Question 2.126: Two cables exert forces on the pipe. Determine the magnitude...
Two cables exert forces on the pipe. Determine the magnitude of the projected component of F _1 along the line of action of F _2.
Force Vector:
\begin{aligned}u _{F_1} &=\cos 30^{\circ} \sin 30^{\circ} i +\cos 30^{\circ} \cos 30^{\circ} j -\sin 30^{\circ} k \\&=0.4330 i +0.75 j -0.5 k\end{aligned}
\begin{aligned}F _1=F_R u _{F_f} &=30(0.4330 i +0.75 j -0.5 k ) lb \\&=\{12.990 i +22.5 j -15.0 k \} lb\end{aligned}
Unit Vector: One can obtain the angle \alpha=135^{\circ} \text { for } \mathbf{F}_2 \text { using Eq. 2-8. }
\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \text {, with }\beta=60^{\circ} \text { and } \gamma=60^{\circ} The unit vector along the line of action of F _2 is
u _{F_2}=\cos 135^{\circ} i +\cos 60^{\circ} j +\cos 60^{\circ} k =-0.7071 i +0.5 j +0.5 k
Projected Component of F _1 Along the Line of Action of F _2 :
\begin{aligned}\left(F_1\right)_{F_2}= F _1 \cdot u _{F_2} &=(12.990 i +22.5 j -15.0 k ) \cdot(-0.7071 i +0.5 j +0.5 k ) \\&=(12.990)(-0.7071)+(22.5)(0.5)+(-15.0)(0.5) \\&=-5.44 lb\end{aligned}
Negative sign indicates that the projected component of acts in the opposite sense of direction to that of u _{F_2}.
\text { The magnitude is }\left(F_1\right)_{F_2}=5.44 lbRelated Answered Questions
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