Question 2.121: Determine the magnitude of the projection of force F = 600 N...
Determine the magnitude of the projection of force F = 600 N along the u axis.
Unit Vectors: The unit vectors and must be determined first. From Fig. a,
u _{O A}=\frac{ r _{O A}}{r_{O A}}=\frac{(-2-0) i +(4-0) j +(4-0) k }{\sqrt{(-2-0)^2+(4-0)^2+(4-0)^2}}=-\frac{1}{3} i +\frac{2}{3} j +\frac{2}{3} k
u _u=\sin 30^{\circ} i +\cos 30^{\circ} j
Thus, the force vectors is given by
F =F u _{O A}=600\left(-\frac{1}{3} i -\frac{2}{3} j +\frac{2}{3} k \right)=\{-200 i +400 j +400 k \} N
Vector Dot Product: The magnitude of the projected component of along the axis is
\begin{aligned}F _u=F \cdot u _u &=(-200 i +400 j +400 k ) \cdot\left(\sin 30^{\circ} i +\cos 30^{\circ} j \right) \\&=(-200)\left(\sin 30^{\circ}\right)+400\left(\cos 30^{\circ}\right)+400(0) \\&=246 N\end{aligned}
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