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Determine the angle \(\theta\) between the force and the line \(A B .\)

\( \vec{r}_{A B}=\vec{r}_{B} - \vec{r}_{A} \)

\( =(0\hat i+0\hat j+4 k)-(0\hat i+3\hat j+0 k) \)

\( =(0 \hat{\imath}-3 \hat{\jmath}+4 K) \)

\( \left|r_{A B}\right|=5 m \)

\( \vec{r}_{A c}=\vec{r}_{c}-\vec{r}_{A} \)

\( =(4 \hat{\imath}+0 \hat{\jmath}+0 k)-(0 i+3 j+0 k) \)

\( =(4\hat i-3\hat j+0 k) m \)

\( \left|r_{A C}\right|=5 \mathrm{m} \)

\(\vec{r}_{A B} \cdot \vec{r}_{A C} =\left|r_{A B}\right|\left|r_{A C}\right| \cos \theta \rightarrow \left(0\hat i -3 \hat j+4 k\right) \cdot(4 \hat i-3\hat j+0 k)=5(5) \cos \theta \)

\( ∴ {\theta}=68 \cdot 9^{\circ} \)

Two cables exert forces on the pipe. Determine the magnitude of the projected component of \(\mathbf{F}_{1}\) along the line of action of \(\mathbf{F}_{2}\) .

\( \vec{F}=\left(30 \cos 30^{\circ} \sin 30^{\circ} i\right)+\left(30 \cos 30 \cos 30\hat j\right) \)

\( -(30 \sin 30 k) \)

\( =(13 \hat{\imath}+22.5 \hat j-15 k) N \)

* unit vector of \(F_{2}\):-

\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma =1 \quad \longrightarrow\alpha=45^{\circ} \) or \( \alpha=135^{\circ} \)

\(\overrightarrow{U} \ {F_{2}}=\cos 135 \hat i+\cos 60^{\circ}\hat  j+\cos 60^{\circ} k \)

           \( =(-0.707\hat  i)+0.5\hat  j+0.6 k ) \)

\( F_∥=F_{1} \cdot \overrightarrow{U} F_{2} = (13\hat  i+22 \cdot 5\hat  j-15 k) \cdot(-0.707\hat i+0.5\hat j+0.6 k) \)

                             \( =-5.44 N \)

\( \left|F_∥\right|=5.44 N \)

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