derive expression for magnitude of resulant of two vectors using triangle law of vector addition
Answers
Explanation:
Consider two vectors P and Q that are represented in the order of magnitude and direction by the sides OA and AB, respectively of the triangle OAB. ... To determine the direction of the resultant vector, let ɸ be the angle between the resultant vector R and P.
Explanation:
♡ ↪☛Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.
Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.
So, we have
R = P + Q
Now, expand A to C and draw BC perpendicular to OC.
From triangle OCB,
OB 2 =C 2 +BC 2
OB 2 =C 2 +BC 2
OB 2 =C 2 +BC 2 or OB 2 =(OA+AC) 2 +BC 2
OB 2 =C 2 +BC 2 or OB 2 =(OA+AC) 2 +BC 2 . . . . . . ( i )
Intriangle ABC,
Intriangle ABC,cosθ= ABAC
Intriangle ABC,cosθ= ABAC
Intriangle ABC,cosθ= ABAC
Intriangle ABC,cosθ= ABAC or , AC = AB cosθ
or , AC = OD cosθ
= Q cosθ
[ AB = OD = Q ]
Also,
Also, cosθ= ABBC
Also, cosθ= ABBC
Also, cosθ= ABBC or , BC = AB sinθ
or , BC = OD sinθ
= Q sinθ [ AB = OD = Q }
= Q sinθ [ AB = OD = Q }Magnitude of resultant:
Substituting value of AC and BC in ( i ), we get
R 2 =(P+Qcosθ) 2 +(Qsinθ) 2
R 2 =(P+Qcosθ) 2 +(Qsinθ) 2
R 2 =(P+Qcosθ) 2 +(Qsinθ) 2 or , R 2 =P 2 +2PQcosθ+Q 2
cos 2θ+Q 2 sin 2 θ
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC = OA+ACBC
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC = OA+ACBC
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC = OA+ACBC or , tanϕ= P+QcosθQsinθ
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC = OA+ACBC or , tanϕ= P+QcosθQsinθ
cos 2θ+Q 2 sin 2 θor , R 2 =P 2+2PQcosθ+Q 2 R = P 2 +2PQcosθ+Q 2 Which is the magnitude of resultant.Direction of resultant :Let ϕ be the angle made by resultant R with P . Then,From triangle OBC,tanϕ= OCBC = OA+ACBC or , tanϕ= P+QcosθQsinθ ϕ=tan −1 ( P+QQcosθQsin )
which is the direction of resultant.
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