Question 2.120: Determine the magnitudes of the components of F = 600 N acti...
Determine the magnitudes of the components of F = 600 N acting along and perpendicular to segment DE of the pipe assembly.
Unit Vectors:The unit vectors u _{E B} and u _{E D} must be determined first. From Fig. a
u _{E B}=\frac{ r _{E B}}{r_{E B}}=\frac{(0-4) i +(2-5) j +[0-(-2)] k }{\sqrt{(0-4)^2+(2-5)^2+[0-(-2)]^2}}=-0.7428 i -0.5571 j +0.3714 k
u _{E D}=- j
Thus, the force vector F is given by
F =F u _{E B}=600(-0.7428 i -0.5571 j +0.3714 k )=[-445.66 i -334.25 j +222.83 k ] N
Vector Dot Product:The magnitude of the component of F parallel to segment DE of the pipe assembly is
\begin{aligned}\left(F_{E D}\right)_{\text {paral }}= F \cdot u _{E D} &=(-445.66 i -334.25 j +222.83 k ) \cdot(- j ) \\&=(-445.66)(0)+(-334.25)(-1)+(222.83)(0) \\&=334.25=334 N\end{aligned}
The component of F perpendicular to segment DE of the pipe assembly is
\left(F_{E D}\right)_{\text {per }}=\sqrt{F^2-\left(F_{E D}\right)_{\text {paral }}{ }^2}=\sqrt{600^2-334.25^2}=498 N
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