Logical Class

Class 11 / Physics / Motion in a Plane / Vector Addition – Analytical Method

VECTOR ADDITION

VECTOR ADDITION

Although the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors by combining their respective components.Consider two vectors A and B in x-y plane with components A_x,A_y and B_x,B_y.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2 % da9iaadgeadaWgaaWcbaGaamiEaaqabaGcceWGPbGbaKaacqGHRaWk % caWGbbWaaSbaaSqaaiaadMhaaeqaaOGabmOAayaajaaaaa!3E94! A = {A_x}\hat i + {A_y}\hat j\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqaiabg2 % da9iaadkeadaWgaaWcbaGaamiEaaqabaGcceWGPbGbaKaacqGHRaWk % caWGcbWaaSbaaSqaaiaadMhaaeqaaOGabmOAayaajaaaaa!3E97! B = {B_x}\hat i + {B_y}\hat j\)

Let R be their sum. We have

R = A + B

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaae % WaaeaacaWGbbWaaSbaaSqaaiaadIhaaeqaaOGabmyAayaajaGaey4k % aSIaamyqamaaBaaaleaacaWG5baabeaakiqadQgagaqcaaGaayjkai % aawMcaamaabmaabaGaamOqamaaBaaaleaacaWG4baabeaakiqadMga % gaqcaiabgUcaRiaadkeadaWgaaWcbaGaamyEaaqabaGcceWGQbGbaK % aaaiaawIcacaGLPaaaaaa!47B4! = \left( {{A_x}\hat i + {A_y}\hat j} \right)\left( {{B_x}\hat i + {B_y}\hat j} \right)\)

Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (4.19a) as convenient to us :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGsb % Gaeyypa0ZaaeWaaeaacaWGbbWaaSbaaSqaaiaadIhaaeqaaOGaey4k % aSIaamOqamaaBaaaleaacaWG4baabeaaaOGaayjkaiaawMcaaiqadM % gagaqcamaabmaabaGaamyqamaaBaaaleaacaWG5baabeaakiabgUca % RiaadkeadaWgaaWcbaGaamyEaaqabaaakiaawIcacaGLPaaaceWGQb % GbaKaaaeaaciGGtbGaaiyAaiaac6gacaWGJbGaamyzaiaadkfacqGH % 9aqpcaWGsbWaaSbaaSqaaiaadIhaaeqaaOGabmyAayaajaGaey4kaS % IaamOuamaaBaaaleaacaWG5baabeaakiqadQgagaqcaaaaaa!53F0! \begin{array}{l} R = \left( {{A_x} + {B_x}} \right)\hat i\left( {{A_y} + {B_y}} \right)\hat j\\ {\mathop{\rm Sin}\nolimits} ceR = {R_x}\hat i + {R_y}\hat j \end{array}\)

we have, R_x = A_x =B_x , R_y =A_y = B_y.

Thus, each component of the resultant vector R is the sum of the corresponding components of A and B.

In three dimensions, we have

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGbb % Gaeyypa0JaamyqamaaBaaaleaacaWG4baabeaakiqadMgagaqcaiab % gUcaRiaadgeadaWgaaWcbaGaamyEaaqabaGcceWGQbGbaKaacqGHRa % WkcaWGbbWaaSbaaSqaaiaadQhaaeqaaOGabm4AayaajaaabaGaamyq % aiabg2da9iaadkeadaWgaaWcbaGaamiEaaqabaGcceWGPbGbaKaacq % GHRaWkcaWGcbWaaSbaaSqaaiaadMhaaeqaaOGabmOAayaajaGaey4k % aSIaamOqamaaBaaaleaacaWG6baabeaakiqadUgagaqcaaqaaiaadk % facqGH9aqpcaWGbbGaey4kaSIaamOqaiabg2da9iaadkfadaWgaaWc % baGaamyAaaqabaGcceWGPbGbaKaacqGHRaWkcaWGsbWaaSbaaSqaai % aadMhaaeqaaOGabmOAayaajaGaey4kaSIaamOuamaaBaaaleaacaWG % 6baabeaakiqadUgagaqcaaqaaiaadEhacaWGPbGaamiDaiaadIgaca % WGsbWaaSbaaSqaaiaadIhaaeqaaOGaeyypa0JaamyqamaaBaaaleaa % caWG4baabeaakiabgUcaRiaadkeadaWgaaWcbaGaamiEaaqabaaake % aacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caWGsbWaaSbaaSqaaiaadMhaaeqaaOGaeyypa0Jaamyqam % aaBaaaleaacaWG5baabeaakiabgUcaRiaadkeadaWgaaWcbaGaamyE % aaqabaaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca % aMc8UaaGPaVlaaykW7caaMc8UaamOuamaaBaaaleaacaWG6baabeaa % kiabg2da9iaadgeadaWgaaWcbaGaamOEaaqabaGccqGHRaWkcaWGcb % WaaSbaaSqaaiaadQhaaeqaaaaaaa!97EE! \begin{array}{l} A = {A_x}\hat i + {A_y}\hat j + {A_z}\hat k\\ B= {B_x}\hat i + {B_y}\hat j + {B_z}\hat k\\ R = A + B = {R_i}\hat i + {R_y}\hat j + {R_z}\hat k\\ with{R_x} = {A_x} + {B_x}\\ \,\,\,\,\,\,\,\,\,{R_y} = {A_y} + {B_y}\\ \,\,\,\,\,\,\,\,\,\,{R_z} = {A_z} + {B_z} \end{array}\)

