Answer on Question #62096-Physics – Mechanics | Relativity
Add these vectors:
1511 at 35 degrees W of N + 8822 at 72 degrees W of S + 333 at 27 degrees S of W
Solution
Let Y be the north direction and X will be the eastern.
A = 1511 ( − sin 35 , cos 35 ) = ( − 866.674 , 1237.74 ) B = 8822 ( − sin 72 , − cos 72 ) = ( − 8390.22 , − 2726.15 ) C = 333 ( − cos 27 , − sin 27 ) = ( − 296.705 , − 151.179 ) A + B + C = ( − 866.674 , 1237.74 ) + ( − 8390.22 , − 2726.15 ) + ( − 296.705 , − 151.179 ) = ( − 9553.6 , − 1639.59 ) . \begin{array}{l}
\boldsymbol{A} = 1511(-\sin 35, \cos 35) = (-866.674, 1237.74) \\
\boldsymbol{B} = 8822(-\sin 72, -\cos 72) = (-8390.22, -2726.15) \\
\boldsymbol{C} = 333(-\cos 27, -\sin 27) = (-296.705, -151.179) \\
\boldsymbol{A} + \boldsymbol{B} + \boldsymbol{C} = (-866.674, 1237.74) + (-8390.22, -2726.15) + (-296.705, -151.179) \\
= (-9553.6, -1639.59).
\end{array} A = 1511 ( − sin 35 , cos 35 ) = ( − 866.674 , 1237.74 ) B = 8822 ( − sin 72 , − cos 72 ) = ( − 8390.22 , − 2726.15 ) C = 333 ( − cos 27 , − sin 27 ) = ( − 296.705 , − 151.179 ) A + B + C = ( − 866.674 , 1237.74 ) + ( − 8390.22 , − 2726.15 ) + ( − 296.705 , − 151.179 ) = ( − 9553.6 , − 1639.59 ) .
The magnitude will be
( − 9553.6 ) 2 + ( − 1639.59 ) 2 = 9693 \sqrt{(-9553.6)^2 + (-1639.59)^2} = 9693 ( − 9553.6 ) 2 + ( − 1639.59 ) 2 = 9693
The direction is
tan − 1 1639.59 9553.6 = 9.7 degrees S of W \tan^{-1} \frac{1639.59}{9553.6} = 9.7 \text{ degrees S of W} tan − 1 9553.6 1639.59 = 9.7 degrees S of W
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