Projection of force on the axis and on the plane. Physics

2024 Author: Angel Austin | [email protected]. Last modified: 2023-12-17 05:14

Power is one of the most important concepts in physics. It causes a change in the state of any objects. In this article, we will consider what this value is, what forces there are, and also show how to find the projection of the force on the axis and on the plane.

Power 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 considers force to be a dynamic characteristic. From the point of view of statics, force in physics is a measure of 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. Its mathematical notation is as follows:

F¯=ma¯

Here F¯ is some external force acting on a body of mass m and resulting in acceleration a¯. The quantitative definition of one newton follows from the formula: 1 N is such a force that leads to a change in the speed of a body with a mass of 1 kg by 1 m / s for every second.

Examples of dynamicmanifestations of force are the acceleration of a car or a freely falling body in the earth's gravitational field.

The static manifestation of force, as noted, is associated with deformation phenomena. The following formulas should be given here:

F=PS

F=-kx

The first expression relates the force F to the pressure P that it exerts on some area S. Through this formula, 1 N can be defined as a pressure of 1 pascal applied to an area of 1 m². For example, a column of atmospheric air at sea level presses on a site of 1 m²with a force of 10⁵N!

The second expression is the classical form of Hooke's law. For example, stretching or compressing a spring by a linear value x leads to the emergence of an opposing force F (in the expression k is the proportionality factor).

What forces are there

It has already been shown above that forces can be static and dynamic. Here we say that in addition to this feature, they can be contact or long-range forces. For example, friction force, 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 a certain carrier field. For example, such are the force of gravity or electromagnetic interaction. Both powers have an infinite range,however, their intensity drops as the square of the distance (Coulomb's laws and gravity).

Power is a vector quantity

Having de alt with the meaning of the considered physical quantity, we can proceed to the study of the issue of force projection on the axis. First of all, we note that this quantity is a vector, that is, it is characterized by a module and direction. We will show how to calculate the force modulus and its direction.

It is known that any vector can be uniquely specified in a given coordinate system if the values of the coordinates of its beginning and end are known. Assume that there is some directed segment MN¯. Then its direction and module can be determined using the following expressions:

MN¯=(x₂-x₁; y₂-y 1_{; z}2_-z1_);

|MN¯|=√((x₂-x₁)²+ (y_{2 -y}1₎2^{+ (z}2_-z1 ₎2^).

Here, coordinates with indices 2 correspond to point N, those with indices 1 correspond to point M. The vector MN¯ is directed from M to N.

For the sake of generality, we have shown how to find the modulus and coordinates (direction) of a vector in three-dimensional space. Similar formulas without the third coordinate are valid for the case on the plane.

Thus, the modulus of force is its absolute value, expressed in newtons. From the point of view of geometry, the modulus is the length of the directed segment.

What is the projection of force onaxis?

It is most convenient to talk about projections of directed segments onto 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¯. Let's place its beginning at the point (0; 0; 0), then the coordinates of the vector can be written as follows:

F¯=((x₁- 0); (y₁- 0); (z₁ - 0))=(x₁; y₁; z₁).

Vector F¯ shows the direction of the force in space in the given coordinate system. Now let's 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 x₁, y₁ and z 1_{, respectively. Note that these coordinates show the modules of force projections (the length of the segments).}

Angles between the force and its projections on the coordinate axes

Calculating these angles is not difficult. All that is required to solve it is knowledge of the properties of trigonometric functions and the ability to apply the Pythagorean theorem.

For example, let's define the angle between the force direction and its projection on the x-axis. The corresponding right triangle will be formed by the hypotenuse (vector F¯) and leg (segment x₁). 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(|x₁|/|F¯|)=arccos(x₁/√(x ₁²+y₁²+z₁ 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(|y₁|/|F¯|)=arccos(y₁/√(x 12^+y12^+z 12^));

γ=arccos(|z₁|/|F¯|)=arccos(z₁/√(x 12^+y12^+z 12^)).

