state and prove parallelogram low of vector addition and subtraction .
Answers
Explanation:
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parallelogram-method
In case of addition of two vectors by parallelogram method as shown in figure, the magnitude of resultant will be given by,
(AC)2 = (AE)2 + (EC)2
or R2 = (P + Q cos θ)2 (Q sin θ)2
or R = √(P2+ Q2 )+ 2PQcos θ
And the direction of resultant from vector P will be given by
tan ? = CE/AE = Qsinθ/(P+Qcosθ)
? = tan-1 [Qsinθ/(P+Qcosθ)]
Special Cases
(a) When θ = 0°, cos θ = 1 , sin θ = 0°
Substituting for cos θ in equation R = √(P2+ Q2 )+ 2PQcos θ, we get,
R = √(P2+ Q2 )+ 2PQcos θ
= √(P+ Q)2
or R = P+Q (maximum)
Substituting for sin θ and cos θ in equation ? = tan-1 [Qsinθ/(P+Qcosθ)], we get,
? = tan-1 [Qsinθ/(P+Qcosθ)]
= tan-1 [(Q×0)/(P+(Q×1))]
= tan-1(0)
= 0°
The resultant of two vectors acting in the same directions is equal to the sum of the two. The direction of resultant coincides with those of the two vectors.
(b) When θ = 180°, cos θ = -1 , sin θ = 0°
Substituting for cos θ in equation R = √(P2+ Q2 )+ 2PQcos θ, we get,
R = √(P2+ Q2 )+ 2PQ(-1)
=√P2+ Q2 – 2PQ
= √(P – Q)2 (minimum)
R = P – Q (minimum)
Substituting for sin θ and cos θ in equation ? = tan-1 [Qsinθ/(P+Qcosθ)], we get,
? = tan-1 [Qsinθ/(P+Qcosθ)]
= tan-1 [(Q×0)/(P+(Q×(-1)))]
= tan-1(0)
= 0°
This magnitude of the resultant of two vectors acting in opposite direction is equal to the difference of magnitudes of the two and represents the minimum value. The direction of the resultant is in the direction of the bigger one.
(c) When θ = 90°, cos θ = 0 , sin θ = 1
Substituting for cos θ in equation R = √(P2+ Q2 )+ 2PQcos θ, we get,
R = √(P2+ Q2 )+ (2PQ×0)
= √P2+ Q2
Substituting for sin θ and cos θ in equation ? = tan-1 [Qsinθ/(P+Qcosθ)], we get,
? = tan-1 [Qsinθ/(P+Qcosθ)]
= tan-1 [(Q×1)/(P+(Q×(0)))]
= tan-1(Q/P)
The resultant of two vectors acting at right angles to each other is equal to the square root of the sum of the squares of the magnitudes of the two vectors. Direction of the resultant depends upon their relative magnitudes.