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Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.
What is a Resultant? Which Law is Followed by Vector Addition?
Šuppose a particle starting from its initial position O [Fig.] undergoes a displacement \(\vec{a}\)’ and reaches the point A. Hence, \(\vec{a}\) = \(\overrightarrow{O A}\). After a further displacement of \(\vec{b}\) in a different direction, the particle reaches the point B. Hence, \(\vec{b}\) = \(\overrightarrow{A B}\). Starting from the point O, the net displacement of the particle is thus \(\overrightarrow{O B}\) = \(\vec{c}\) (say). Vector \(\vec{c}\) is called the sum or the resultant of \(\vec{a}\) and \(\vec{b}\), expressed as \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\).
This sum will obviously be a vector sum, because by adding only the magnitudes of \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) cannot be found. The magnitudes of \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) can be determined from the lengths of the sides OA, AB and OB respectively. This method of finding the resultant of two displacements also holds good for finding the sum of any two vectors. Figure illustrates one such case where forces \(\vec{P}\) and \(\vec{Q}\) pull a boat simultaneously. As a result, the boat moves along the resultant \(\vec{R}\), and \(\vec{P}\) + \(\vec{Q}\) = \(\vec{R}\).
Resultant: A single vector, which represents the result of the summation of a number of vectors both in magnitude and in direction, is called the resultant of those vectors,
In Fig., \(\overrightarrow{O B}\) is the resultant.
To find the sum of \(\vec{P}\) and \(\vec{Q}\) [Fig.], draw the vectors \(\vec{P}\) as shown in Fig. and name it as AB i.e., \(\overrightarrow{A B}\) = \(\vec{P}\).
Now move the vector \(\vec{Q}\) parallel to itself such that its tail coincides with the tip B of the vector \(\vec{P}\). Mark the tip of the vector \(\vec{Q}\) as C i.e., \(\overrightarrow{B C}\) = \(\vec{Q}\). Join the tail of the vector \(\vec{P}\) to the tip of the vector \(\vec{Q}\) i.e., \(\overrightarrow{A C}\) = \(\vec{R}\) where \(\vec{R}\) represents the sum of the vectors \(\vec{P}\) and \(\vec{Q}\).Thus, \(\vec{R}\) = \(\vec{P}\) + \(\vec{Q}\) [Fig.].
Vector addition follows either of the following equivalent laws:
- law of triangle of vectors,
- law of parallelogram of vectors and
- polygon law of vectors.
Magnitude in vector addition: Let us consider two vectors \(\vec{a}\) and \(\vec{b}\) and whose resultant is found to be \(\vec{c}\) i.e., \(\vec{a}\) + \(\vec{b}\) = \(\vec{c}\).
Now to think as, that the magnitude of \(\vec{c}\) is actually the sum of the magnitude of \(\vec{a}\) and magnitude of \(\vec{b}\) is incorrect. This is because, as we look at the figure, we can see that |\(\vec{c}\)| < |\(\vec{a}\)| + |\(\vec{b}\)|.
It means that, the magnitude of resultant vector not only depends on magnitude of a and b but also depends on the angle Fig. between the two vectors a and b. To find out the relationship between \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and the angle between the vectors, let us study the following law of vector addition.
Law of Triangle of Vectors
Geometrical method: In Fig., \(\vec{a}\) + \(\vec{b}\) = \(\vec{c}\) or \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\). Vectors \(\overrightarrow{O A}\), \(\overrightarrow{A B}\) and their resultant \(\overrightarrow{O B}\), therefore represent the three sides of the ΔOAB. The following triangle law of vector addition can be obtained by considering the magnitudes and directions of the vectors.
Statement : When the magnitudes and directions of two vectors, are represented by two adjacent sides of a triangle taken in order, the third side taken in the opposite order, represents the magnitude and direction of the resultant of the two vectors.
Two adjacent sides taken in order and the third side taken in the opposite order—means that if the former are in clock wise direction in the triangle, the third side, representing the resultant, should be in anticlockwise direction.
