The two mooring cables exert forces on the stern of a ship as shown. Represent each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant.

#### Solution:

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We will first write the position vectors for each cable. To do so, we will write the locations of points A, B and C in Cartesian vector form.

Using the diagram, the locations of the points are:

B:(50i+50j-30k) ft

C:(0i+0j+0k) ft

Let us now write our position vectors:

r_{CB}\,=\,\left\{(50-0)i+(50-0)j+(-30-0)k\right\}=\left\{50i+50j-30k\right\} ft

The next step is to find the magnitude of each position vector.

magnitude of r_{CB}\,=\,\sqrt{(50)^2+(50)^2+(-30)^2}=76.81

We can now find the unit vector for each position vector as follows:

u_{CB}\,=\,\left(\dfrac{50}{76.81}i+\dfrac{50}{76.81}j-\dfrac{30}{76.81}k\right)

Let us now express each force in Cartesian vector form. From the question, we know F_A=200 lb and F_B=150 lb.

F_{CA}=\left\{169i+33.8j-101.4k\right\} lb

F_{CB}=150\left(\dfrac{50}{76.81}i+\dfrac{50}{76.81}j-\dfrac{30}{76.81}k\right)

F_{CB}=\left\{97.6i+97.6j-58.6k\right\} lb

We can now find the resultant force by adding the two forces together.

F_R=\left\{169i+33.8j-101.4k\right\}+\left\{97.6i+97.6j-58.6k\right\}

F_R=\left\{266.6i+131.4j-160k\right\} lb

The magnitude of this resultant force is:

The coordinate direction angles can be calculated by taking the cosine inverse of each component of the resultant force divided by the magnitude.

\beta=\cos^{-1}\left(\dfrac{131.4}{337.5}\right)=67.1^0

\gamma=\cos^{-1}\left(\dfrac{-160}{337.5}\right)=118^0

#### Final Answers:

F_{CB}=\left\{97.6i+97.6j-58.6k\right\} lb

F_R=\left\{266.6i+131.4j-160k\right\} lb

magnitude of F_R=337.5 lb

\alpha=37.8^0

\beta=67.1^0

\gamma=118^0