Question

Derive an expression for magnitude and direction of resultant of two concurrent vectors

Answer 1

Answer:

Since we know that concurrent vectors have the same origin and cross a single point, also a vectorial magnitude is that which has a number, direction and sense.

Here is the derivation for resultant of two concurrent vectors.

Let two concurrent vectors be P and Q. Let the resultant be R.

R=P+Q

Let us consider a triangle OCB

In triangle OCB,

OB²=OC²+BC²

OB²=(OA+AC)²+BC²

cos θ = AC=AB cos θ

AC=OD cos θ=Q

Also,

cosθ = AC/AB

or AC = AB cosθ

or, AC = OD cosθ = Q cosθ since, [AB = AD =Q]

BC=AB sin θ

BC=OD sinθ=Q sin θ

Substitute the values in the resultant:

R²=(P+Q cosθ)²+(Q sin θ)²

R²=P²+Q²cos²θ+2PQ cos θ+ Q² sin²θ

R²=P²+Q²(cos²θ+ sin²θ)+2PQ cos θ= P²+Q²+2PQ cos θ

R=√(P²+Q²+2PQ cos θ)

Hope this helps.

Explanation:

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