This method can be extended to addition and subtraction of any number of vectors. For example, if vectors a, b and c are given as

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGHb % Gaeyypa0JaamyyamaaBaaaleaacaWG4baabeaakiqadMgagaqcaiab % gUcaRiaadggadaWgaaWcbaGaamyEaaqabaGcceWGQbGbaKaacqGHRa % WkcaWGHbWaaSbaaSqaaiaadQhaaeqaaOGabm4AayaajaaabaGaamOy % aiabg2da9iaadkgadaWgaaWcbaGaamiEaaqabaGcceWGPbGbaKaacq % GHRaWkcaWGIbWaaSbaaSqaaiaadMhaaeqaaOGabmOAayaajaGaey4k % aSIaamOyamaaBaaaleaacaWG6baabeaakiqadUgagaqcaaqaaiaado % gacqGH9aqpcaWGJbWaaSbaaSqaaiaadIhaaeqaaOGabmyAayaajaGa % ey4kaSIaam4yamaaBaaaleaacaWG5baabeaakiqadQgagaqcaiabgU % caRiaadogadaWgaaWcbaGaamOEaaqabaGcceWGRbGbaKaaaaaa!5CFB! \begin{array}{l} a = {a_x}\hat i + {a_y}\hat j + {a_z}\hat k\\ b = {b_x}\hat i + {b_y}\hat j + {b_z}\hat k\\ c = {c_x}\hat i + {c_y}\hat j + {c_z}\hat k \end{array}\)

then, a vector T = a + b – c has components :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGub % WaaSbaaSqaaiaadIhaaeqaaOGaeyypa0JaamyyamaaBaaaleaacaWG % 4baabeaakiabgUcaRiaadkgadaWgaaWcbaGaamiEaaqabaGccqGHsi % slcaWGJbWaaSbaaSqaaiaadIhaaeqaaaGcbaGaaGPaVlaadsfadaWg % aaWcbaGaamyEaaqabaGccqGH9aqpcaWGHbWaaSbaaSqaaiaadMhaae % qaaOGaey4kaSIaamOyamaaBaaaleaacaWG5baabeaakiabgkHiTiaa % dogadaWgaaWcbaGaamyEaaqabaaakeaacaWGubWaaSbaaSqaaiaadQ % haaeqaaOGaeyypa0JaamyyamaaBaaaleaacaWG6baabeaakiabgUca % RiaadkgadaWgaaWcbaGaamOEaaqabaGccqGHsislcaWGJbWaaSbaaS % qaaiaadQhaaeqaaaaaaa!5918! \begin{array}{l} {T_x} = {a_x} + {b_x} - {c_x}\\ \,{T_y} = {a_y} + {b_y} - {c_y}\\ {T_z} = {a_z} + {b_z} - {c_z} \end{array}\)

EXAMPLE 2

Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle \(\theta\) between them.





Fig.4.10.

ANSWER

Let OP and OQ represent the two vectors A and B making an angle \(\theta\) (Fig. 4.10). Then, using the parallelogram method of vector addition, OS represents the resultant vector R :

R = A + B

SN is normal to OP and PM is normal to OS. From the geometry of the figure,

OS² = ON² + SN²

butON = OP + PN = A + B cos\(\theta\)

SN = B sin \(\theta\)

OS² = (A + B cos \(\theta\))² + (B sin \(\theta\))²

or,R² = A² + B² + 2AB cos\(\theta\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maakaaabaGaamyqamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa % dkeadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyqaiaadk % eaciGGJbGaai4BaiaacohacqaH4oqCaSqabaaaaa!43F7! R = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \)

In \(\Delta\) OSN, SN = OS sin\(\alpha\) = R sin\(\alpha\) and in\(\Delta\) PSN, SN = PS sin \(\theta\) = B sin \(\theta\)

Therefore, R sin \(\alpha\) = B sin \(\theta\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Baiaadk % hacaGGSaWaaSaaaeaacaWGsbaabaGaci4CaiaacMgacaGGUbGaeqiU % dehaaiabg2da9maalaaabaGaamOuaaqaaiGacohacaGGPbGaaiOBai % abeg7aHbaaaaa!446A! or,\frac{R}{{\sin \theta }} = \frac{R}{{\sin \alpha }}\)

Similarly

PM = A sin\(\alpha\)= B sin\(\beta\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Baiaadk % hacaGGSaWaaSaaaeaacaWGbbaabaGaci4CaiaacMgacaGGUbGaeqOS % digaaiabg2da9maalaaabaGaamOqaaqaaiGacohacaGGPbGaaiOBai % abeg7aHbaaaaa!4434! or,\frac{A}{{\sin \beta }} = \frac{B}{{\sin \alpha }}\)

