Examples of Component in the following topics:

 Another way of adding vectors is to add the components.
 To add vectors, merely express both of them in terms of their horizontal and vertical components and then add the components together.
 This can be seen by adding the horizontal components of the two vectors ($4+4$) and the two vertical components ($3+3$).
 These additions give a new vector with a horizontal component of 8 ($4+4$) and a vertical component of 6 ($3+3$).
 To subtract vectors by components, simply subtract the two horizontal components from each other and do the same for the vertical components.

 It is often useful in analyzing vectors to break them into their component parts.
 For twodimensional vectors, these components are horizontal and vertical.
 For three dimensional vectors, the magnitude component is the same, but the direction component is expressed in terms of $x$, $y$ and $z$.
 This is the horizontal component of the vector.
 Together, the two components and the vector form a right triangle.

 Trigonometry is also used in determining the horizontal and vertical components of forces and objects.
 Free body diagrams are very helpful in visually identifying which components are unknown and where the moments are applied.
 Sometimes people need to analyze the horizontal and vertical components of forces and object orientation.
 To make the problem easier, the force F will be expressed in terms of its horizontal and vertical components .
 Explain why trigonometry is useful in determining horizontal and vertical components of forces

 We can always represent one, say $\mathbf{b}$ , in terms of its components parallel and perpendicular to the other.
 The length of the component of $\mathbf{b}$ along $\mathbf{a}$ is $\\mathbf{b}\ \cos \theta$ which is also $\mathbf{b}^T \mathbf{a}/\\mathbf{a}\$ .
 Now suppose we want to construct a vector in the direction of $\mathbf{a}$ but whose length is the component of $\mathbf{b}$ along $\\mathbf{b}\$ .
 We can always represent one, say $\mathbf{b}$ , in terms of its components parallel and perpendicular to the other.
 The length of the component of $\mathbf{b}$ along $\mathbf{a}$ is $\\mathbf{b}\ \cos \theta$ which is also $\mathbf{b}^T \mathbf{a}/\\mathbf{a}\$ .

 In order to analytically add these vectors, you need to remember the relationship between the magnitude and direction of the vector and its components on the x and y axis of the coordinate system:
 These components are shown above.
 The first two equations are for when the magnitude and direction are known, but you are looking for the components.
 The last two equations are for when the components are known, and you are looking for the magnitude and direction.

 Then we consider only the component of v that is perpendicular to the field when making our calculations, so that the equations of motion become:
 The component of the velocity parallel to the field is unaffected, since the magnetic force is zero for motion parallel to the field.
 The component of velocity parallel to the lines is unaffected, and so the charges spiral along the field lines.
 Cosmic rays are a component of background radiation; consequently, they give a higher radiation dose at the poles than at the equator .
 Energetic electrons and protons, components of cosmic rays, from the Sun and deep outer space often follow the Earth's magnetic field lines rather than cross them.

 You can also accomplish scalar multiplication through the use of a vector's components.
 Once you have the vector's components, multiply each of the components by the scalar to get the new components and thus the new vector.

 The components of the normal force $N$ in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively.
 In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.
 Only the normal force has a horizontal component, and so this must equal the centripetal force—that is:
 Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction.
 From the figure, we see that the vertical component of the normal force is $N\cos\theta$, and the only other vertical force is the car's weight.

 To solve a two dimensional elastic collision problem, decompose the velocity components of the masses along perpendicular axes.
 The general approach to solving a two dimensional elastic collision problem is to choose a coordinate system in which the velocity components of the masses can be decomposed along perpendicular axes .
 By defining the xaxis to be along the direction of the incoming particle, we save ourselves time in breaking that velocity vector into its x and y components.
 The components of velocities along the xaxis have the form $v \cdot cos \theta $, where θ is the angle between the velocity vector of the particle of interest and the xaxis.
 In finding Eq. 3, it was taken into consideration that the incoming particle had no component of velocity along the yaxis.

 A moving car for which the net x and y force components are zero