Question 5.88: Determine the x and z components of reaction at the journal ...
Determine the x and z components of reaction at the journal bearing A and the tension in cords BC and BD necessary for equilibrium of the rod.
\mathbf{F}_1=\{-800 \mathbf{k}\} \mathbf{N}
\mathbf{F}_2=\{350 \mathbf{j}\} \mathbf{N}
\mathbf{F}_{B C}=F_{B C} \frac{(-3 \mathbf{j}+4 \mathbf{k})}{5}
=\left\{-0.6 F_{B C} \mathbf{j}+0.8 F_{B C} \mathbf{k}\right\} \mathbf{N}
\mathbf{F}_{B D}=F_{B D} \frac{(3 \mathbf{j}+4 \mathbf{k})}{5}=\left\{0.6 F_{B D} \mathbf{j}+0.8 F_{B D} \mathbf{k}\right\} \mathrm{N}
\Sigma F_x=0 ; \quad A_x=0
\Sigma F_y=0 ; \quad 350-0.6 F_{B C}+0.6 F_{B D}=0
\Sigma F_z=0 ; \quad A_z-800+0.8 F_{B C}+0.8 F_{B D}=0
\Sigma M_x=0 ; \quad\left(M_A\right)_x+0.8 F_{B D}(6)+0.8 F_{B C}(6)-800(6)=0
\Sigma M_y=0 ; \quad 800(2)-0.8 F_{B C}(2)-0.8 F_{B D}(2)=0
\Sigma M_z=0 ; \quad\left(M_A\right)_z-0.6 F_{B C}(2)+0.6 F_{B D}(2)=0
F_{B D}=208 \mathrm{~N}
F_{B C}=792 \mathrm{~N}
A_z=0
\left(M_A\right)_x=0
\left(M_A\right)_z=700 \mathrm{~N} \cdot \mathrm{m}
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