Combining Eqs. (4.24b) and (4.24c), we get

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGsbaabaGaci4CaiaacMgacaGGUbGaeqiUdehaaiabg2da9maalaaa % baGaamyqaaqaaiGacohacaGGPbGaaiOBaiabek7aIbaacqGH9aqpda % WcaaqaaiaadkeaaeaaciGGZbGaaiyAaiaac6gacqaHXoqyaaaaaa!4814! \frac{R}{{\sin \theta }} = \frac{A}{{\sin \beta }} = \frac{B}{{\sin \alpha }}\)

Using Eq. (4.24d), we get:

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbGaeqySdeMaeyypa0ZaaSaaaeaacaWGcbaabaGaamOuaaaa % ciGGZbGaaiyAaiaac6gacqaH4oqCaaa!41AF! \sin \alpha = \frac{B}{R}\sin \theta \)

where R is given by Eq. (4.24a).

or,\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiDaiaacg % gacaGGUbGaeqySdeMaeyypa0ZaaSaaaeaacaWGtbGaamOtaaqaaiaa % d+eacaWGqbGaey4kaSIaamiuaiaad6eaaaGaeyypa0ZaaSaaaeaaca % WGcbGaci4CaiaacMgacaGGUbGaeqiUdehabaGaamyqaiabgUcaRiaa % dkeaciGGJbGaai4BaiaacohacqaH4oqCaaaaaa!4EBD! \tan \alpha = \frac{{SN}}{{OP + PN}} = \frac{{B\sin \theta }}{{A + B\cos \theta }}\)

Equation (4.24a) gives the magnitude of the resultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) is known as the Law of cosines and Eq. (4.24d) as the Law of sines.

EXAMPLE 4

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.

ANSWER

The vector v_b representing the velocity of the motorboat and the vector vc representing the water current are shown in Fig. 4.11 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.



fig.4.11.

We can obtain the magnitude of R using the Law of cosine :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGsb % Gaeyypa0ZaaOaaaeaacaWG2bWaa0baaSqaaiaadkgaaeaacaaIYaaa % aOGaey4kaSIaamODamaaDaaaleaacaWGJbaabaGaaGOmaaaakiabgU % caRiaaikdacaWG2bWaaSbaaSqaaiaadkgaaeqaaOGaamODamaaBaaa % leaacaWGJbaabeaakiGacogacaGGVbGaai4CaiaaigdacaaIYaGaaG % imamaaCaaaleqabaGaaGimaaaaaeqaaaGcbaGaeyypa0ZaaOaaaeaa % caaIYaGaaGynamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdaca % aIWaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiabgEna0kaa % ikdacaaI1aGaey41aqRaaGymaiaaicdadaqadaqaaiabgkHiTiaaig % dacaGGVaGaaGOmaaGaayjkaiaawMcaaaWcbeaakiabgwKianaalaaa % baGaamODamaaBaaaleaacaWGJbaabeaaaOqaaiaadkfaaaGaci4Cai % aacMgacaGGUbGaeqiUdehaaaa!673B! \begin{array}{l} R = \sqrt {v_b^2 + v_c^2 + 2{v_b}{v_c}\cos {{120}^0}} \\ = \sqrt {{{25}^2} + {{10}^2} + 2 \times 25 \times 10\left( { - 1/2} \right)} \cong \frac{{{v_c}}}{R}\sin \theta \end{array}\)

To obtain the direction, we apply the Law of sines

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiaadkfaaeaaciGGZbGaaiyAaiaac6gacqaH4oqCaaGaeyypa0Za % aSaaaeaacaWG2bWaaSbaaSqaaiaadogaaeqaaaGcbaGaci4CaiaacM % gacaGGUbGaeqy1dygaaiaad+gacaWGYbGaaiilaiGacohacaGGPbGa % aiOBaiabew9aMjabg2da9maalaaabaGaamODamaaBaaaleaacaWGJb % aabeaaaOqaaiaadkfaaaGaci4CaiaacMgacaGGUbGaeqiUdehabaGa % eyypa0ZaaSaaaeaacaaIXaGaaGimaiabgEna0kGacohacaGGPbGaai % OBaiaaigdacaaIYaGaam4BamaaCaaaleqabaGaaGimaaaaaOqaaiaa % ikdacaaIXaGaaiOlaiaaiIdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaG % imamaakaaabaGaaG4maaWcbeaaaOqaaiaaikdacqGHxdaTcaaIYaGa % aGymaiaac6cacaaI4aaaaiabgwKiajaaicdacaGGUaGaaG4maiaaiM % dacaaI3aaabaGaeqy1dyMaeyypa0JaaGOmaiaaiodacaGGUaGaaGin % amaaCaaaleqabaGaaGimaaaaaaaa!757B! \begin{array}{l} \frac{R}{{\sin \theta }} = \frac{{{v_c}}}{{\sin \phi }}or,\sin \phi = \frac{{{v_c}}}{R}\sin \theta \\ = \frac{{10 \times \sin 12{o^0}}}{{21.8}} = \frac{{10\sqrt 3 }}{{2 \times 21.8}} \cong 0.397\\ \phi = {23.4^0} \end{array}\)

VECTOR ADDITION