Note that in all formulas there are modules in the numerators, which eliminates the appearance of obtuse corners. Between the force and its axial projections, the angles are always less than or equal to 90^o.

Force and its projections on the coordinate plane

The definition of the force projection onto the plane does not differ from that for the axis, only in this case the perpendicular should be lowered not 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 points of intersection of the perpendiculars dropped from the end of the vector to the named planes are:

(x₁; y₁; 0) for xy;

(x₁; 0; z₁) for xz;

(0; y₁; z₁) for zy.

If each of the marked points is connected to the origin, then we get the projection of the force F¯ onto the corresponding plane. What is the modulus of force, we know. To find the modulus of each projection, you need to apply the Pythagorean theorem. Let's denote the projections on the plane as F_xy, F_xz and F_zy. Then the equalities will be valid for their modules:

F_xy=√(x₁²+y₁ ²);

F_xz=√(x₁²+ z₁ ²);

F_zy=√(y₁²+ z₁ ²).

Angles between projections onto the plane and force vector

In the paragraph above, formulas were given for the modules 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 right-angled triangles. Therefore, as in the case of projections on the axis, you can use the definition of trigonometric functions to calculate the angles in question. You can write the following equalities:

α=arccos(F_xy/|F¯|)=arccos(√(x₁^{2 +y}12^{) /√(x}12 +y¹₂+z¹₂));

β=arccos(F_xz/|F¯|)=arccos(√(x₁^{2 +z}12^)/√(x12 +y¹₂+z¹₂));

γ=arccos(F_zy/|F¯|)=arccos(√(y₁²+z₁²)/√(x₁²+y₁² +z₁²)).

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 force projections used?

The above formulas for force projections on the coordinate axes and on the plane are not only of theoretical interest. They are often used in solving physical problems. The very process of finding projections is called the decomposition of the force into its components. The latter are vectors, the sum of which should give the original force vector. In the general case, it is possible to decompose the force into arbitrary components, however, for solving problems, it is convenient to use projections onto perpendicular axes and planes.

Problems where the concept of force projections is applied can be very different. For example, the same Newton's second law assumes that the external force F¯ acting on the body must be directed in the same way as the velocity vector v¯. If their directions differ by some angle, then, in order for the equality to remain valid, one should substitute into it not the force F¯ itself, but its projection onto the direction v¯.

Next, we will give a couple of examples, where we will show how to use the recordedformulas.

The task of determining force projections on the plane and on the coordinate axes

Assume that there is some force F¯, which is represented by a vector having the following end and start coordinates:

(2; 0; 1);

(-1; 4; -1).

It is necessary to determine the modulus of the force, as well as all its projections on the coordinate axes and planes, and the angles between F¯ and each of its projections.

Let's start solving 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:

|F¯|=√(9 + 16 + 4)=√29 ≈ 5, 385 N.

Projections onto the coordinate axes are equal to the corresponding coordinates of the vector F¯. Let's calculate the angles between them and the F¯ direction. 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, it is possible to calculate the modules of force projections on the coordinate plane. 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 closest to the xy coordinate plane.

Problem with a sliding bar on an inclined plane

Now let's solve a physical problem where it will be necessary to apply the concept of force projection. 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 bar will move down the plane if it is known that the coefficient of sliding friction is 0.7.

First, let's make the equation of motion of the body. Since only two forces will act on it (the projection of gravity onto the plane and the friction force), the equation will take the form:

F_g- F_f=ma=>

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 gravity F_g along the surface of the plane will be equal to:

F_g=mgsin(45^o)=39, 81/√2 ≈ 20, 81 N.

This force projection seeks to unsettlewooden block and give it acceleration.

According to the definition, the force of sliding friction is:

F_f=ΜN

Where Μ=0, 7 (see the condition of the problem). The reaction force of the support N is equal to the projection of the force of gravity on the axis perpendicular to the inclined plane, that is: