Question #1b250

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Question #1b250

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Answer 1 · 2017-01-18T00:20:11

We need to use law of parallelogram of vector addition. For two vectors $\to P and \to Q$ $vecP and vec Q$ having an angle $θ$ $theta$ between the two the resultant vector $\to R$ $vecR$ is given by
$∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{{∣ ∣ ∣ \to P ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \to Q ∣ ∣ ∣}^{2} + 2 ∣ ∣ ∣ \to P ∣ ∣ ∣ ∣ ∣ ∣ \to Q ∣ ∣ ∣ cos θ}$ $|vecR|=sqrt(|vecP|^2+|vecQ|^2+2|vecP||vecQ|costheta)$
and angle of the resultant $α = {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{∣ ∣ ∣ \to Q ∣ ∣ ∣ sin θ}{∣ ∣ ∣ \to P ∣ ∣ ∣ + ∣ ∣ ∣ \to Q ∣ ∣ ∣ cos θ} ⎞ ⎟ ⎟ ⎟ ⎠$ $alpha=tan^-1((|vecQ|sintheta)/(|vecP|+|vecQ|cos theta))$

Inserting given values we get
$∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{{(6000)}^{2} + {(6000)}^{2} + 2 \times 6000 \times 6000 \times cos 60}$ $|vecR|=sqrt((6000)^2+(6000)^2+2xx6000xx6000xxcos60)$
$\Rightarrow ∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{{(6000)}^{2} + {(6000)}^{2} + 2 \times 6000 \times 6000 \times 0.5}$ $=>|vecR|=sqrt((6000)^2+(6000)^2+2xx6000xx6000xx0.5)$
$\Rightarrow ∣ ∣ ∣ \to R ∣ ∣ ∣ = 6000 \sqrt{3}$ $=>|vecR|=6000sqrt3$
$\Rightarrow ∣ ∣ ∣ \to R ∣ ∣ ∣ = 10392.3 N$ $=>|vecR|=10392.3N$ rounded to one decimal place.

and $α = {tan}^{- 1} (\frac{6000 \times sin 60}{6000 + 6000 cos 60})$ $alpha=tan^-1((6000xxsin60)/(6000+6000cos 60))$
$\Rightarrow α = {tan}^{- 1} ⎛ ⎜ ⎝ \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} ⎞ ⎟ ⎠$ $=>alpha=tan^-1((sqrt3/2)/(1+1/2))$
$\Rightarrow α = {tan}^{- 1} (\frac{1}{\sqrt{3}})$ $=>alpha=tan^-1(1/sqrt3)$
$\Rightarrow α = 30^{\circ}$ $=>alpha=30^@$ measured as per measurement of $∠ θ$ $angle theta$

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Question #1b250

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