Explanations & Calculations
By horizontal resolution, → X ⃗ = 30 m sin 30 + 50 m = 15 m + 50 m = 65 m \qquad\qquad
\begin{aligned}
\to \\
\small \vec{X}&=\small 30m\sin30+50m\\
&=\small 15m+50m\\
&=\small 65m
\end{aligned} → X = 30 m sin 30 + 50 m = 15 m + 50 m = 65 m
↑ Y ⃗ = 30 m cos 30 − 40 m = 15 3 m − 40 m = 5 ( 3 3 − 8 ) m = − 14.02 m \qquad\qquad
\begin{aligned}
\uparrow\\
\small \vec{Y}&=\small 30m\cos30-40m\\
&=\small 15\sqrt3m-40m\\
&=\small 5(3\sqrt3-8)m\\
&=\small -14.02m
\end{aligned} ↑ Y = 30 m cos 30 − 40 m = 15 3 m − 40 m = 5 ( 3 3 − 8 ) m = − 14.02 m
This negative sign implies that the Y resolution is not to the right, it is to the left. Now the resultant can be found by R 2 = X 2 + Y 2 = ( 65 m ) 2 + ( − 14.02 m ) 2 = 4421.56 m 2 R = 4421.56 m 2 = 66.49 m ≈ 66 m \qquad\qquad
\begin{aligned}
\small R^2&=\small X^2+Y^2\\
&=\small (65m)^2+(-14.02m)^2\\
&=\small 4421.56m^2\\
\small R&=\small \sqrt{4421.56m^2}\\
&=\small 66.49m\\
&\approx\small 66m
\end{aligned} R 2 R = X 2 + Y 2 = ( 65 m ) 2 + ( − 14.02 m ) 2 = 4421.56 m 2 = 4421.56 m 2 = 66.49 m ≈ 66 m
The direction of the resultant is θ = tan − 1 ( Y X ) = tan − 1 ( − 14.02 m 65 m ) = − 12.1 7 0 \qquad\qquad
\begin{aligned}
\small \theta &=\small \tan^{-1}\Big(\frac{Y}{X}\Big)\\
&=\small \tan^{-1}\Big(\frac{-14.02m}{65m}\Big)\\
&=\small -12.17^0
\end{aligned} θ = tan − 1 ( X Y ) = tan − 1 ( 65 m − 14.02 m ) = − 12.1 7 0
Then the direction is 12.1 7 0 \small 12.17^0 12.1 7 0 North of west.
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