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Determine the force components acting on the ball- and-socket at  $A$ , the reaction at the roller  $B$  and the tension  on the cord  $C D$  needed for cquilibrium of the quarter  circular plate.

$\stackrel{+}\swarrow\sum F_{x}=0 \longrightarrow \quad A_ x=0$

$\sum F_{y}=0 \longrightarrow A_ y=0$

$\sum M_{x}=0$

$-200(3 \sin 60)-200(3)+B_ z(3)=0$

$∴ B _z=373.2 N \uparrow$

$\sum M_ y=0$

$350(2)+200\left(3 cos 60^{\circ}\right)-A_{z}(3)=0$

$∴ A_z=333.3 N \uparrow$

$+\uparrow\sum F_z=0\longrightarrow 333.3 A_z+B_z+F_d=350-200-200=0\longrightarrow\ ∴ F_{CD}=43.5 N\uparrow$

Member  $A B$  is supported at  $B$  by a cable and at  $A$  by  a smooth fixed square rod which fits loosely through the  square hole of the collar. Determine the tension in cable  $B C$ if the force  $\mathbf{F}=\{-4.5 \mathbf{k}\} \mathrm{kN}$ .

$\overline{r}_{B C}=\overline{r}_{C}-\overline{r}_{B}$

$=(0 i-2.4\hat j+1.8 k)$

$=(3.6 i-1.2\hat j+0 k)$

$=(-3.6\hat i-1.2\hat j+1.8 k) m$

$\left|\overrightarrow{r}_{B C}\right|=4 \cdot 2 m$

$\vec{U}_{B C}=\frac{\vec{r}_{B C}}{\left|\vec{r}_{B C}\right|}$      $=\left(-\frac{3.6}{4.2}\hat i-\frac{1.2}{4.2} \hat{j}+\frac{1.8}{4.2} k\right)$

$=\left(-\frac{6}{7} \hat{i}-\frac{2}{7} \hat{\jmath}+\frac{3}{7} k\right)$

$\vec{F}_{D C} =\left(-\frac{6}{7} F_{B C} \hat{\imath}\right. -\frac{2}{7} {F}_{DC} \hat{j} +\frac{3}{7}F_{BC}k)$

$+\uparrow\sum F_ z=0\quad \longrightarrow\quad\frac{3}{7} F_{B C}-F=0\longrightarrow\frac{3}{7} F_{B C}=F$

$∴ F_{B C}=10.5 \mathrm{kN}$