moment of force. In simple terms, the scalar magnitude of the moment of force is the product of the modulus of the force itself by the radius vector from the axis of rotation to the point of application of force. In other words, it makes sense to consider the moment of force only with respect to any axis of rotation of the system. The scalar mathematical form of recording the moment of force looks like this:
M = F * d.
Where d is the shoulder of power.
It follows from the written expression that if the force F is applied to any point of the axis of rotation at any angle to it, then its moment of force will be zero.
The physical meaning of the quantity M lies in the ability of the force F to make a turn. This ability increases with increasing distance between the point of application of force and the axis of rotation.
The second condition for the equilibrium of the system
As you might guess, the second condition for the equilibrium of bodies is connected with the moment of force. First, we give the corresponding mathematical formula, and then we analyze it in more detail. So, the condition for the absence of rotation in the system is written as follows:
∑ i = 1 n M i = 0.
That is, the sum of the moments of all forces must be zero relative to each axis of rotation in the system.
The moment of force is a vector quantity, however, to determine the rotational equilibrium, it is important to know only the sign of the given moment M i . It should be remembered that if a force tends to rotate in the direction of the clock, then it creates a negative moment. On the contrary, rotation against the direction of the arrow leads to the appearance of a positive moment M i .
Methodology for determining the equilibrium of the system
Two conditions for the equilibrium of bodies were given above. Obviously, in order for the body to not move and be at rest, it is necessary to simultaneously fulfill both conditions.
When solving equilibrium problems, one should consider a system of written two equations. The solution to this system will give an answer to any task on statics.
Sometimes the first condition, reflecting the lack of translational motion, may not provide any useful information, then solving the problem comes down to analyzing the condition of the moments.
When considering the problems of static on the equilibrium conditions of bodies, the center of gravity of the body plays an important role, since it is through it that the axis of rotation passes. If the sum of the moments of forces relative to the center of gravity is zero, then the rotation of the system will not be observed.
Problem solving example
It is known that two loads were put on the ends of a weightless board. The mass of the right cargo is twice as much as the mass of the left. It is necessary to determine the position of the support under the board, in which this system would be in equilibrium.
We denote the length of the board with the letter l, and the distance from its left end to the support with the letter x. It is clear that this system does not experience any translational motion, therefore, the first condition for solving the problem does not need to be applied.
The weight of each load creates a moment of force relative to the support, both of which have a different sign. In our notation, the second equilibrium condition will have the form:
P 1 * x = P 2 * (Lx).
Here P 1 and P 2 are the weights of the left and right weights, respectively. Dividing both sides of the equality by P 1 , and using the condition of the problem, we obtain:
x = P 2 / P 1 * (Lx) =>
x = 2 * L - 2 * x =>
x = 2/3 * L.
Thus, in order for the system to be in equilibrium, the support should be located 2/3 of the length of the board from its left end (1/3 from the right end).