Corollary: With reference to the ΔOAB in Fig.,
\(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\) or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) – \(\overrightarrow{O B}\) = 0
or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B O}\) = 0
Hence, if three vectors are completely represented by the three sides of a triangle taken in order (all clockwise or all anticlockwise), then the resultant of the vectors will be zero.
Analytical method:
Let the two vectors \(\vec{a}\) and \(\vec{b}\) be represented both in magnitude and direction by the sides \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) of Δ OAB taken in the same order [Fig.]. Then according to the triangle law of vector addition, the resultant \(\vec{c}\) is given by the side OB taken in the reverse order, as shown in the figure.
Magnitude of the resultant \(\vec{c}\): In Fig., BE is the perpendicular drawn from B on the extension of OA. The angle between the vectors \(\vec{a}\) and \(\vec{b}\), ∠BAE = α. It is to be noted that to measure the angle between two vectors, their initial points are superimposed on each other without changing the directions of the vectors.
Now from right angled ΔAEB we have
\(\frac{B E}{A B}\) = sinα or, BE = ABsinα or, BE = bsinα
and \(\frac{A E}{A B}\) = cosα or, AE = ABcosα or, AE = bcosα
Applying Pythagoras theorem in right angled ΔOEB we get,
(OB)2 = (OE)2 + (EB)2 = (OA + AE)2 + (EB)2
or, c2 = (a + bcosα)2 + (bsinα)2
= a2 + 2abcosα + b2cos2α + b2sin2α
or, c2 = a2 + 2abcosα + b2(cos2α + sin2α)
or, c2 = a2 + b2 + 2abcosα [∵ cos2α + sin2α = 1]
or, c = \(\sqrt{a^2+b^2+2 a b \cos \alpha}\) ……… (1)
Direction of the resultant \(\vec{c}\) : Let the resultant vector (\(\vec{c}\)) makes an angle θ with the first vector \(\vec{a}\). Then from right angled ΔOEB we get,
tanθ = \(\frac{B E}{O E}\) = \(\frac{B E}{O A+A E}\)
or, tanθ = \(\frac{b \sin \alpha}{a+b \cos \alpha}\) …… (2)
Therefore, if the magnitudes a and b, and the value of the angle α are known, then from equation (1) and (2), the magnitude of the resultant vector and its direction can be determined.
Law of Parallelogram of Vectors
Geometrical method: The law of parallelogram of vector addition is just an alternative form of the law of triangle of vector addition. In Fig., let \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{A B}\) = \(\vec{a}\). Therefore, from the triangle law of vector addition, the vector \(\overrightarrow{O B}\) represents the magnitude and direction of the resultant \(\vec{c}\). Hence, \(\vec{a}\) + \(\vec{b}\) = \(\vec{c}\) or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\) or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) – \(\overrightarrow{O B}\) = 0
or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B O}\) = 0
Hence, If three vectors are completely represented by the three sides of a triangle taken in order (all clockwise or all anticlockwise), then the resultant of the vectors will be zero.
Analytical method:
Let the two vectors \(\vec{a}\) and \(\vec{b}\) be represented both in magnitude and direction by the sides \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) of ΔOAB taken in the same order [Fig.]. Then according to the triangle law of vector addition, the resultant \(\vec{c}\) is given by the side OB taken in the reverse order, as shown in the figure.
Magnitude of the resultant \(\vec{c}\) : In Fig., BE is the perpendicular drawn from B on the extension of OA. The angle between the vectors \(\vec{a}\) and \(\vec{b}\), ∠BAE = α. It is to be noted that to measure the angle between two vectors, their initial points are superimposed on each other without changing the directions of the vectors.
