Contents
Physics Topics are often described using mathematical equations, making them precise and quantifiable.
Vector Resolution: Finding the Components of a Vector
A vector can be resolved into many components just as many vectors can be added to give a single resultant vector.
Definition: When a vector is split into two or more vectors in such a manner that, the original vector becomes the resultant of the resolved parts or components of the vector, this splitting is called resolution of vectors.
Apparently it may seem that resolution of vectors is just the opposite of addition of vectors. But it may be noted that, when two vectors are added, it gives only one resultant vector, On the other hand, in resolution of vectors, different sets of components can be formed. Resolution of vectors into two or three components is an elegant technique to solve most of the problems related to vectors. Splitting into more than three components is very rare in practice.
Resolution in Two Dimensions
Resolution of vectors into two components: Let the magnitude and direction of a given vector \(\vec{R}\) be represented by \(\overrightarrow{O A}\) [Fig.]. OM and ON are inclined to OA by angles α and β respectively, in such a way that OA, OM and ON lie on the same plane. The vector \(\vec{R}\) is to be resolved into components along OM and ON.
From the figure, AC parallel to OM and AB parallel to ON are drawn to complete the parallelogram. Now, from the law of parallelogram of vectors, \(\overrightarrow{O B}\) + \(\overrightarrow{O C}\) = \(\overrightarrow{O A}\)
Thus, the two components of \(\overrightarrow{O A}\) are \(\overrightarrow{O B}\) and \(\overrightarrow{O C}\) and their magnitudes are a and b respectively.
∴ OB = a and OC = BA = b
Applying trigonometric rules to ΔOAB, we get,
\(\frac{O B}{\sin \angle O A B}\) = \(\frac{B A}{\sin \angle A O B}\) = \(\frac{O A}{\sin \angle A B O}\)
or, \(\frac{a}{\sin \beta}\) = \(\frac{b}{\sin \alpha}\) = \(\frac{R}{\sin [\pi-\{\alpha+\beta\}]}\)
or, \(\frac{a}{\sin \beta}\) = \(\frac{b}{\sin \alpha}\) = \(\frac{R}{\sin (\alpha+\beta)}\) …… (1)
Since α and β can have many sets of values, a and b can also have many values. Hence, a vector can be resolved into different pairs of components.
The two most useful components of a vector are the two mutually perpendicular components, when α + β = 90°.
Then from equation (1), we get
a = \(\frac{R \sin \left(90^{\circ}-\alpha\right)}{\sin 90^{\circ}}\) = Rcosα
and b = \(\frac{R \sin \alpha}{\sin 90^{\circ}}\) = Rsinα ….. (2)
Again, depending on α, different pairs of orthogonal components are possible.
Resolution of Vectors into Rectangular or Orthogonal Components
In Fig., \(\vec{R}\) has been resolved into two components along two mutually perpendicular axes OX and OY.
Let \(\overrightarrow{O B}\) = \(\vec{a}\), \(\overrightarrow{O C}\) = \(\vec{b}\) and ∠AOB = α; i.e., the component \(\vec{a}\) is inclined to the vector \(\vec{R}\) at an angle α.
According to the figure, cosα = \(\frac{O B}{O A}\) = \(\frac{a}{R}\)
or, a = Rcosα ….. (1)
and sinα = \(\frac{A B}{O A}\) = \(\frac{b}{R}\) or, b = Rsinα ……. (2)
Hence, the component of \(\vec{R}\) along a direction which makes an angle α with \(\vec{R}\) is Rcosα and the other component is Rsinα.
Special cases: To determine the component of a vector \(\vec{R}\) along its own direction, we put α = 0° in equation (1) and get a = Rcos0° = R.
Again, by putting α = 0° in equation (2) [or, α = 90° in equation (1)), we get the other component, i.e., the component in the direction perpendicular to \(\vec{R}\) as, b = Rsin0° = 0.
Hence, we may conclude that,
- the component (or resolved part) of a vector along its own direction is the vector itself.
- there is no component in a direction perpendicular to the vector.
Practical example-pull or push: A body can be set into motion along a horizontal plane by pushing it from the back or by pulling it towards the front. When the body is pushed, the applied force F1 usually acts downwards at an angle with the horizontal [Fig.].
Horizontal motion of the body is due to the horizontal component Fx of the applied force F1. The vertical component Fy, acting downwards, adds to the weight of the body and hence, pushing becomes difficult.
On the other hand, while pulling, the applied force F2 acts upwards at an angle with the horizontal. In this case also, the horizontal component F’x of F2 produces the horizontal motion of the body. The vertical component F’y acting upwards, effectively reduces the weight of the body. Hence pulling becomes easier.
Thus, it is easier to pull a body than to push it.
Numerical Examples
Example 1.
A force of 30 dyn is inclined to the y-axis at an angle of 60°. Find the components of the force along x and y axes respectively.
Solution:
Given, F = 30 dyn and θ = 60
Hence, component of F along y -axis,
Fy = F cos θ = F cos60° = 30 × \(\frac{1}{2}\) = 15 dyn and component of F along x -axis
Fx = Fsin60° = 30 × \(\frac{\sqrt{3}}{2}\) = 15\(\sqrt{3}\)dyn.
Example 2.
The value of the sultant of two mutually perpendicular forces is 80 dyn. The resultant makes an angle 60° with one of the forces. Find the magnitudes of the forces.
Solution:
Let the forces be \(\vec{P}\), \(\vec{Q}\) and the resultant that makes an angle 60° with \(\vec{P}\) be \(\vec{R}\).
Here, R = 80 dyn
∴ P = R cosθ = 80 cos60° = 80 × \(\frac{\sqrt{1}}{2}\) = 40 dyn
and Q = R sin60° = 80 sin60° = 80 × \(\frac{\sqrt{3}}{2}\) = 40\(\sqrt{3}\) dyn.