Strength is one of the important concepts in physics. It is the reason for the change in the state of any objects. In this article, we will consider what this quantity is, what forces are, and also show how to find the projection of the force on the axis and on the plane.
Strength and its physical meaning
In physics, force is a vector quantity that shows the change in the momentum of a body per unit of time. This definition assumes strength as a dynamic characteristic. From the point of view of statics, force in physics is a measure of the elastic or plastic deformation of bodies.
The international SI system expresses force in Newtons (N). What is 1 Newton, the easiest way to understand the example of the second law of classical mechanics. The mathematical record is as follows:
F¯ = m * a¯
Here F¯ is some external force acting on a body of mass m and leading to acceleration a¯. The quantification of one Newton follows from the formula: 1 N is such a force that leads to a change in the speed of a body weighing 1 kg per 1 m / s for every second.
Examples of the dynamic manifestation of force are the acceleration of a car or a freely falling body in a gravitational earth field.
The static manifestation of force, as was noted, is associated with the phenomena of deformation. The following formulas should be given here:
F = p * s
F = -k * x
The first expression connects the force F with the pressure P that it exerts on a certain area S. Through this formula, 1 N can be defined as the pressure of 1 pascal applied to the area of 1 m 2 . For example, a column of atmospheric air at sea level presses on a platform of 1 m 2 with a force of 10 5 N!
The second expression is a classic form of notation for Hooke's law. For example, stretching or compressing a spring by a linear quantity x leads to the occurrence of an opposing force F (in the expression k is the proportionality coefficient).
What are the forces
It has already been shown above that forces can be static and dynamic. Here we say that in addition to this feature of them, they can be contact forces or long-range ones. For example, friction, support reactions are contact forces. The reason for their appearance is the validity of the Pauli principle. The latter states that two electrons cannot occupy the same state. That is why the touch of two atoms leads to their repulsion.
Long-range forces appear as a result of the interaction of bodies through some carrier field. For example, such are the force of gravity or electromagnetic interaction. Both forces have an infinite radius of action, however, their intensity drops like a square of distance (Coulomb laws and universal gravitation).
Strength is a vector quantity
Having figured out the meaning of the physical quantity under consideration, we can proceed to the study of the projection of the force on the axis. First of all, we note that this value is vector, that is, it is characterized by the modulus and direction. We show how to calculate the modulus of force and its direction.
It is known that any vector can be specified uniquely in a given coordinate system if the coordinates of its beginning and end are known. Suppose that there is some directed segment MN¯. Then its direction and module can be determined using the following expressions:
MN¯ = (x 2 -x 1 ; y 2 -y 1 ; z 2 -z 1 );
| MN¯ | = √ ((x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 + (z 2 -z 1 ) 2 ).
Here, the coordinates with indices 2 correspond to point N, with indices 1 to point M. Vector MN¯ is directed from M to N.
For generality, we have shown how to find the module and coordinates (direction) of a vector in three-dimensional space. Similar formulas without a third coordinate are valid for the case on the plane.
Thus, the modulus of force is its absolute value, expressed in newtons. In terms of geometry, the modulus is the length of the directional segment.
What is the projection of force on the axis?
It is most convenient to talk about projections of directed segments on coordinate axes and planes if you first place the corresponding vector at the origin, that is, at the point (0; 0; 0). Suppose we have some force vector F¯. We place its beginning at the point (0; 0; 0), then the coordinates of the vector can be written as follows:
F¯ = ((x 1 - 0); (y 1 - 0); (z 1 - 0)) = (x 1 ; y 1 ; z 1 ).
The vector F¯ shows the direction of the force in space in a given coordinate system. Now we draw perpendicular segments from the end of F¯ to each of the axes. The distance from the point of intersection of the perpendicular with the corresponding axis to the origin is called the projection of the force on the axis. It is not difficult to guess that in the case of the force F¯, its projections on the x, y and z axes will be equal to x 1 , y 1 and z 1 , respectively. Note that these coordinates show the modules of the projections of the force (the length of the segments).
Angles between the force and its projections on the coordinate axes
The calculation of these angles is not a difficult task. All that is required to solve it is a knowledge of the properties of trigonometric functions and the ability to apply the Pythagorean theorem.
For example, we define the angle between the direction of the force and its projection onto the x axis. The corresponding right-angled triangle will be formed by the hypotenuse (vector F¯) and leg (segment x 1 ). The second leg is the distance from the end of the vector F¯ to the x axis. The angle α between F¯ and the x axis is calculated by the formula:
α = arccos (| x 1 | / | F¯ |) = arccos (x 1 / √ (x 1 2 + y 1 2 + z 1 2 )).
As you can see, to determine the angle between the axis and the vector, it is necessary and sufficient to know the coordinates of the end of the directed segment.
For angles with other axes (y and z), you can write similar expressions:
β = arccos (| y 1 | / | F¯ |) = arccos (y 1 / √ (x 1 2 + y 1 2 + z 1 2 ));
γ = arccos (| z 1 | / | F¯ |) = arccos (z 1 / √ (x 1 2 + y 1 2 + z 1 2 )).
Note that in all formulas there are modules in the numerators, which eliminates the appearance of obtuse angles. Between the force and its axial projections, the angles are always less than or equal to 90 o .
Strength and its projection on the coordinate plane
The definition of the projection of the force onto the plane does not differ from that for the axis, only in this case the perpendicular should not be lowered onto the axis, but onto the plane.
In the case of a spatial rectangular coordinate system, we have three mutually perpendicular planes xy (horizontal), yz (frontal vertical), xz (lateral vertical). The intersection points of the perpendiculars omitted from the end of the vector to the named planes are:
(x 1 ; y 1 ; 0) for xy;
(x 1 ; 0; z 1 ) for xz;
(0; y 1 ; z 1 ) for zy.