Now from right angled ΔAEB we have
\(\frac{B E}{A B}\) = sinα or, BE = ABsinα or BE = bsinα
and \(\frac{A E}{A B}\) = cosα or, AE = ABcosα or BE = bcosα
Applying Pythagoras theorem in right angled ΔOEB we get,
(OB)2 = (OE)2 + (EB)2 = (OA + AE)2 + (EB)2
or, c2 = (a + bcosα)2 + (bsinα)2
= a2 + 2abcosα + b2cos2α + b2sin2α
or, c2 = a2 + 2abcosα + b2(cos2α + sin2α)
or, c2 = a2 + b2 + 2abcosα [∵ cos2α + sin2α = 1]
or, c = \(\sqrt{a^2+b^2+2 a b \cos \alpha}\) ….. (1)
Direction of the resultant \(\vec{c}\): Let the resultant vector (\(\vec{c}\)) makes an angle θ with the first vector \(\vec{a}\). Then from right angled ΔOEB we get,
tanθ = \(\frac{B E}{O E}\) = \(\frac{B E}{O A+A E}\)
or, tanθ = \(\frac{b \sin \alpha}{a+b \cos \alpha}\) …. (2)
Therefore, if the magnitudes a and b, and the value of the angle α are known, then from equation (1) and (2), the magnitude of the resultant vector and its direction can be determined.
Law of Parallelogram of Vectors
Geometrical method: The law of parallelogram of vector addition is just an alternative form of the law of triangle of vector addition. In Fig., let \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{A B}\) = \(\vec{b}\). Therefore, from the triangle law of vector addition, the vector \(\overrightarrow{O B}\) represents the magnitude and direction of the resultant \(\vec{c}\). Hence, \(\vec{a}\) + \(\vec{b}\) = \(\vec{c}\) or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\).
Now the parallelogram OABD is completed. AB and OD being the opposite sides of the parallelogram, are equal and parallel.
Hence, \(\overrightarrow{A B}\) = \(\overrightarrow{O D}\) = \(\vec{b}\). So \(\overrightarrow{O A}\) + \(\overrightarrow{O D}\) = \(\overrightarrow{O B}\)
Statement: If the magnitudes and directions of two vectors are represented by two adjacent sides of a parallelogram, the diagonal drawn from the point of origin of (he
two vectors represents the magnitude and direction of the resultant.
Analytical method:
Magnitude of resultant \(\vec{c}\) : Let the point of origin of the vectors \(\vec{a}\) and \(\vec{b}\) be O [Fig.]. The two vectors are represented by the line segments OA and OD. The parallelogram OABD is completed by drawing DB parallel to OA and AB parallel to OD. Then the diagonal OB from the point O represents the magnitude and direction is the resultant of the vectors \(\vec{a}\) and \(\vec{b}\).
Here, \(\vec{a}\) = \(\overrightarrow{O A}\), \(\vec{b}\) = \(\overrightarrow{O D}\) = \(\overrightarrow{A B}\) and \(\vec{c}\) = \(\overrightarrow{O B}\)
In Fig., perpendicular BE is drawn from B on the extension of OA . Then ∠BAE = ∠DOA = α
Here, length of the side OA = |\(\vec{a}\)| = a
length of the side OD = length of the side AB = |\(\vec{b}\)| = b
and length of the side OB = |\(\vec{c}\)| = c
From right angled ΔBAE we have,
\(\frac{B E}{A B}\) = sinα or, BE = ABsinα
∴ BE = bsinα
Again, \(\frac{A E}{A B}\) = cosα or, AB = ABcosα
∴ AE= bcosα
Using Pythagoras theorem in right angled ΔOEB, we get,
(OB)2 = (OE)2 + (BE)2 = (OA + AE)2 + (BE)2
or, c2 =(a + bcosα)2 + (bsinα)2
or, c2 = a2 + b2cos2α + 2abcosα + b2sin2α
c2 = a2 + b2(cos2α + sin2α) + 2abcosα
c2 = a2 + b2 + 2abcosα [∵ cos2α + sin2α = 1]
or, c = \(\sqrt{a^2+b^2+2 a b \cos \alpha}\) …… (1)
Direction of resultant \(\vec{c}\): Let the resultant \(\vec{c}\) makes an angle θ with the direction of \(\vec{a}\). Then from right angled ΔOEB we get,
tanθ = \(\frac{B E}{O E}\) = \(\frac{B E}{O A+A E}\) or, tanθ = \(\frac{b \sin \alpha}{a+b \cos \alpha}\) …… (2)
Equations (1) and (2) are identical to those in the previous section.
It may be noted that, two vectors and their resultant are always confined on a plane. This is not the case for three or more vectors and their resultant. The vectors and their resultant are in general, distributed in 3-dimensional space.