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Determine the length of the rod and the position vector directed from  $A$ to $B$. What is the angle $\theta$ ?

$\vec{r}_{A B}=\vec{r}_{B}-\vec{r}_{A}$

$=(0 i+1 \hat{j}+2 k)-(2 \hat{\imath}+{0} \hat{j}+0 \hat{k})$

$=-2 i+1 j+2 k$

$\left|r_{A B}\right|=\sqrt{(-2)^{2}+(1)^{2}+(2)^{2}}=\sqrt{4+1+4}=\sqrt{9}=3 m$

$A \cdot B=|A||B| \cos \theta$

$(-2 i+1 j+2 k) \cdot(-1 i+0 j+0 k)=(3)(1) \cos \theta$

$2=3 \cos \theta \rightarrow \theta=48 \cdot 2^{\circ}$

The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole.

$\vec{r}_{A C}=\vec{r}_{C}-\vec{r}_{A}$

$=(-1 \hat{i}+4 \hat{j}+0 k)-(0 \hat{\imath}+{0} {j}+4{k})$

$=(-1 \hat{i}+4 \hat{\jmath}-4 \hat{k}) m$

$| \ r_{AC}|=5.74m$

$\vec{u}_{A C}=\frac{\vec{r}_{A C}}{\left|r_{A C}\right|}=(-0.174 \hat{i}+0.696 \hat{\jmath}-0.696{k}) N$

$\vec{F}_{A}=F_{A} \cdot \vec{U}_{A C} =(-43.5 \hat{i}+174 \hat{\jmath}-174{k}) N$

$\vec{r}_{B D}=\vec{r}_{D}-\vec{r}_{B}$

$=(2 \hat{\imath}-3 \hat{j}+0 k)-(0 \hat{i}+0 \hat{j}+5.5 k)=(2 \hat{\imath}-3 \hat{j}-5.5 k) m$

$\left|r_{B D}\right|=6.58 \mathrm{m}$

$\vec{u}_{B D}=\frac{\vec{r}_{BD}}{\left|r_{BD}\right|}=(0.304 \hat{i}-0.456 \hat{\jmath}-0.836{k}) N$

$∴ \vec{F}_{B}=F_{B} \cdot \vec{U}_{B D} =(53.2\hat{i}-79.8\hat{j}-146 k) N$