Two vectors A and B are inclined to each other at an angle Q. Using parallelogram law of vector addition find the magnitude and direction of their resultant.
Answers
R=A=B=x
R2=A2+B2+ABcos thetha
X2=x2+x2+x.xcos thetha
X2=2x2+x2 cos thetha
X2-2x2=x2cos thetha
-x2=x2cos thetha
-x2/x2=cos thetha
-1/2=cos thetha
(Cos -1) =cos 60
They have = 60 answer
Concept
In mathematics, the simplest form of the parallelogram of force belongs to basic geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram is equal to the sum of the squares of the lengths of the two diagonals.
Given
Two vectors A and B inclined to each other at angle Q
Find
Using parallelogram law of vector addition find the magnitude and direction of resultant of vector A and B
Solution
The steps are as follow:
- As shown in the figure, P and Q act simultaneously at one point and are two vectors represented in both magnitude and direction by the two adjacent sides OA and OD of the parallelogram OABD.
- Let θ be the resulting vector as the angle between P and Q and R.
- According to the parallelogram of force of vector addition, the diagonal OB represents the result of P and Q.
So, R=P+Q
From triangle OCB
OB^2 =OC^2 + BC^2
OB ^2=(OA+AC)^2+BC^2. .........(1)
In ΔABC
cosθ= AC/AA
AC=ABcosθ
AC=ODcosθ=Qcosθ
as [AB=OD=Q]
Also
cosθ= BC/AB
BC=ABsinθ
BC=ODsinθ=Qsinθ
[as AB=OD=Q]
Magnitude of Resultant substituting AC & BC in equation (1) we get
R^2 =(P+Qcosθ)^2+(Qsinθ)^2
∴R= P^2+2PQcosθ+Q^2
From Δ ABC
tanϕ= BC/OO= BC/(OA+AC)
tanϕ= Qsinθ/(P+Qcosθ)
∴ϕ=tan^−1(Qsinθ/(P+Qcosθ))
Hence the magnitude and direction of vector is derived
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