If we connect each of the marked points to the origin, then we get the projection of the force F¯ on the corresponding plane. What is the modulus of force, we know. To find the module of each projection, it is necessary to apply the Pythagorean theorem. We denote projections on the plane as F xy , F xz and F zy . Then for their modules the following equalities are valid:
F xy = √ (x 1 2 + y 1 2 );
F xz = √ (x 1 2 + z 1 2 );
F zy = √ (y 1 2 + z 1 2 ).
Angles between projections onto a plane and a force vector
In the section above, formulas were given for the moduli of projections onto the plane of the considered vector F¯. These projections, together with the segment F¯ and the distance from its end to the plane, form rectangular triangles. Therefore, as in the case of projections onto the axis, one can use the definition of trigonometric functions to calculate the angles under consideration. We can write the following equalities:
α = arccos (F xy / | F¯ |) = arccos (√ (x 1 2 + y 1 2 ) / √ (x 1 2 + y 1 2 + z 1 2 ));
β = arccos (F xz / | F¯ |) = arccos (√ (x 1 2 + z 1 2 ) / √ (x 1 2 + y 1 2 + z 1 2 ));
γ = arccos (F zy / | F¯ |) = arccos (√ (y 1 2 + z 1 2 ) / √ (x 1 2 + y 1 2 + z 1 2 )).
It is important to understand that the angle between the direction of the force F¯ and its corresponding projection onto the plane is equal to the angle between F¯ and this plane. If we consider this problem from the point of view of geometry, then we can say that the directed segment F¯ is inclined with respect to the planes xy, xz and zy.
Where are power projections used?
The above formulas for the projections of force on the coordinate axis and on the plane are not only of theoretical interest. They are often used in solving physical problems. The process of finding projections is called the decomposition of force into its components. The latter are vectors, the sum of which should give the original force vector. In the general case, the force can be decomposed into arbitrary components, however, for solving problems it is convenient to use precisely projections on perpendicular axes and planes.
Tasks where the concept of projections of forces are applied can be very different. For example, the same Newton’s second law assumes that the external force F¯ acting on the body should be directed in the same way as the velocity vector v¯. If their directions differ by a certain angle then, so that the equality remains valid, then substitute into it not the force F¯ itself, but its projection onto the direction v¯.
Next, we give a couple of examples where we show how to use the written formulas.
The task of determining the projections of force on the plane and on the coordinate axis
Suppose that there is some force F¯, which is represented by a vector having the following coordinates of the end and the beginning:
(2; 0; 1);
(-1; 4; -1).
It is necessary to determine the modulus of force, as well as all its projections on the coordinate axes and planes and the angles between F¯ and each of its projections.
We begin to solve the problem by calculating the coordinates of the vector F¯. We have:
F¯ = (-1; 4; -1) - (2; 0; 1) = (-3; 4; -2).
Then the modulus of force will be equal to:
| F¯ | = √ (9 + 16 + 4) = √29 ≈ 5.385 N.
The projections on the coordinate axis are equal to the corresponding coordinates of the vector F¯. We calculate the angles between them and the direction F¯. We have:
α = arccos (| -3 | /5,385) ≈ 56.14 o ;
β = arccos (| 4 | /5,385) ≈ 42.03 o ;
γ = arccos (| -2 | /5,385) ≈ 68.20 o .
Since the coordinates of the vector F¯ are known, the moduli of the projections of the force on the coordinate plane can be calculated. Using the above formulas, we get:
F xy = √ (9 +16) = 5 N;
F xz = √ (9 + 4) = 3.606 N;
F zy = √ (16 + 4) = 4.472 N.
Finally, it remains to calculate the angles between the found projections on the plane and the force vector. We have:
α = arccos (F xy / | F¯ |) = arccos (5 / 5,385) ≈ 21.8 o ;
β = arccos (F xz / | F¯ |) = arccos (3,606 / 5,385) ≈ 48.0 o ;
γ = arccos (F zy / | F¯ |) = arccos (4.472 / 5.385) ≈ 33.9 o .
Thus, the vector F¯ is the most inclined to the coordinate plane xy.
The task with a sliding bar on an inclined plane
Now we solve the physical problem, where it will be necessary to apply the concept of projection of force. Let a wooden inclined plane be given. The angle of its inclination to the horizon is 45 o . On the plane is a wooden block having a mass of 3 kg. It is necessary to determine with what acceleration this block will move down the plane, if it is known that the coefficient of sliding friction is 0.7.
To begin with, we will draw up the equation of motion of the body. Since only two forces will act on it (the projection of gravity onto a plane and friction force), the equation will take the form:
F g - F f = m * a =>
a = (F g - F f ) / m.
Here F g , F f is the projection of gravity and friction, respectively. That is, the task is reduced to calculating their values.
Since the angle at which the plane is inclined to the horizon is 45 o , it is easy to show that the projection of the gravity F g along the surface of the plane will be equal to:
F g = m * g * sin (45 o ) = 3 * 9.81 / √2 ≈ 20.81 N.
This projection of force seeks to bring a wooden block out of a state of rest and give it acceleration.
According to the definition, the sliding friction force is equal to:
F f = μ * N
Where μ = 0.7 (see the condition of the problem). The reaction force of the support N is equal to the projection of gravity on the axis perpendicular to the inclined plane, that is:
N = m * g * cos (45 o )
Then the friction force is equal to:
F f = μ * m * g * cos (45 o ) = 0.7 * 3 * 9.81 / √2 ≈ 14.57 N.
Substitute the found forces in the equation of motion, we obtain:
a = (F g - F f ) / m = (20.81 - 14.57) / 3 = 2.08 m / s 2 .
Thus, the bar will descend along an inclined plane, increasing its speed by 2.08 m / s for every second.