Question

A force 2i^ + 4j^ acts in xy plane. Angle made by this force with +x axis will be_
answer is tan-¹ (2)
please explain the steps

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Answer 1

A given force is acting, in x-y plane:

$\rm \vec{F} = 2\hat{i} + 4\hat{j}$

》The components of it are 2 and 4 in x and y-axis respectively.

$\rm tan\theta = \dfrac{perpendicular}{base}$

$\implies \rm tan\theta = \dfrac{dy}{dx}$

$\implies \rm tan\theta = \dfrac{4}{2}$

$\implies \rm tan\theta = 2$

$\implies \rm \theta = tan^{-1}(2)$

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Answer 2

A given force is acting, in x-y plane:

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^The components of it are 2 and 4 in x and y-axis respectively.

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^The components of it are 2 and 4 in x and y-axis respectively. \rm tan\theta = \dfrac{perpendicular}{base}tanθ=baseperpendicular

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^The components of it are 2 and 4 in x and y-axis respectively. \rm tan\theta = \dfrac{perpendicular}{base}tanθ=baseperpendicular \implies \rm tan\theta = \dfrac{dy}{dx}⟹tanθ=dxdy

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^The components of it are 2 and 4 in x and y-axis respectively. \rm tan\theta = \dfrac{perpendicular}{base}tanθ=baseperpendicular \implies \rm tan\theta = \dfrac{dy}{dx}⟹tanθ=dxdy\implies \rm tan\theta = \dfrac{4}{2}⟹tanθ=24

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^The components of it are 2 and 4 in x and y-axis respectively. \rm tan\theta = \dfrac{perpendicular}{base}tanθ=baseperpendicular \implies \rm tan\theta = \dfrac{dy}{dx}⟹tanθ=dxdy\implies \rm tan\theta = \dfrac{4}{2}⟹tanθ=24\implies \rm tan\theta = 2⟹tanθ=2

A given force is acting, in x-y plane:\rm \vec{F} = 2\hat{i} + 4\hat{j}F=2i^+4j^The components of it are 2 and 4 in x and y-axis respectively. \rm tan\theta = \dfrac{perpendicular}{base}tanθ=baseperpendicular \implies \rm tan\theta = \dfrac{dy}{dx}⟹tanθ=dxdy\implies \rm tan\theta = \dfrac{4}{2}⟹tanθ=24\implies \rm tan\theta = 2⟹tanθ=2\implies \rm \theta = tan^{-1}(2)⟹θ=tan−1(2)

Explanation:

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