AnalytIcal (Mathematical) Method of Vector Addition Physics Notes
Analytical (Mathematical) Method of Vector Addition:
1. Triangle Law of Vector Addition: Let us consider the two vectors \(\vec{P}\) and \(\vec{Q}\), inclined at angle θ, be acting on a particle at the same time. Let them be represented in magnitude and direction by two sides
\(\vec{OA}\) and \(\vec{AB}\) of triangle OAB, taken in the same order, figure. Then according to the triangle law of vectors addition, the resultant \(\vec{R}\) is represented by the side \(\vec{OB}\) of the triangle, taken in opposite order with the omagnitude and direction.
Addition of vectors
To find out the magnitude of the resultant:
From the point B draw BN perpendicular to OA, meeting at point N, when produced. Let ∠BAN = 0 then, from the right angled triangle ONB, we have
OB2 = (ON)2 + (NB)2
= (OA + AN)2 + (NB)2
(OB)2 = (OA)2 + (AN)2 + 2OA AN + (NB)2
Now, (AN)2+ (NB)2 = (AB)2
⇒ (OB)2 = (OA)2 + (AN)2 + 2(OA)(AN) + (NB)2 …(i)
Using the Pythagoras theorem in right angled Δ ANB,
(AB)2 = (AN)2 + (NB)2 …(ii)
But, cos θ = \(\frac{A N}{A B}\)
AN = AB cos θ …(iii)
From equations (i), (ii) and (iii) we get
(OB)2 = (OA)2 + 2 (OA)(ABcosθ) + (AB)2
or, R2 = P2 + 2PQcosθ + Q2
or, R2 = P2 + Q2 + 2PQcosθ
or, R = P2 + Q2 + 2PQcosθ …(iv)
The above expression is known as the law of cosines.
To determine the direction of the resultant vector
From the right-angled Δ ONB
tan α = \(\frac{N B}{O N}\)
or, tan α = \(\frac{N B}{O A+A N}\) ……(v)
Special Cases:
1. When \(\vec{P}\) and \(\vec{Q}\) are in the same direction, then θ = 0°.
From equation (iv), we have
2. When P and are perpendicular to each other, then θ = 90°,
From equation (iv), we have
3. When \(\vec{p}\) and \(\vec{Q}\) are in opposite directions. Then,
θ = 180°,
From equation (iv),
⇒ R = \(\)
or, R = P – Q………(xiii)
Here, the value of R is minimum.
From equation (viii), we get
α = tan-1 \(\left(\frac{Q \sin 180^{\circ}}{P+Q \cos 180^{\circ}}\right)\)
or, α = 0° when P > Q
and, α = 180° when P < Q ….(xiv)
It may be pointed out that the magnitude of the resultant of two vectors is maximum, when they act along the same direction and minimum, when they act along the opposite direction.
2. Parallelogram Law:
Proof: The proof for the resultant vector in parallelogram addition is as follows:
Let the two vectors \(\vec{P}\) & \(\vec{Q}\) inclined to each other at an angle 0 be represented in magnitude as well as in the direction by the sides \(\vec{OA}\) and \(\vec{OC}\) of the parallelogram OABC [figure], Then according to the parallelogram law, the resultant of \(\vec{P}\) and \(\vec{Q}\) is represented both in magnitude and direction by the diagonal \(\vec{OB}\) of the parallelogram.
To find out the magnitude: From point B draw a perpendicular BN on OA, meeting OA at point N.
From the right-angled Δ ONB
OB2 = ON2 + NB2
But ON = OA + AN
From Δ ANB
It is known as the law of cosines.
The direction of the Resultant:
Suppose that the resultant R makes an angle a with the direction of vector P. Then from the right-angled triangle ONB
tan α = \(\frac{B N}{O N}=\frac{B N}{O A+A N}=\frac{Q \sin \theta}{P+Q \cos \theta}\)
1. It should also be noted that while finding the resultant of two vectors by the parallelogram law, the two vectors \(\vec{P}\) and \(\vec{Q}\) have to be co-initial vectors.
2. The magnitude of the resultant of two vectors is maximum, when angle between them is 00 and is minimum when they act is oppsiste directions
i.e., θ=180° 0
Rmax = P + Q
Rmin = P – Q
3. The resultant of two equal vectors can be zero if they act in opposite direction.
4. The number of three unequal vectors can be zero. If they are coplanar and they can be represented by three sides of a triangle taken in one order.
5. If \(\vec{P}+\vec{Q}=\vec{P}-\vec{Q}\), then vector \(\vec{Q}\) must be zero vector.
6. If \(|\vec{P}+\vec{Q}|=|\vec{P}-\vec{Q}|\) then vectors \(\vec{P}\) and \(\vec{Q}\) must be at right angle to each other.